Chapter 13 Probability

Given:

$P(A) = \frac{7}{13}$

$P(B) = \frac{9}{13}$

$P(A \cap B) = \frac{4}{13}$

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To Find:

$P(A|B)$

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Solution:

The formula for conditional probability $P(A|B)$ is given by:

Substitute the given values of $P(A \cap B)$ and $P(B)$ into formula (i):

$P(A|B) = \frac{\frac{4}{13}}{\frac{9}{13}}$

Simplify the expression:

$P(A|B) = \frac{4}{13} \times \frac{13}{9}$

Cancel out the common factor 13 from the numerator and denominator:

$P(A|B) = \frac{4}{\cancel{13}} \times \frac{\cancel{13}}{9}$

$P(A|B) = \frac{4}{9}$

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Final Answer:

Thus, $P(A|B) = \frac{4}{9}$.

Given:

A family has two children.

We are given that at least one of the children is a boy.

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To Find:

The probability that both children are boys, given that at least one is a boy.

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Solution:

Let the sample space S for a family with two children be represented by the genders of the children in order of birth (Boy B, Girl G). Assuming each outcome is equally likely, the possible outcomes are:

$S = \{BB, BG, GB, GG\}$

The total number of outcomes in the sample space is 4.

Let A be the event that "both children are boys".

$A = \{BB\}$

Let B be the event that "at least one child is a boy".

$B = \{BB, BG, GB\}$

We are asked to find the probability of event A occurring given that event B has occurred, which is $P(A|B)$.

First, find the intersection of events A and B, denoted by $A \cap B$. This is the event that both children are boys AND at least one child is a boy.

$A \cap B = \{BB\}$

Now, calculate the probabilities of $A \cap B$ and B:

$P(A \cap B) = \frac{\text{Number of outcomes in } A \cap B}{\text{Total number of outcomes}} = \frac{1}{4}$

$P(B) = \frac{\text{Number of outcomes in B}}{\text{Total number of outcomes}} = \frac{3}{4}$

Using the formula for conditional probability:

Substitute the calculated probabilities into formula (i):

$P(A|B) = \frac{1/4}{3/4}$

Simplify the fraction:

$P(A|B) = \frac{1}{4} \times \frac{4}{3}$

Cancel out the common factor 4:

$P(A|B) = \frac{1}{\cancel{4}} \times \frac{\cancel{4}}{3}$

$P(A|B) = \frac{1}{3}$

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Final Answer:

The probability that both children are boys given that at least one of them is a boy is $\frac{1}{3}$.

Given:

Ten cards numbered from 1 to 10 are in a box.

A card is drawn randomly.

It is known that the number on the drawn card is more than 3.

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To Find:

The probability that the drawn card is an even number, given that its number is more than 3.

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Solution:

The sample space S of drawing a card from 1 to 10 is:

$S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$

The total number of outcomes in the sample space is 10.

Let A be the event that "the drawn card is an even number".

$A = \{2, 4, 6, 8, 10\}$

Let B be the event that "the number on the drawn card is more than 3".

$B = \{4, 5, 6, 7, 8, 9, 10\}$

We need to find the probability of event A given that event B has occurred, i.e., $P(A|B)$.

First, find the intersection of events A and B, denoted by $A \cap B$. This is the event that the drawn card is both even AND more than 3.

$A \cap B = \{4, 6, 8, 10\}$

Now, calculate the probabilities of $A \cap B$ and B:

$P(A \cap B) = \frac{\text{Number of outcomes in } A \cap B}{\text{Total number of outcomes}} = \frac{4}{10}$

$P(B) = \frac{\text{Number of outcomes in B}}{\text{Total number of outcomes}} = \frac{7}{10}$

Using the formula for conditional probability:

Substitute the calculated probabilities into formula (i):

$P(A|B) = \frac{\frac{4}{10}}{\frac{7}{10}}$

Simplify the expression:

$P(A|B) = \frac{4}{10} \times \frac{10}{7}$

Cancel out the common factor 10:

$P(A|B) = \frac{4}{\cancel{10}} \times \frac{\cancel{10}}{7}$

$P(A|B) = \frac{4}{7}$

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Alternate Solution (Using reduced sample space):

Since it is known that the number on the drawn card is more than 3, our new sample space is restricted to the outcomes in event B:

$B = \{4, 5, 6, 7, 8, 9, 10\}$

The number of outcomes in this reduced sample space is 7.

Within this reduced sample space B, the outcomes that are even numbers are the elements of $A \cap B$:

$A \cap B = \{4, 6, 8, 10\}$

The number of favorable outcomes within the reduced sample space is 4.

The probability of drawing an even number given the number is more than 3 is the number of favorable outcomes in $A \cap B$ divided by the total number of outcomes in the reduced sample space B.

$P(A|B) = \frac{\text{Number of outcomes in } A \cap B}{\text{Number of outcomes in B}}$

$P(A|B) = \frac{4}{7}$

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Final Answer:

The probability that the drawn card is an even number given that it is more than 3 is $\frac{4}{7}$.

Given:

Total number of students in the school = 1000

Number of girls = 430

Percentage of girls studying in Class XII = 10%

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To Find:

The probability that a student studies in Class XII given that the chosen student is a girl.

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Solution:

Let G be the event that the randomly chosen student is a girl.

Let C12 be the event that the randomly chosen student studies in Class XII.

We are given that the chosen student is a girl, so the sample space is restricted to the set of girls in the school.

The number of students in this reduced sample space (number of girls) is 430.

We are interested in the probability that the student from this reduced sample space (girls) studies in Class XII.

The number of girls who study in Class XII is 10% of the total number of girls:

Number of girls in Class XII = $10\%$ of 430

Number of girls in Class XII = $0.10 \times 430 = 43$

The event that the chosen student is a girl AND studies in Class XII is the intersection of events G and C12, denoted by $G \cap C12$. The number of students in this event is 43.

The required probability is the conditional probability $P(\text{C12|G})$, which can be calculated as the number of favorable outcomes (girls in Class XII) divided by the total number of outcomes in the reduced sample space (total girls).

$P(\text{C12|G}) = \frac{\text{Number of girls in Class XII}}{\text{Total number of girls}}$

$P(\text{C12|G}) = \frac{43}{430}$

Simplify the fraction:

$P(\text{C12|G}) = \frac{\cancel{43}^{1}}{\cancel{430}_{10}}$

$P(\text{C12|G}) = \frac{1}{10}$

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Final Answer:

The probability that a student chosen randomly studies in Class XII given that the chosen student is a girl is $\frac{1}{10}$ or $0.1$.

Given:

A die is thrown three times.

Event A: 4 on the third throw.

Event B: 6 on the first and 5 on the second throw.

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To Find:

The probability of A given that B has already occurred, i.e., $P(A|B)$.

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Solution:

The sample space S for throwing a die three times consists of $6 \times 6 \times 6 = 216$ outcomes. Each outcome is an ordered triplet $(x, y, z)$ where x, y, z are the results of the first, second, and third throws, respectively, and $x, y, z \in \{1, 2, 3, 4, 5, 6\}$.

Event A is "4 on the third throw". This means the third component of the triplet must be 4, while the first two can be any number from 1 to 6.

$A = \{(x, y, 4) : x, y \in \{1, 2, 3, 4, 5, 6\}\}$

The number of outcomes in A is $6 \times 6 \times 1 = 36$.

$P(A) = \frac{\text{Number of outcomes in A}}{\text{Total number of outcomes}} = \frac{36}{216}$

Event B is "6 on the first and 5 on the second throw". This means the first component is 6, the second is 5, and the third can be any number from 1 to 6.

$B = \{(6, 5, z) : z \in \{1, 2, 3, 4, 5, 6\}\}$

The number of outcomes in B is $1 \times 1 \times 6 = 6$.

$P(B) = \frac{\text{Number of outcomes in B}}{\text{Total number of outcomes}} = \frac{6}{216}$

The intersection of A and B, $A \cap B$, is the event where both A and B occur. This means the outcome must have 6 on the first throw, 5 on the second throw, AND 4 on the third throw.

$A \cap B = \{(6, 5, 4)\}$

The number of outcomes in $A \cap B$ is 1.

$P(A \cap B) = \frac{\text{Number of outcomes in } A \cap B}{\text{Total number of outcomes}} = \frac{1}{216}$

The probability of A given B is calculated using the formula for conditional probability:

Substitute the calculated probabilities into formula (i):

$P(A|B) = \frac{\frac{1}{216}}{\frac{6}{216}}$

Simplify the expression:

$P(A|B) = \frac{1}{216} \times \frac{216}{6}$

Cancel out the common factor 216:

$P(A|B) = \frac{1}{\cancel{216}} \times \frac{\cancel{216}}{6}$

$P(A|B) = \frac{1}{6}$

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Alternate Solution (Using reduced sample space):

Since event B is known to have occurred, our new sample space is the set of outcomes in B.

$B = \{(6, 5, 1), (6, 5, 2), (6, 5, 3), (6, 5, 4), (6, 5, 5), (6, 5, 6)\}$

The number of outcomes in this reduced sample space B is 6.

Event A is "4 on the third throw". Within this reduced sample space B, we look for the outcomes that satisfy event A.

The outcomes in B that have 4 on the third throw is just one outcome:

$\{(6, 5, 4)\}$

This is the event $A \cap B$. The number of favorable outcomes within the reduced sample space is 1.

The probability of A given B is the number of favorable outcomes in $A \cap B$ divided by the total number of outcomes in the reduced sample space B.

$P(A|B) = \frac{\text{Number of outcomes in } A \cap B}{\text{Number of outcomes in B}}$

$P(A|B) = \frac{1}{6}$

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Final Answer:

The probability of getting 4 on the third throw given that there was a 6 on the first and a 5 on the second throw is $\frac{1}{6}$.

Given:

A die is thrown twice.

The sum of the numbers appearing is observed to be 6.

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To Find:

The conditional probability that the number 4 has appeared at least once, given that the sum is 6.

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Solution:

Let S be the sample space of throwing a die twice. The total number of outcomes in S is $6 \times 6 = 36$. Each outcome is an ordered pair $(x, y)$, where x is the result of the first throw and y is the result of the second throw.

Let B be the event that "the sum of the numbers appearing is 6".

The outcomes in B are the pairs $(x, y)$ such that $x+y=6$.

$B = \{(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)\}$

The number of outcomes in B is 5.

Let A be the event that "the number 4 has appeared at least once".

The outcomes in A are the pairs where at least one component is 4.

$A = \{(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (1, 4), (2, 4), (3, 4), (5, 4), (6, 4)\}$

We are given that event B has already occurred, so we need to find the probability of A given B, denoted by $P(A|B)$.

First, find the intersection of events A and B, denoted by $A \cap B$. This is the event where the sum is 6 AND the number 4 has appeared at least once.

From the list of outcomes in B, we identify those that contain the number 4:

$B = \{(1, 5), \textbf{(2, 4)}, (3, 3), \textbf{(4, 2)}, (5, 1)\}$

$A \cap B = \{(2, 4), (4, 2)\}$

The number of outcomes in $A \cap B$ is 2.

The probability of $A \cap B$ is $\frac{\text{Number of outcomes in } A \cap B}{\text{Total number of outcomes}} = \frac{2}{36}$.

The probability of B is $\frac{\text{Number of outcomes in B}}{\text{Total number of outcomes}} = \frac{5}{36}$.

Using the formula for conditional probability:

Substitute the probabilities into formula (i):

$P(A|B) = \frac{\frac{2}{36}}{\frac{5}{36}}$

Simplify the complex fraction:

$P(A|B) = \frac{2}{36} \times \frac{36}{5}$

$P(A|B) = \frac{2}{\cancel{36}} \times \frac{\cancel{36}}{5}$

$P(A|B) = \frac{2}{5}$

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Alternate Solution (Using reduced sample space):

Given that the sum of the numbers appearing is 6, our reduced sample space is the set of outcomes in event B:

$B = \{(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)\}$

The number of outcomes in this reduced sample space is 5.

Within this reduced sample space B, we identify the outcomes where the number 4 has appeared at least once. These are the outcomes in $A \cap B$:

$\{(2, 4), (4, 2)\}$

The number of favorable outcomes within the reduced sample space is 2.

The conditional probability $P(A|B)$ is the number of favorable outcomes in the reduced sample space divided by the total number of outcomes in the reduced sample space:

$P(A|B) = \frac{\text{Number of outcomes in } A \cap B}{\text{Number of outcomes in B}}$

$P(A|B) = \frac{2}{5}$

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Final Answer:

The conditional probability that the number 4 has appeared at least once, given that the sum is 6, is $\frac{2}{5}$.

Given:

Experiment: Toss a coin. If Head (H), toss coin again. If Tail (T), throw a die.

Event E: The die shows a number greater than 4.

Event F: There is at least one tail.

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To Find:

The conditional probability $P(E|F)$.

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Solution:

Let S be the sample space for this experiment. The possible outcomes are:

• If the first toss is Head (probability $1/2$), the second toss can be Head or Tail (each with probability $1/2$). This gives outcomes HH and HT.
• If the first toss is Tail (probability $1/2$), a die is thrown (outcomes 1 to 6, each with probability $1/6$). This gives outcomes T1, T2, T3, T4, T5, T6.

The sample space is $S = \{HH, HT, T1, T2, T3, T4, T5, T6\}$.

The probabilities of these outcomes are:

$P(HH) = P(H \text{ on 1st}) \times P(H \text{ on 2nd}|H \text{ on 1st}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$

$P(HT) = P(H \text{ on 1st}) \times P(T \text{ on 2nd}|H \text{ on 1st}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$

$P(Ti) = P(T \text{ on 1st}) \times P(i \text{ on die}|T \text{ on 1st}) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}$, for $i \in \{1, 2, 3, 4, 5, 6\}$.

Event E is that 'the die shows a number greater than 4'. This corresponds to outcomes where the first toss was Tail and the die showed 5 or 6.

$E = \{T5, T6\}$

$P(E) = P(T5) + P(T6) = \frac{1}{12} + \frac{1}{12} = \frac{2}{12} = \frac{1}{6}$

Event F is that 'there is at least one tail'. This includes the outcome HT from the first branch and all outcomes from the second branch.

$F = \{HT, T1, T2, T3, T4, T5, T6\}$

$P(F) = P(HT) + P(T1) + P(T2) + P(T3) + P(T4) + P(T5) + P(T6) = \frac{1}{4} + 6 \times \frac{1}{12} = \frac{1}{4} + \frac{6}{12} = \frac{1}{4} + \frac{1}{2} = \frac{1}{4} + \frac{2}{4} = \frac{3}{4}$

The intersection of events E and F, $E \cap F$, is the event that 'the die shows a number greater than 4' AND 'there is at least one tail'. The outcomes that satisfy both conditions are the ones from the second branch (first toss was T) where the die is 5 or 6, and also contain a Tail (which all outcomes from the second branch do). The outcome HT has a tail but no die throw, so it's not in E. The outcomes T5 and T6 have a die throw > 4 and have a tail.

$E \cap F = \{T5, T6\}$

$P(E \cap F) = P(T5) + P(T6) = \frac{1}{12} + \frac{1}{12} = \frac{2}{12} = \frac{1}{6}$

The conditional probability $P(E|F)$ is given by the formula:

Substitute the calculated probabilities into formula (i):

$P(E|F) = \frac{1/6}{3/4}$

Simplify the expression:

$P(E|F) = \frac{1}{6} \times \frac{4}{3} = \frac{4}{18} = \frac{2}{9}$

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Final Answer:

The conditional probability of the die showing a number greater than 4, given that there is at least one tail, is $\frac{2}{9}$.

Given:

$P(E) = 0.6$

$P(F) = 0.3$

$P(E \cap F) = 0.2$

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To Find:

$P(E|F)$ and $P(F|E)$

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Solution:

To find $P(E|F)$, we use the formula for conditional probability:

Substitute the given values into formula (i):

$P(E|F) = \frac{0.2}{0.3}$

$P(E|F) = \frac{2}{3}$

To find $P(F|E)$, we use the formula for conditional probability:

Substitute the given values into formula (ii):

$P(F|E) = \frac{0.2}{0.6}$

$P(F|E) = \frac{2}{6}$

$P(F|E) = \frac{1}{3}$

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Final Answer:

$P(E|F) = \frac{2}{3}$

$P(F|E) = \frac{1}{3}$

Given:

$P(B) = 0.5$

$P(A \cap B) = 0.32$

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To Find:

$P(A|B)$

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Solution:

To compute $P(A|B)$, we use the formula for conditional probability:

Substitute the given values into formula (i):

$P(A|B) = \frac{0.32}{0.5}$

To simplify, we can write the decimals as fractions or multiply both numerator and denominator by 100:

$P(A|B) = \frac{0.32 \times 100}{0.5 \times 100} = \frac{32}{50}$

Simplify the fraction by dividing both numerator and denominator by their greatest common divisor, which is 2:

$P(A|B) = \frac{\cancel{32}^{16}}{\cancel{50}_{25}}$

$P(A|B) = \frac{16}{25}$

Alternatively, using decimals directly:

$P(A|B) = \frac{0.32}{0.5} = \frac{0.32}{0.50}$

$P(A|B) = 0.64$

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Final Answer:

$P(A|B) = \frac{16}{25}$ or $0.64$.

Given:

$P(A) = 0.8$

$P(B) = 0.5$

$P(B|A) = 0.4$

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To Find:

(i) $P(A \cap B)$

(ii) $P(A|B)$

(iii) $P(A \cup B)$

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Solution:

(i) Find $P(A \cap B)$:

We use the formula for conditional probability $P(B|A)$:

Rearranging formula (1) to solve for $P(A \cap B)$:

$P(A \cap B) = P(B|A) \times P(A)$

Substitute the given values:

$P(A \cap B) = 0.4 \times 0.8$

$P(A \cap B) = 0.32$

(ii) Find $P(A|B)$:

We use the formula for conditional probability $P(A|B)$:

Substitute the value of $P(A \cap B)$ calculated in (i) and the given value of $P(B)$ into formula (2):

$P(A|B) = \frac{0.32}{0.5}$

$P(A|B) = \frac{32}{50}$

$P(A|B) = \frac{\cancel{32}^{16}}{\cancel{50}_{25}}$

$P(A|B) = \frac{16}{25}$ or $0.64$

(iii) Find $P(A \cup B)$:

We use the formula for the probability of the union of two events:

Substitute the given values of $P(A)$ and $P(B)$, and the calculated value of $P(A \cap B)$ into formula (3):

$P(A \cup B) = 0.8 + 0.5 - 0.32$

$P(A \cup B) = 1.3 - 0.32$

$P(A \cup B) = 0.98$

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Final Answer:

(i) $P(A \cap B) = 0.32$

(ii) $P(A|B) = \frac{16}{25}$ or $0.64$

(iii) $P(A \cup B) = 0.98$

Given:

$2P(A) = \frac{5}{13}$

$P(B) = \frac{5}{13}$

$P(A|B) = \frac{2}{5}$

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To Find:

$P(A \cup B)$

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Solution:

From the given information $2P(A) = \frac{5}{13}$, we can find $P(A)$:

$P(A) = \frac{1}{2} \times \frac{5}{13} = \frac{5}{26}$

We are given $P(B) = \frac{5}{13}$ and $P(A|B) = \frac{2}{5}$.

The formula for conditional probability $P(A|B)$ is:

We can rearrange formula (1) to find $P(A \cap B)$:

$P(A \cap B) = P(A|B) \times P(B)$

Substitute the given values:

$P(A \cap B) = \frac{2}{5} \times \frac{5}{13}$

$P(A \cap B) = \frac{2}{\cancel{5}} \times \frac{\cancel{5}}{13}$

$P(A \cap B) = \frac{2}{13}$

Now, we use the formula for the probability of the union of two events:

Substitute the values of $P(A)$, $P(B)$, and $P(A \cap B)$ into formula (2):

$P(A \cup B) = \frac{5}{26} + \frac{5}{13} - \frac{2}{13}$

To add and subtract these fractions, find a common denominator, which is 26:

$P(A \cup B) = \frac{5}{26} + \frac{5 \times 2}{13 \times 2} - \frac{2 \times 2}{13 \times 2}$

$P(A \cup B) = \frac{5}{26} + \frac{10}{26} - \frac{4}{26}$

Combine the numerators:

$P(A \cup B) = \frac{5 + 10 - 4}{26}$

$P(A \cup B) = \frac{15 - 4}{26}$

$P(A \cup B) = \frac{11}{26}$

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Final Answer:

$P(A \cup B) = \frac{11}{26}$

Given:

$P(A) = \frac{6}{11}$

$P(B) = \frac{5}{11}$

$P(A \cup B) = \frac{7}{11}$

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To Find:

(i) $P(A \cap B)$

(ii) $P(A|B)$

(iii) $P(B|A)$

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Solution:

(i) Find $P(A \cap B)$:

We use the formula for the probability of the union of two events:

Rearranging formula (1) to solve for $P(A \cap B)$:

$P(A \cap B) = P(A) + P(B) - P(A \cup B)$

Substitute the given values:

$P(A \cap B) = \frac{6}{11} + \frac{5}{11} - \frac{7}{11}$

$P(A \cap B) = \frac{6 + 5 - 7}{11}$

$P(A \cap B) = \frac{11 - 7}{11}$

$P(A \cap B) = \frac{4}{11}$

(ii) Find $P(A|B)$:

We use the formula for conditional probability $P(A|B)$:

Substitute the value of $P(A \cap B)$ calculated in (i) and the given value of $P(B)$ into formula (2):

$P(A|B) = \frac{\frac{4}{11}}{\frac{5}{11}}$

$P(A|B) = \frac{4}{11} \times \frac{11}{5}$

$P(A|B) = \frac{4}{\cancel{11}} \times \frac{\cancel{11}}{5}$

$P(A|B) = \frac{4}{5}$

(iii) Find $P(B|A)$:

We use the formula for conditional probability $P(B|A)$:

Substitute the value of $P(A \cap B)$ calculated in (i) and the given value of $P(A)$ into formula (3):

$P(B|A) = \frac{\frac{4}{11}}{\frac{6}{11}}$

$P(B|A) = \frac{4}{11} \times \frac{11}{6}$

$P(B|A) = \frac{4}{\cancel{11}} \times \frac{\cancel{11}}{6}$

$P(B|A) = \frac{4}{6}$

$P(B|A) = \frac{\cancel{4}^{2}}{\cancel{6}_{3}}$

$P(B|A) = \frac{2}{3}$

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Final Answer:

(i) $P(A \cap B) = \frac{4}{11}$

(ii) $P(A|B) = \frac{4}{5}$

(iii) $P(B|A) = \frac{2}{3}$

Given:

A coin is tossed three times.

The sample space $S$ for tossing a coin three times is:

$S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$

The total number of outcomes in $S$ is 8. Assuming the coin is fair, each outcome has a probability of $\frac{1}{8}$.

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To Find:

$P(E|F)$ for each given pair of events (E, F).

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Solution:

We use the formula for conditional probability: $P(E|F) = \frac{P(E \cap F)}{P(F)}$, provided $P(F) \neq 0$.

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(i) E : head on third toss , F : heads on first two tosses

Event E: Head on the third toss.

$E = \{HHH, HTH, THH, TTH\}$

Event F: Heads on the first two tosses.

$F = \{HHH, HHT\}$

The intersection of E and F is the event where both occur: Heads on the first two tosses AND Head on the third toss.

$E \cap F = \{HHH\}$

Number of outcomes in $E \cap F$ is 1.

Number of outcomes in F is 2.

$P(E \cap F) = \frac{1}{8}$

$P(F) = \frac{2}{8} = \frac{1}{4}$

Using the conditional probability formula:

$P(E|F) = \frac{P(E \cap F)}{P(F)} = \frac{1/8}{1/4}$

$P(E|F) = \frac{1}{8} \times \frac{4}{1} = \frac{4}{8} = \frac{1}{2}$

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(ii) E : at least two heads , F : at most two heads

Event E: At least two heads (exactly 2 or 3 heads).

$E = \{HHT, HTH, THH, HHH\}$

Event F: At most two heads (exactly 0, 1 or 2 heads).

$F = \{HTT, THT, TTH, TTT, HHT, HTH, THH\}$

The intersection of E and F is the event where there are at least two heads AND at most two heads. This means exactly two heads.

$E \cap F = \{HHT, HTH, THH\}$

Number of outcomes in $E \cap F$ is 3.

Number of outcomes in F is 7.

$P(E \cap F) = \frac{3}{8}$

$P(F) = \frac{7}{8}$

Using the conditional probability formula:

$P(E|F) = \frac{P(E \cap F)}{P(F)} = \frac{3/8}{7/8}$

$P(E|F) = \frac{3}{\cancel{8}} \times \frac{\cancel{8}}{7} = \frac{3}{7}$

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(iii) E : at most two tails , F : at least one tail

Event E: At most two tails (exactly 0, 1 or 2 tails). This is the complement of getting exactly 3 tails (TTT).

$E = \{HHH, HHT, HTH, THH, HTT, THT, TTH\}$

Event F: At least one tail (exactly 1, 2 or 3 tails). This is the complement of getting exactly 0 tails (HHH).

$F = \{HHT, HTH, THH, HTT, THT, TTH, TTT\}$

The intersection of E and F is the event where there are at most two tails AND at least one tail. This means exactly one tail or exactly two tails.

$E \cap F = \{HHT, HTH, THH, HTT, THT, TTH\}$

Number of outcomes in $E \cap F$ is 6.

Number of outcomes in F is 7.

$P(E \cap F) = \frac{6}{8} = \frac{3}{4}$

$P(F) = \frac{7}{8}$

Using the conditional probability formula:

$P(E|F) = \frac{P(E \cap F)}{P(F)} = \frac{3/4}{7/8}$

$P(E|F) = \frac{3}{4} \times \frac{8}{7} = \frac{3}{\cancel{4}^{1}} \times \frac{\cancel{8}^{2}}{7} = \frac{3 \times 2}{1 \times 7} = \frac{6}{7}$

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Final Answer:

(i) $P(E|F) = \frac{1}{2}$

(ii) $P(E|F) = \frac{3}{7}$

(iii) $P(E|F) = \frac{6}{7}$

Given:

Two coins are tossed once.

The sample space S is:

$S = \{HH, HT, TH, TT\}$

The total number of outcomes is 4. Each outcome has a probability of $\frac{1}{4}$.

---

To Find:

$P(E|F)$ for each given pair of events (E, F).

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Solution:

We use the formula for conditional probability: $P(E|F) = \frac{P(E \cap F)}{P(F)}$, provided $P(F) \neq 0$.

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(i) E : tail appears on one coin, F : one coin shows head

Event E: Tail appears on one coin (exactly one tail).

$E = \{HT, TH\}$

Event F: One coin shows head (exactly one head).

$F = \{HT, TH\}$

The intersection of E and F is the event where both occur: exactly one tail AND exactly one head.

$E \cap F = \{HT, TH\}$

Number of outcomes in $E \cap F$ is 2.

Number of outcomes in F is 2.

$P(E \cap F) = \frac{2}{4} = \frac{1}{2}$

$P(F) = \frac{2}{4} = \frac{1}{2}$

Using the conditional probability formula:

$P(E|F) = \frac{P(E \cap F)}{P(F)} = \frac{1/2}{1/2}$

$P(E|F) = 1$

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(ii) E : no tail appears, F : no head appears

Event E: No tail appears (both are heads).

$E = \{HH\}$

Event F: No head appears (both are tails).

$F = \{TT\}$

The intersection of E and F is the event where both occur: no tail appears AND no head appears.

$E \cap F = \emptyset$ (The empty set)

Number of outcomes in $E \cap F$ is 0.

Number of outcomes in F is 1.

$P(E \cap F) = \frac{0}{4} = 0$

$P(F) = \frac{1}{4}$

Using the conditional probability formula:

$P(E|F) = \frac{P(E \cap F)}{P(F)} = \frac{0}{1/4}$

$P(E|F) = 0$

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Final Answer:

(i) $P(E|F) = 1$

(ii) $P(E|F) = 0$

Given:

A die is thrown three times.

Event E: 4 appears on the third toss.

Event F: 6 and 5 appears respectively on first two tosses.

---

To Find:

The probability of E given that F has already occurred, i.e., $P(E|F)$.

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Solution:

The sample space S for throwing a die three times consists of $6 \times 6 \times 6 = 216$ outcomes. Each outcome is an ordered triplet $(d_1, d_2, d_3)$ where $d_1, d_2, d_3 \in \{1, 2, 3, 4, 5, 6\}$.

Event E is "4 appears on the third toss".

$E = \{(d_1, d_2, 4) : d_1 \in \{1, \dots, 6\}, d_2 \in \{1, \dots, 6\}\}$

The number of outcomes in E is $6 \times 6 \times 1 = 36$.

Event F is "6 and 5 appears respectively on first two tosses".

$F = \{(6, 5, d_3) : d_3 \in \{1, \dots, 6\}\}$

The number of outcomes in F is $1 \times 1 \times 6 = 6$.

The intersection of E and F, $E \cap F$, is the event where both E and F occur. This means the first throw is 6, the second is 5, and the third is 4.

$E \cap F = \{(6, 5, 4)\}$

The number of outcomes in $E \cap F$ is 1.

The probability of $E \cap F$ is $P(E \cap F) = \frac{\text{Number of outcomes in } E \cap F}{\text{Total number of outcomes}} = \frac{1}{216}$.

The probability of F is $P(F) = \frac{\text{Number of outcomes in F}}{\text{Total number of outcomes}} = \frac{6}{216}$.

The probability of E given F is calculated using the formula for conditional probability:

Substitute the calculated probabilities into formula (i):

$P(E|F) = \frac{1/216}{6/216}$

Simplify the expression:

$P(E|F) = \frac{1}{216} \times \frac{216}{6} = \frac{1}{\cancel{216}} \times \frac{\cancel{216}}{6}$

$P(E|F) = \frac{1}{6}$

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Alternate Solution (Using reduced sample space):

Since event F is known to have occurred, the relevant sample space is reduced to the outcomes in F.

$F = \{(6, 5, 1), (6, 5, 2), (6, 5, 3), (6, 5, 4), (6, 5, 5), (6, 5, 6)\}$

The number of outcomes in this reduced sample space is 6.

Within this reduced sample space, we identify the outcomes that satisfy event E (4 on the third toss). The only outcome in F that has 4 on the third toss is $(6, 5, 4)$.

This corresponds to the outcomes in $E \cap F = \{(6, 5, 4)\}$.

The number of favorable outcomes within the reduced sample space is 1.

The conditional probability $P(E|F)$ is the number of favorable outcomes in the reduced sample space divided by the total number of outcomes in the reduced sample space F.

$P(E|F) = \frac{\text{Number of outcomes in } E \cap F}{\text{Number of outcomes in F}}$

$P(E|F) = \frac{1}{6}$

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Final Answer:

The probability of E given F is $\frac{1}{6}$.

Given:

Mother (M), father (Fth), and son (S) line up at random.

Event E: Son on one end.

Event F: Father in middle.

---

To Find:

The probability of E given that F has already occurred, i.e., $P(E|F)$.

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Solution:

The sample space S consists of all possible arrangements of the three people. The permutations of 3 distinct items are $3! = 3 \times 2 \times 1 = 6$. The sample space is:

$S = \{(M, Fth, S), (M, S, Fth), (Fth, M, S), (Fth, S, M), (S, M, Fth), (S, Fth, M)\}$

Each outcome is equally likely with probability $\frac{1}{6}$.

Event E is "son on one end". The outcomes where the son (S) is either in the first or third position are:

$E = \{(S, M, Fth), (S, Fth, M), (M, Fth, S), (Fth, M, S)\}$

The number of outcomes in E is 4.

Event F is "father in middle". The outcomes where the father (Fth) is in the second position are:

$F = \{(M, Fth, S), (S, Fth, M)\}$

The number of outcomes in F is 2.

The intersection of E and F, $E \cap F$, is the event where both E and F occur. This means the father is in the middle AND the son is on one end.

From the outcomes in F, we check if the son is on one of the ends:

$(M, Fth, S)$: Son is on the right end (3rd position).

$(S, Fth, M)$: Son is on the left end (1st position).

So, $E \cap F = \{(M, Fth, S), (S, Fth, M)\}$

The number of outcomes in $E \cap F$ is 2.

The probability of $E \cap F$ is $P(E \cap F) = \frac{\text{Number of outcomes in } E \cap F}{\text{Total number of outcomes}} = \frac{2}{6} = \frac{1}{3}$.

The probability of F is $P(F) = \frac{\text{Number of outcomes in F}}{\text{Total number of outcomes}} = \frac{2}{6} = \frac{1}{3}$.

The probability of E given F is calculated using the formula for conditional probability:

Substitute the calculated probabilities into formula (i):

$P(E|F) = \frac{1/3}{1/3}$

Simplify the expression:

$P(E|F) = 1$

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Alternate Solution (Using reduced sample space):

Since event F is known to have occurred, the relevant sample space is reduced to the outcomes in F.

$F = \{(M, Fth, S), (S, Fth, M)\}$

The number of outcomes in this reduced sample space is 2.

Within this reduced sample space F, we identify the outcomes that satisfy event E (son on one end).

• In $(M, Fth, S)$, the son is on the end.
• In $(S, Fth, M)$, the son is on the end.

Both outcomes in F satisfy event E. These are the outcomes in $E \cap F = \{(M, Fth, S), (S, Fth, M)\}$.

The number of favorable outcomes within the reduced sample space is 2.

The conditional probability $P(E|F)$ is the number of favorable outcomes in the reduced sample space divided by the total number of outcomes in the reduced sample space F.

$P(E|F) = \frac{\text{Number of outcomes in } E \cap F}{\text{Number of outcomes in F}}$

$P(E|F) = \frac{2}{2} = 1$

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Final Answer:

The probability of the son being on one end given that the father is in the middle is 1.

Given:

A black die and a red die are rolled.

The sample space S consists of all ordered pairs $(b, r)$, where $b$ is the outcome on the black die and $r$ is the outcome on the red die, with $b, r \in \{1, 2, 3, 4, 5, 6\}$.

The total number of outcomes is $6 \times 6 = 36$. Each outcome is equally likely.

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(a) Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5.

To Find:

$P(A|B)$ where A is the event 'sum is greater than 9' and B is the event 'black die resulted in a 5'.

Solution:

Event A: Sum greater than 9. The outcomes are pairs $(b, r)$ such that $b+r > 9$.

$A = \{(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)\}$

Event B: Black die resulted in a 5. The outcomes are pairs $(b, r)$ such that $b=5$.

$B = \{(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)\}$

The intersection of A and B, $A \cap B$, is the event where the black die is 5 AND the sum is greater than 9.

From the outcomes in B, we find those whose sum is greater than 9:

$5+1=6$, $5+2=7$, $5+3=8$, $5+4=9$, $5+5=10$, $5+6=11$.

$A \cap B = \{(5, 5), (5, 6)\}$

The number of outcomes in $A \cap B$ is 2.

The number of outcomes in B is 6.

The probabilities are:

$P(A \cap B) = \frac{2}{36}$

$P(B) = \frac{6}{36}$

Using the formula for conditional probability:

Substitute the probabilities into formula (a.i):

$P(A|B) = \frac{2/36}{6/36} = \frac{2}{\cancel{36}} \times \frac{\cancel{36}}{6} = \frac{2}{6} = \frac{1}{3}$

Alternate Solution (Using reduced sample space):

Given that the black die resulted in a 5 (Event B), the reduced sample space consists of the outcomes in B:

$B = \{(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)\}$

The number of outcomes in the reduced sample space is 6.

Within this space, the outcomes where the sum is greater than 9 (Event A) are those in $A \cap B$:

$\{(5, 5), (5, 6)\}$

The number of favorable outcomes in the reduced sample space is 2.

The conditional probability $P(A|B)$ is $\frac{\text{Number of favorable outcomes in } A \cap B}{\text{Number of outcomes in B}} = \frac{2}{6} = \frac{1}{3}$.

Final Answer for (a):

The conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5, is $\frac{1}{3}$.

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(b) Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.

To Find:

$P(C|D)$ where C is the event 'sum is 8' and D is the event 'red die resulted in a number less than 4'.

Solution:

Event C: Sum is 8. The outcomes are pairs $(b, r)$ such that $b+r = 8$.

$C = \{(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)\}$

Event D: Red die resulted in a number less than 4. The outcomes are pairs $(b, r)$ such that $r \in \{1, 2, 3\}$.

$D = \{(b, r) : b \in \{1, \dots, 6\}, r \in \{1, 2, 3\}\}$

The number of outcomes in D is $6 \times 3 = 18$.

The intersection of C and D, $C \cap D$, is the event where the sum is 8 AND the red die is less than 4.

From the outcomes in C, we find those where the red die is 1, 2, or 3:

$(2, 6)$ (red die is 6, not < 4)

$(3, 5)$ (red die is 5, not < 4)

$(4, 4)$ (red die is 4, not < 4)

$(5, 3)$ (red die is 3, is < 4)

$(6, 2)$ (red die is 2, is < 4)

$C \cap D = \{(5, 3), (6, 2)\}$

The number of outcomes in $C \cap D$ is 2.

The number of outcomes in D is 18.

The probabilities are:

$P(C \cap D) = \frac{2}{36}$

$P(D) = \frac{18}{36}$

Using the formula for conditional probability:

Substitute the probabilities into formula (b.i):

$P(C|D) = \frac{2/36}{18/36} = \frac{2}{\cancel{36}} \times \frac{\cancel{36}}{18} = \frac{2}{18} = \frac{1}{9}$

Alternate Solution (Using reduced sample space):

Given that the red die resulted in a number less than 4 (Event D), the reduced sample space consists of the outcomes in D:

$D = \{(b, r) : b \in \{1, \dots, 6\}, r \in \{1, 2, 3\}\}$

The number of outcomes in the reduced sample space is $6 \times 3 = 18$.

Within this space, the outcomes where the sum is 8 (Event C) are those in $C \cap D$:

$\{(5, 3), (6, 2)\}$

The number of favorable outcomes in the reduced sample space is 2.

The conditional probability $P(C|D)$ is $\frac{\text{Number of favorable outcomes in } C \cap D}{\text{Number of outcomes in D}} = \frac{2}{18} = \frac{1}{9}$.

Final Answer for (b):

The conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4, is $\frac{1}{9}$.

Given:

A fair die is rolled.

Sample Space $S = \{1, 2, 3, 4, 5, 6\}$. Total number of outcomes = 6.

Events: $E = \{1, 3, 5\}$, $F = \{2, 3\}$, $G = \{2, 3, 4, 5\}$

---

To Find:

(i) $P(E|F)$ and $P(F|E)$

(ii) $P(E|G)$ and $P(G|E)$

(iii) $P((E \cup F)|G)$ and $P((E \cap F)|G)$

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Solution:

The probabilities of the individual events are:

$P(E) = \frac{|E|}{|S|} = \frac{3}{6} = \frac{1}{2}$

$P(F) = \frac{|F|}{|S|} = \frac{2}{6} = \frac{1}{3}$

$P(G) = \frac{|G|}{|S|} = \frac{4}{6} = \frac{2}{3}$

The formula for conditional probability is $P(A|B) = \frac{P(A \cap B)}{P(B)}$.

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(i) Find $P(E|F)$ and $P(F|E)$:

First, find the intersection of E and F:

$E \cap F = \{1, 3, 5\} \cap \{2, 3\} = \{3\}$

The probability of $E \cap F$ is:

$P(E \cap F) = \frac{|E \cap F|}{|S|} = \frac{1}{6}$

Calculate $P(E|F)$:

$P(E|F) = \frac{1/6}{2/6} = \frac{1}{\cancel{6}} \times \frac{\cancel{6}}{2} = \frac{1}{2}$

Calculate $P(F|E)$:

$P(F|E) = \frac{1/6}{3/6} = \frac{1}{\cancel{6}} \times \frac{\cancel{6}}{3} = \frac{1}{3}$

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(ii) Find $P(E|G)$ and $P(G|E)$:

First, find the intersection of E and G:

$E \cap G = \{1, 3, 5\} \cap \{2, 3, 4, 5\} = \{3, 5\}$

The probability of $E \cap G$ is:

$P(E \cap G) = \frac{|E \cap G|}{|S|} = \frac{2}{6} = \frac{1}{3}$

Calculate $P(E|G)$:

$P(E|G) = \frac{2/6}{4/6} = \frac{2}{\cancel{6}} \times \frac{\cancel{6}}{4} = \frac{2}{4} = \frac{1}{2}$

Calculate $P(G|E)$:

$P(G|E) = \frac{2/6}{3/6} = \frac{2}{\cancel{6}} \times \frac{\cancel{6}}{3} = \frac{2}{3}$

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(iii) Find $P((E \cup F)|G)$ and $P((E \cap F)|G)$:

First, find the union of E and F:

$E \cup F = \{1, 3, 5\} \cup \{2, 3\} = \{1, 2, 3, 5\}$

Now, find the intersection of $(E \cup F)$ and G:

$(E \cup F) \cap G = \{1, 2, 3, 5\} \cap \{2, 3, 4, 5\} = \{2, 3, 5\}$

The probability of $(E \cup F) \cap G$ is:

$P((E \cup F) \cap G) = \frac{|(E \cup F) \cap G|}{|S|} = \frac{3}{6} = \frac{1}{2}$

Calculate $P((E \cup F)|G)$:

$P((E \cup F)|G) = \frac{3/6}{4/6} = \frac{3}{\cancel{6}} \times \frac{\cancel{6}}{4} = \frac{3}{4}$

Next, find the intersection of E and F (calculated in part i):

$E \cap F = \{3\}$

Now, find the intersection of $(E \cap F)$ and G:

$(E \cap F) \cap G = \{3\} \cap \{2, 3, 4, 5\} = \{3\}$

The probability of $(E \cap F) \cap G$ is:

$P((E \cap F) \cap G) = \frac{|(E \cap F) \cap G|}{|S|} = \frac{1}{6}$

Calculate $P((E \cap F)|G)$:

$P((E \cap F)|G) = \frac{1/6}{4/6} = \frac{1}{\cancel{6}} \times \frac{\cancel{6}}{4} = \frac{1}{4}$

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Final Answer:

(i) $P(E|F) = \frac{1}{2}$, $P(F|E) = \frac{1}{3}$

(ii) $P(E|G) = \frac{1}{2}$, $P(G|E) = \frac{2}{3}$

(iii) $P((E \cup F)|G) = \frac{3}{4}$, $P((E \cap F)|G) = \frac{1}{4}$

Given:

A family has two children. Each child is equally likely to be a boy (B) or a girl (G).

The sample space S, considering the order of birth (e.g., oldest first, youngest second), is:

$S = \{BB, BG, GB, GG\}$

The total number of outcomes is 4. Each outcome is equally likely with probability $\frac{1}{4}$.

Let A be the event that "both children are girls".

$A = \{GG\}$

$P(A) = \frac{1}{4}$

---

To Find:

(i) The conditional probability that both are girls given that the youngest is a girl.

(ii) The conditional probability that both are girls given that at least one is a girl.

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Solution:

(i) Find P(Both girls | Youngest is girl):

Let $F_1$ be the event that "the youngest is a girl".

The outcomes where the youngest child is a girl are:

$F_1 = \{BG, GG\}$

The number of outcomes in $F_1$ is 2.

$P(F_1) = \frac{2}{4} = \frac{1}{2}$

The intersection of A and $F_1$, $A \cap F_1$, is the event that "both children are girls" AND "the youngest is a girl".

$A \cap F_1 = \{GG\} \cap \{BG, GG\} = \{GG\}$

The number of outcomes in $A \cap F_1$ is 1.

$P(A \cap F_1) = \frac{1}{4}$

The conditional probability $P(A|F_1)$ is given by:

Substitute the probabilities into formula (i.1):

$P(A|F_1) = \frac{1/4}{1/2} = \frac{1}{4} \times \frac{2}{1} = \frac{2}{4} = \frac{1}{2}$

(ii) Find P(Both girls | At least one is girl):

Let $F_2$ be the event that "at least one child is a girl".

The outcomes where at least one child is a girl are:

$F_2 = \{BG, GB, GG\}$

The number of outcomes in $F_2$ is 3.

$P(F_2) = \frac{3}{4}$

The intersection of A and $F_2$, $A \cap F_2$, is the event that "both children are girls" AND "at least one is a girl".

$A \cap F_2 = \{GG\} \cap \{BG, GB, GG\} = \{GG\}$

The number of outcomes in $A \cap F_2$ is 1.

$P(A \cap F_2) = \frac{1}{4}$

The conditional probability $P(A|F_2)$ is given by:

Substitute the probabilities into formula (ii.1):

$P(A|F_2) = \frac{1/4}{3/4} = \frac{1}{\cancel{4}} \times \frac{\cancel{4}}{3} = \frac{1}{3}$

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Final Answer:

(i) The conditional probability that both are girls given that the youngest is a girl is $\frac{1}{2}$.

(ii) The conditional probability that both are girls given that at least one is a girl is $\frac{1}{3}$.

Given:

• Easy True/False questions: 300
• Difficult True/False questions: 200
• Easy Multiple Choice questions: 500
• Difficult Multiple Choice questions: 400

Total number of questions in the question bank = $300 + 200 + 500 + 400 = 1400$.

---

To Find:

The probability that a randomly selected question is easy, given that it is a multiple choice question.

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Solution:

Let E be the event that the selected question is an easy question.

Let M be the event that the selected question is a multiple choice question.

We want to find the conditional probability $P(E|M)$.

The number of outcomes in the sample space (total number of questions) is 1400.

The number of questions that are multiple choice (Event M) is the sum of easy multiple choice and difficult multiple choice questions:

Number of multiple choice questions = $500 + 400 = 900$

$P(M) = \frac{\text{Number of multiple choice questions}}{\text{Total number of questions}} = \frac{900}{1400} = \frac{9}{14}$

The intersection of E and M, $E \cap M$, is the event that the question is easy AND it is a multiple choice question.

Number of outcomes in $E \cap M$ = Number of easy multiple choice questions = 500.

$P(E \cap M) = \frac{\text{Number of easy multiple choice questions}}{\text{Total number of questions}} = \frac{500}{1400} = \frac{5}{14}$

Using the formula for conditional probability:

Substitute the probabilities into formula (i):

$P(E|M) = \frac{5/14}{9/14}$

$P(E|M) = \frac{5}{\cancel{14}} \times \frac{\cancel{14}}{9} = \frac{5}{9}$

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Alternate Solution (Using reduced sample space):

Given that the selected question is a multiple choice question (Event M), our reduced sample space consists only of multiple choice questions. The total number of questions in this reduced space is $500 + 400 = 900$.

Within this reduced sample space, the questions that are easy (Event E) are the easy multiple choice questions, which number 500.

The conditional probability $P(E|M)$ is the number of favorable outcomes (easy multiple choice) in the reduced sample space divided by the total number of outcomes (multiple choice) in the reduced sample space.

$P(E|M) = \frac{\text{Number of easy multiple choice questions}}{\text{Total number of multiple choice questions}}$

$P(E|M) = \frac{500}{900} = \frac{5}{9}$

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Final Answer:

The probability that the question is easy given that it is a multiple choice question is $\frac{5}{9}$.

Given:

Two dice are thrown.

The numbers appearing on the two dice are different.

---

To Find:

The probability that the sum of the numbers on the dice is 4, given that the numbers are different.

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Solution:

Let S be the sample space when two dice are thrown. The total number of outcomes is $6 \times 6 = 36$. Each outcome is an ordered pair $(d_1, d_2)$ where $d_1, d_2 \in \{1, 2, 3, 4, 5, 6\}$.

Let A be the event that "the two numbers appearing on the dice are different".

The outcomes where the numbers are the same are $(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)$. There are 6 such outcomes.

The outcomes where the numbers are different are the total outcomes minus those where they are the same.

Number of outcomes in A = $36 - 6 = 30$.

$P(A) = \frac{\text{Number of outcomes in A}}{\text{Total number of outcomes}} = \frac{30}{36} = \frac{5}{6}$.

Let B be the event that "the sum of the numbers on the dice is 4".

The outcomes in B are the pairs $(d_1, d_2)$ such that $d_1 + d_2 = 4$.

$B = \{(1, 3), (2, 2), (3, 1)\}$

The number of outcomes in B is 3.

We are asked to find the probability of event B occurring given that event A has occurred, i.e., $P(B|A)$.

First, find the intersection of events A and B, denoted by $A \cap B$. This is the event where the sum is 4 AND the numbers are different.

From the outcomes in B, we select those where the two numbers are different:

$B = \{(1, 3), (2, 2), (3, 1)\}$

The outcomes $(1,3)$ and $(3,1)$ have different numbers. The outcome $(2,2)$ has the same numbers.

$A \cap B = \{(1, 3), (3, 1)\}$

The number of outcomes in $A \cap B$ is 2.

The probability of $A \cap B$ is $P(A \cap B) = \frac{\text{Number of outcomes in } A \cap B}{\text{Total number of outcomes}} = \frac{2}{36} = \frac{1}{18}$.

Using the formula for conditional probability:

Substitute the calculated probabilities into formula (i):

$P(B|A) = \frac{1/18}{5/6}$

$P(B|A) = \frac{1}{18} \times \frac{6}{5}$

$P(B|A) = \frac{1}{\cancel{18}_{3}} \times \frac{\cancel{6}^{1}}{5}$

$P(B|A) = \frac{1 \times 1}{3 \times 5} = \frac{1}{15}$

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Alternate Solution (Using reduced sample space):

Given that the two numbers appearing are different (Event A), the relevant sample space is reduced to the outcomes in A. The number of outcomes in this reduced sample space is $|A| = 30$.

Within this reduced sample space A, we identify the outcomes that satisfy event B (the sum of numbers is 4).

The outcomes from the original sample space where the sum is 4 are $\{(1, 3), (2, 2), (3, 1)\}$.

Out of these, only $(1, 3)$ and $(3, 1)$ consist of different numbers. The outcome $(2,2)$ is excluded from the reduced sample space A.

So, the favorable outcomes in the reduced sample space A are $\{(1, 3), (3, 1)\}$. The number of favorable outcomes is 2.

The conditional probability $P(B|A)$ is the number of favorable outcomes in the reduced sample space divided by the total number of outcomes in the reduced sample space.

$P(B|A) = \frac{\text{Number of outcomes in } A \cap B}{\text{Number of outcomes in A}}$

$P(B|A) = \frac{2}{30} = \frac{1}{15}$

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Final Answer:

The probability of the sum of numbers on the dice being 4, given that the numbers are different, is $\frac{1}{15}$.

Given:

Experiment: Throw a die. If the outcome is a multiple of 3 ({3, 6}), throw the die again. If the outcome is any other number ({1, 2, 4, 5}), toss a coin.

Event E: The coin shows a tail.

Event F: At least one die shows a 3.

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To Find:

The conditional probability $P(E|F)$.

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Solution:

Let's define the sample space S for this experiment and determine the probability of each outcome. The outcomes depend on the result of the first die throw ($d_1$).

• If $d_1 \in \{1, 2, 4, 5\}$ (probability $\frac{4}{6} = \frac{2}{3}$), toss a coin (H or T, each with probability $\frac{1}{2}$). The outcomes are $(d_1, C)$ where $C \in \{H, T\}$.
• If $d_1 \in \{3, 6\}$ (probability $\frac{2}{6} = \frac{1}{3}$), throw a second die ($d_2 \in \{1, 2, 3, 4, 5, 6\}$, each with probability $\frac{1}{6}$). The outcomes are $(d_1, d_2)$.

The sample space S is the set of all possible outcomes:

$S = \{(1, H), (1, T), (2, H), (2, T), (4, H), (4, T), (5, H), (5, T), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)\}$

The probabilities of these outcomes are:

• For $(d_1, C)$ where $d_1 \in \{1, 2, 4, 5\}$: $P((d_1, C)) = P(d_1 \text{ on 1st}) \times P(C \text{ on coin}|d_1) = \frac{1}{6} \times \frac{1}{2} = \frac{1}{12}$
• For $(d_1, d_2)$ where $d_1 \in \{3, 6\}$: $P((d_1, d_2)) = P(d_1 \text{ on 1st}) \times P(d_2 \text{ on 2nd}|d_1) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$

Event E: 'the coin shows a tail'. This occurs when the first die was not a multiple of 3, and the coin toss was Tail.

$E = \{(1, T), (2, T), (4, T), (5, T)\}$

The probability of E is the sum of the probabilities of these outcomes:

$P(E) = P(1, T) + P(2, T) + P(4, T) + P(5, T) = 4 \times \frac{1}{12} = \frac{4}{12} = \frac{1}{3}$

Event F: 'at least one die shows a 3'. This means either the result of the first die throw is 3, or the result of the second die throw (if the die was thrown again) is 3.

• Outcomes where the first die is 3: $(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)$. These are in F.
• Outcomes where the first die is 6, and the second die is 3: $(6, 3)$. This is in F.
• Outcomes where the first die is not a multiple of 3 ({1, 2, 4, 5}): A coin is tossed, not a die. So, a 3 cannot appear in the second part of these outcomes.

So, event F consists of the following outcomes:

$F = \{(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (6, 3)\}$

The probability of F is the sum of the probabilities of these outcomes:

$P(F) = P(3, 1) + P(3, 2) + P(3, 3) + P(3, 4) + P(3, 5) + P(3, 6) + P(6, 3)$

$P(F) = 6 \times \frac{1}{36} + 1 \times \frac{1}{36} = \frac{6}{36} + \frac{1}{36} = \frac{7}{36}$

The intersection of E and F, $E \cap F$, is the event where both E and F occur. Outcomes in E come from the branch where the first die was {1, 2, 4, 5} and a coin was tossed. Outcomes in F come exclusively from the branch where the first die was {3, 6} and a second die was thrown.

There are no outcomes common to E and F.

$E \cap F = \emptyset$ (The empty set)

The probability of the intersection is $P(E \cap F) = 0$.

The conditional probability of E given F is calculated using the formula:

Substitute the calculated probabilities into formula (i):

$P(E|F) = \frac{0}{7/36}$

$P(E|F) = 0$

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Alternate Solution (Using reduced sample space):

Given that event F ('at least one die shows a 3') has occurred, our relevant sample space is restricted to the outcomes in F.

$F = \{(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (6, 3)\}$

The total probability of this reduced sample space is $P(F) = 7/36$.

We are looking for the probability of event E ('the coin shows a tail') within this reduced sample space. Event E is $E = \{(1, T), (2, T), (4, T), (5, T)\}$.

We need to find the outcomes that are in both E and F, which is $E \cap F$. As determined earlier, $E \cap F = \emptyset$.

The probability of the intersection within the original sample space is $P(E \cap F) = 0$.

The conditional probability $P(E|F)$ is the probability of the intersection divided by the probability of the conditioning event (F):

$P(E|F) = \frac{P(E \cap F)}{P(F)} = \frac{0}{7/36} = 0$

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Final Answer:

The conditional probability of the coin showing a tail, given that at least one die shows a 3, is 0.

Given:

$P(A) = \frac{1}{2}$

$P(B) = 0$

---

To Find:

$P(A|B)$

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Solution:

The formula for conditional probability $P(A|B)$ is given by:

For $P(A|B)$ to be defined, the probability of the conditioning event, $P(B)$, must be greater than 0.

In this question, we are given $P(B) = 0$.

Since the denominator in the formula for $P(A|B)$ is $P(B)$, and $P(B) = 0$, the expression $\frac{P(A \cap B)}{0}$ involves division by zero.

Division by zero is undefined.

Therefore, $P(A|B)$ is not defined when $P(B) = 0$.

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Final Answer:

The correct answer is (C) not defined.

Given:

A and B are events such that $P(A|B) = P(B|A)$.

---

To Determine:

Which of the given statements is true.

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Solution:

The formula for conditional probability is:

For $P(A|B)$ and $P(B|A)$ to be defined, we must have $P(B) > 0$ and $P(A) > 0$.

Given that $P(A|B) = P(B|A)$, substituting the formulas (1) and (2), we get:

$\frac{P(A \cap B)}{P(B)} = \frac{P(A \cap B)}{P(A)}$

Multiply both sides by $P(A) P(B)$ (since $P(A)>0, P(B)>0$):

$P(A \cap B) P(A) = P(A \cap B) P(B)$

Rearrange the equation:

$P(A \cap B) P(A) - P(A \cap B) P(B) = 0$

$P(A \cap B) (P(A) - P(B)) = 0$

This equation holds if and only if either $P(A \cap B) = 0$ or $P(A) - P(B) = 0$.

So, the condition $P(A|B) = P(B|A)$ implies that $P(A \cap B) = 0$ or $P(A) = P(B)$ (assuming $P(A)>0, P(B)>0$).

Let's examine the given options in light of this result:

• (A) $A \subset B$ but $A \neq B$: This generally implies $P(A) < P(B)$ and $P(A \cap B) = P(A)$. If $P(A) > 0$, then $P(A \cap B) > 0$ and $P(A) \neq P(B)$, which contradicts the condition ($P(A \cap B)=0$ or $P(A)=P(B)$).
• (B) $A = B$: This implies $P(A) = P(B)$ and $P(A \cap B) = P(A)$. If $P(A) > 0$, this satisfies $P(A) = P(B)$, so the original equality holds. However, the original equality does not imply $A=B$ (e.g., if $P(A)=P(B)$ but $A \neq B$, and $P(A \cap B)>0$).
• (C) $A \cap B = \phi$: This implies $P(A \cap B) = 0$. If $P(A)>0$ and $P(B)>0$, the condition $P(A \cap B)=0$ or $P(A)=P(B)$ is satisfied (specifically, $P(A \cap B)=0$ is true). So, if $A \cap B = \phi$, the original equality holds. However, the original equality does not imply $A \cap B = \phi$ (e.g., if $P(A)=P(B)$ and $A \cap B \neq \phi$).
• (D) $P(A) = P(B)$: If $P(A) = P(B) > 0$, the condition $P(A \cap B)=0$ or $P(A)=P(B)$ is satisfied (specifically, $P(A)=P(B)$ is true). Thus, if $P(A)=P(B)>0$, the original equality $P(A|B) = P(B|A)$ holds. Conversely, as shown by $P(A \cap B) (P(A) - P(B)) = 0$, if $P(A \cap B) \neq 0$, then it must be true that $P(A) = P(B)$. While the equality can also hold when $P(A \cap B)=0$ and $P(A) \neq P(B)$, among the given options, $P(A) = P(B)$ is the relationship that holds when the intersection is non-zero, and is the most likely intended answer in this type of question.

The condition $P(A|B) = P(B|A)$ (when defined) is equivalent to "$P(A \cap B) = 0$ or $P(A) = P(B)$". Out of the given options, $P(A) = P(B)$ is the one that, if true (and $P(A)=P(B)>0$), guarantees the original equality holds. Conversely, the original equality doesn't strictly imply any single option to be true in all cases, but option (D) is the most direct probabilistic relationship implied when the intersection is non-zero.

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Final Answer:

The correct answer is (D) P(A) = P(B).

Given:

Number of black balls = 10

Number of white balls = 5

Total number of balls in the urn = $10 + 5 = 15$.

Two balls are drawn one after the other without replacement.

---

To Find:

The probability that both drawn balls are black.

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Solution:

Let A be the event that the first ball drawn is black.

Let B be the event that the second ball drawn is black.

We want to find the probability of the event that both balls drawn are black, which is $P(A \cap B)$.

The probability that the first ball drawn is black is:

$P(A) = \frac{\text{Number of black balls}}{\text{Total number of balls}} = \frac{10}{15}$

Since the first ball drawn was black and it is not replaced, there are now $15 - 1 = 14$ balls left in the urn, and $10 - 1 = 9$ of them are black.

The probability that the second ball drawn is black, given that the first ball was black, is the conditional probability $P(B|A)$:

$P(B|A) = \frac{\text{Number of remaining black balls}}{\text{Total number of remaining balls}} = \frac{9}{14}$

Using the multiplication rule for probabilities (for dependent events):

Substitute the calculated probabilities into formula (i):

$P(A \cap B) = \frac{10}{15} \times \frac{9}{14}$

Simplify the expression by cancelling common factors:

$P(A \cap B) = \frac{\cancel{10}^{2}}{\cancel{15}^{3}} \times \frac{\cancel{9}^{3}}{\cancel{14}^{7}}$

$P(A \cap B) = \frac{2}{3} \times \frac{3}{7}$

$P(A \cap B) = \frac{2}{\cancel{3}} \times \frac{\cancel{3}}{7}$

$P(A \cap B) = \frac{2}{7}$

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Final Answer:

The probability that both drawn balls are black is $\frac{2}{7}$.

Given:

A pack of 52 well shuffled cards.

Three cards are drawn successively without replacement.

---

To Find:

The probability that the first two cards are kings and the third card is an ace.

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Solution:

Let A be the event that the first card drawn is a king.

Let B be the event that the second card drawn is a king.

Let C be the event that the third card drawn is an ace.

We want to find the probability of the event $A \cap B \cap C$, which is the probability that the first card is a king, the second is a king, and the third is an ace.

There are 4 kings in a pack of 52 cards.

The probability that the first card drawn is a king is:

$P(A) = \frac{\text{Number of kings}}{\text{Total number of cards}} = \frac{4}{52} = \frac{1}{13}$

Given that the first card was a king and it was not replaced, there are now 51 cards left in the pack, and $4 - 1 = 3$ of them are kings.

The probability that the second card drawn is a king, given that the first card was a king, is the conditional probability $P(B|A)$:

$P(B|A) = \frac{\text{Number of remaining kings}}{\text{Total number of remaining cards}} = \frac{3}{51} = \frac{1}{17}$

Given that the first two cards were kings and they were not replaced, there are now 50 cards left in the pack. The number of aces is still 4 (since the first two cards were kings, not aces).

The probability that the third card drawn is an ace, given that the first two cards were kings, is the conditional probability $P(C|A \cap B)$:

$P(C|A \cap B) = \frac{\text{Number of aces}}{\text{Total number of remaining cards}} = \frac{4}{50} = \frac{2}{25}$

Using the multiplication rule for probabilities for three dependent events:

Substitute the calculated probabilities into formula (i):

$P(A \cap B \cap C) = \frac{4}{52} \times \frac{3}{51} \times \frac{4}{50}$

Simplify the expression:

$P(A \cap B \cap C) = \frac{\cancel{4}^{1}}{\cancel{52}^{13}} \times \frac{\cancel{3}^{1}}{\cancel{51}^{17}} \times \frac{\cancel{4}^{2}}{\cancel{50}^{25}}$

$P(A \cap B \cap C) = \frac{1}{13} \times \frac{1}{17} \times \frac{2}{25}$

$P(A \cap B \cap C) = \frac{1 \times 1 \times 2}{13 \times 17 \times 25}$

$P(A \cap B \cap C) = \frac{2}{13 \times (17 \times 25)}$

$17 \times 25 = 17 \times (20 + 5) = 340 + 85 = 425$

$13 \times 425 = 13 \times (400 + 25) = 5200 + 325 = 5525$

$P(A \cap B \cap C) = \frac{2}{5525}$

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Final Answer:

The probability that the first two cards are kings and the third card is an ace is $\frac{2}{5525}$.

Given:

A die is thrown.

Event E: The number appearing is a multiple of 3.

Event F: The number appearing is even.

---

To Find:

Whether events E and F are independent.

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Solution:

The sample space S when a die is thrown is:

$S = \{1, 2, 3, 4, 5, 6\}$

The total number of outcomes is $|S| = 6$.

Event E: The number appearing is a multiple of 3. The multiples of 3 in the sample space are 3 and 6.

$E = \{3, 6\}$

The number of outcomes in E is $|E| = 2$.

The probability of event E is $P(E) = \frac{|E|}{|S|} = \frac{2}{6} = \frac{1}{3}$.

Event F: The number appearing is even. The even numbers in the sample space are 2, 4, and 6.

$F = \{2, 4, 6\}$

The number of outcomes in F is $|F| = 3$.

The probability of event F is $P(F) = \frac{|F|}{|S|} = \frac{3}{6} = \frac{1}{2}$.

For events E and F to be independent, the following condition must be satisfied:

First, let's find the intersection of events E and F, $E \cap F$. This is the event where the number is a multiple of 3 AND the number is even. The common outcomes are only 6.

$E \cap F = \{3, 6\} \cap \{2, 4, 6\} = \{6\}$

The number of outcomes in $E \cap F$ is $|E \cap F| = 1$.

The probability of $E \cap F$ is $P(E \cap F) = \frac{|E \cap F|}{|S|} = \frac{1}{6}$.

Now, let's calculate the product $P(E) \times P(F)$:

$P(E) \times P(F) = \frac{1}{3} \times \frac{1}{2} = \frac{1}{6}$

Comparing $P(E \cap F)$ and $P(E) \times P(F)$:

$P(E \cap F) = \frac{1}{6}$

$P(E) \times P(F) = \frac{1}{6}$

Since $P(E \cap F) = P(E) \times P(F)$, the events E and F are independent.

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Final Answer:

Yes, the events E and F are independent.

Given:

An unbiased die is thrown twice.

Event A: Odd number on the first throw.

Event B: Odd number on the second throw.

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To Check:

The independence of events A and B.

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Solution:

The sample space S when an unbiased die is thrown twice consists of ordered pairs $(d_1, d_2)$ where $d_1, d_2 \in \{1, 2, 3, 4, 5, 6\}$.

The total number of outcomes is $|S| = 6 \times 6 = 36$. Each outcome is equally likely.

Event A: Odd number on the first throw. This means $d_1 \in \{1, 3, 5\}$ and $d_2$ can be any number from 1 to 6.

$A = \{(d_1, d_2) : d_1 \in \{1, 3, 5\}, d_2 \in \{1, 2, 3, 4, 5, 6\}\}$

The number of outcomes in A is $|A| = 3 \times 6 = 18$.

The probability of event A is $P(A) = \frac{|A|}{|S|} = \frac{18}{36} = \frac{1}{2}$.

Event B: Odd number on the second throw. This means $d_1$ can be any number from 1 to 6, and $d_2 \in \{1, 3, 5\}$.

$B = \{(d_1, d_2) : d_1 \in \{1, 2, 3, 4, 5, 6\}, d_2 \in \{1, 3, 5\}\}$

The number of outcomes in B is $|B| = 6 \times 3 = 18$.

The probability of event B is $P(B) = \frac{|B|}{|S|} = \frac{18}{36} = \frac{1}{2}$.

For events A and B to be independent, the condition $P(A \cap B) = P(A) \times P(B)$ must be satisfied.

First, find the intersection of events A and B, $A \cap B$. This is the event where the first throw is odd AND the second throw is odd.

$A \cap B = \{(d_1, d_2) : d_1 \in \{1, 3, 5\}, d_2 \in \{1, 3, 5\}\}$

The outcomes in $A \cap B$ are:

$A \cap B = \{(1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5)\}$

The number of outcomes in $A \cap B$ is $|A \cap B| = 3 \times 3 = 9$.

The probability of $A \cap B$ is $P(A \cap B) = \frac{|A \cap B|}{|S|} = \frac{9}{36} = \frac{1}{4}$.

Now, let's calculate the product $P(A) \times P(B)$:

$P(A) \times P(B) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$

Comparing $P(A \cap B)$ and $P(A) \times P(B)$:

$P(A \cap B) = \frac{1}{4}$

$P(A) \times P(B) = \frac{1}{4}$

Since $P(A \cap B) = P(A) \times P(B)$, the events A and B are independent.

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Final Answer:

Yes, the events A and B are independent.

Given:

Three coins are tossed simultaneously.

Event E: Three heads or three tails.

Event F: At least two heads.

Event G: At most two heads.

---

To Determine:

Which pairs of events among (E,F), (E,G), and (F,G) are independent or dependent.

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Solution:

The sample space S for tossing three coins simultaneously is:

$S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$

The total number of outcomes is $|S| = 8$. Each outcome is equally likely, with probability $\frac{1}{8}$.

Let's define the events E, F, and G by listing their outcomes:

Event E: Three heads or three tails.

$E = \{HHH, TTT\}$

$|E| = 2$, $P(E) = \frac{2}{8} = \frac{1}{4}$

Event F: At least two heads (exactly 2 or 3 heads).

$F = \{HHT, HTH, THH, HHH\}$

$|F| = 4$, $P(F) = \frac{4}{8} = \frac{1}{2}$

Event G: At most two heads (exactly 0, 1, or 2 heads). This is the complement of getting exactly 3 heads (HHH).

$G = \{HHT, HTH, THH, HTT, THT, TTH, TTT\}$

$|G| = 7$, $P(G) = \frac{7}{8}$

We check for independence using the condition $P(A \cap B) = P(A) \times P(B)$ for each pair of events.

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Check for pair (E,F):

Find the intersection $E \cap F$. This is the event where there are three heads or three tails AND at least two heads.

$E \cap F = \{HHH, TTT\} \cap \{HHT, HTH, THH, HHH\} = \{HHH\}$

$|E \cap F| = 1$, $P(E \cap F) = \frac{1}{8}$

Calculate $P(E) \times P(F)$:

$P(E) \times P(F) = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}$

Compare $P(E \cap F)$ and $P(E) \times P(F)$:

Since $P(E \cap F) = P(E) \times P(F)$, the events E and F are independent.

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Check for pair (E,G):

Find the intersection $E \cap G$. This is the event where there are three heads or three tails AND at most two heads.

$E \cap G = \{HHH, TTT\} \cap \{HHT, HTH, THH, HTT, THT, TTH, TTT\} = \{TTT\}$

$|E \cap G| = 1$, $P(E \cap G) = \frac{1}{8}$

Calculate $P(E) \times P(G)$:

$P(E) \times P(G) = \frac{1}{4} \times \frac{7}{8} = \frac{7}{32}$

Compare $P(E \cap G)$ and $P(E) \times P(G)$:

Since $P(E \cap G) \neq P(E) \times P(G)$ ($\frac{1}{8} \neq \frac{7}{32}$), the events E and G are dependent.

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Check for pair (F,G):

Find the intersection $F \cap G$. This is the event where there are at least two heads AND at most two heads. This means exactly two heads.

$F \cap G = \{HHT, HTH, THH, HHH\} \cap \{HHT, HTH, THH, HTT, THT, TTH, TTT\} = \{HHT, HTH, THH\}$

$|F \cap G| = 3$, $P(F \cap G) = \frac{3}{8}$

Calculate $P(F) \times P(G)$:

$P(F) \times P(G) = \frac{1}{2} \times \frac{7}{8} = \frac{7}{16}$

Compare $P(F \cap G)$ and $P(F) \times P(G)$:

Since $P(F \cap G) \neq P(F) \times P(G)$ ($\frac{3}{8} \neq \frac{7}{16}$), the events F and G are dependent.

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Final Answer:

The pair (E,F) is independent.

The pairs (E,G) and (F,G) are dependent.

Given:

E and F are independent events.

By the definition of independent events, this means $P(E \cap F) = P(E) \times P(F)$.

---

To Prove:

The events E and F′ (the complement of F) are independent.

This means we need to show that $P(E \cap F') = P(E) \times P(F')$.

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Proof:

We know that the event E can be written as the union of two mutually exclusive (disjoint) events: the intersection of E and F ($E \cap F$) and the intersection of E and the complement of F ($E \cap F'$).

That is, $E = (E \cap F) \cup (E \cap F')$.

Since $E \cap F$ and $E \cap F'$ are disjoint, the probability of their union is the sum of their probabilities:

Rearranging equation (i) to solve for $P(E \cap F')$:

$P(E \cap F') = P(E) - P(E \cap F)$

Since E and F are given to be independent, we know that $P(E \cap F) = P(E) \times P(F)$. Substitute this into equation (ii):

Factor out $P(E)$ from the right side of equation (iii):

We know that the probability of the complement of an event F is $P(F') = 1 - P(F)$. Substitute this into equation (iv):

Equation (v) shows that the probability of the intersection of E and F′ is equal to the product of their individual probabilities. By the definition of independent events, this means that E and F′ are independent.

Thus, if E and F are independent events, then so are the events E and F′.

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Conclusion:

The proof is complete.

Given:

A and B are two independent events.

By definition, this means $P(A \cap B) = P(A) \times P(B)$.

---

To Prove:

The probability of occurrence of at least one of A and B is $1 - P(A') P(B')$.

In other words, we need to prove that $P(A \cup B) = 1 - P(A') P(B')$.

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Proof:

The event "at least one of A and B occurs" is represented by the union of events A and B, denoted as $A \cup B$.

The complement of the event $A \cup B$ is the event that neither A nor B occurs. Using De Morgan's laws, this is $(A \cup B)' = A' \cap B'$.

The probability of the union of two events can be expressed in terms of the probability of its complement:

Substitute the result from De Morgan's law into equation (1):

We are given that A and B are independent events. A property of independent events is that if A and B are independent, then their complements A′ and B′ are also independent.

Substitute this result into equation (2):

Equation (3) shows that the probability of occurrence of at least one of A and B is equal to $1 - P(A') P(B')$.

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Conclusion:

Thus, if A and B are two independent events, the probability of occurrence of at least one of A and B is given by $1 – P(A′) P(B′)$.

The proof is complete.

Given:

$P(A) = \frac{3}{5}$

$P(B) = \frac{1}{5}$

Events A and B are independent.

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To Find:

$P(A \cap B)$

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Solution:

By the definition of independent events, if A and B are independent, then the probability of their intersection is the product of their individual probabilities.

Substitute the given values of $P(A)$ and $P(B)$ into the formula:

$P(A \cap B) = \frac{3}{5} \times \frac{1}{5}$

Multiply the fractions:

$P(A \cap B) = \frac{3 \times 1}{5 \times 5}$

$P(A \cap B) = \frac{3}{25}$

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Final Answer:

The probability of the intersection of events A and B is $\frac{3}{25}$.

Given:

A pack of 52 playing cards.

Two cards are drawn at random and without replacement.

---

To Find:

The probability that both drawn cards are black.

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Solution:

In a standard deck of 52 playing cards, there are 26 black cards (13 spades and 13 clubs) and 26 red cards (13 hearts and 13 diamonds).

Let A be the event that the first card drawn is black.

Let B be the event that the second card drawn is black.

We want to find the probability of the event $A \cap B$, which is the probability that the first card is black AND the second card is black.

The probability that the first card drawn is black is:

$P(A) = \frac{\text{Number of black cards}}{\text{Total number of cards}} = \frac{26}{52} = \frac{1}{2}$

Since the first card drawn was black and it is not replaced, there are now $52 - 1 = 51$ cards left in the pack. The number of black cards remaining is $26 - 1 = 25$.

The probability that the second card drawn is black, given that the first card was black, is the conditional probability $P(B|A)$:

$P(B|A) = \frac{\text{Number of remaining black cards}}{\text{Total number of remaining cards}} = \frac{25}{51}$

Using the multiplication rule for probabilities (for dependent events):

Substitute the calculated probabilities into formula (i):

$P(A \cap B) = \frac{26}{52} \times \frac{25}{51}$

Simplify the expression:

$P(A \cap B) = \frac{\cancel{26}^{1}}{\cancel{52}^{2}} \times \frac{25}{51}$

$P(A \cap B) = \frac{1}{2} \times \frac{25}{51}$

$P(A \cap B) = \frac{1 \times 25}{2 \times 51}$

$P(A \cap B) = \frac{25}{102}$

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Final Answer:

The probability that both drawn cards are black is $\frac{25}{102}$.

Given:

A box contains 15 oranges.

Number of good oranges = 12

Number of bad oranges = 3

Three oranges are randomly selected without replacement.

The box is approved if all three selected oranges are good.

---

To Find:

The probability that the box will be approved for sale.

---

Solution:

Let $G_1$ be the event that the first orange drawn is good.

Let $G_2$ be the event that the second orange drawn is good.

Let $G_3$ be the event that the third orange drawn is good.

For the box to be approved, all three oranges must be good. We want to find the probability of the event $G_1 \cap G_2 \cap G_3$.

The probability that the first orange drawn is good is:

$P(G_1) = \frac{\text{Number of good oranges}}{\text{Total number of oranges}} = \frac{12}{15}$

Given that the first orange was good and it is not replaced, there are now $15 - 1 = 14$ oranges left in the box, and $12 - 1 = 11$ of them are good.

The probability that the second orange drawn is good, given that the first orange was good, is the conditional probability $P(G_2|G_1)$:

$P(G_2|G_1) = \frac{\text{Number of remaining good oranges}}{\text{Total number of remaining oranges}} = \frac{11}{14}$

Given that the first two oranges were good and they are not replaced, there are now $14 - 1 = 13$ oranges left in the box, and $11 - 1 = 10$ of them are good.

The probability that the third orange drawn is good, given that the first two oranges were good, is the conditional probability $P(G_3|G_1 \cap G_2)$:

$P(G_3|G_1 \cap G_2) = \frac{\text{Number of remaining good oranges}}{\text{Total number of remaining oranges}} = \frac{10}{13}$

Using the multiplication rule for probabilities for three dependent events:

Substitute the calculated probabilities into formula (i):

$P(G_1 \cap G_2 \cap G_3) = \frac{12}{15} \times \frac{11}{14} \times \frac{10}{13}$

Simplify the expression by cancelling common factors:

$P(G_1 \cap G_2 \cap G_3) = \frac{\cancel{12}^{4}}{\cancel{15}^{5}} \times \frac{11}{14} \times \frac{\cancel{10}^{2}}{13}$

$P(G_1 \cap G_2 \cap G_3) = \frac{4}{5} \times \frac{11}{14} \times \frac{2}{13}$

$P(G_1 \cap G_2 \cap G_3) = \frac{\cancel{4}^{2}}{5} \times \frac{11}{\cancel{14}^{7}} \times \frac{2}{13}$

$P(G_1 \cap G_2 \cap G_3) = \frac{2}{5} \times \frac{11}{7} \times \frac{2}{13}$

$P(G_1 \cap G_2 \cap G_3) = \frac{2 \times 11 \times 2}{5 \times 7 \times 13}$

$P(G_1 \cap G_2 \cap G_3) = \frac{44}{35 \times 13}$

$35 \times 13 = 35 \times (10 + 3) = 350 + 105 = 455$

$P(G_1 \cap G_2 \cap G_3) = \frac{44}{455}$

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Final Answer:

The probability that the box will be approved for sale is $\frac{44}{455}$.

Given:

A fair coin is tossed and an unbiased die is thrown.

Event A: Head appears on the coin.

Event B: 3 on the die.

---

To Check:

Whether events A and B are independent.

---

Solution:

The sample space S for the combined experiment of tossing a coin and throwing a die consists of ordered pairs $(C, D)$, where C is the result of the coin toss (H or T) and D is the result of the die throw (1, 2, 3, 4, 5, or 6).

$S = \{(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)\}$

The total number of outcomes is $|S| = 2 \times 6 = 12$. Each outcome is equally likely, with probability $\frac{1}{12}$.

Event A: Head appears on the coin.

$A = \{(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6)\}$

The number of outcomes in A is $|A| = 6$.

The probability of event A is $P(A) = \frac{|A|}{|S|} = \frac{6}{12} = \frac{1}{2}$.

Event B: 3 on the die.

$B = \{(H, 3), (T, 3)\}$

The number of outcomes in B is $|B| = 2$.

The probability of event B is $P(B) = \frac{|B|}{|S|} = \frac{2}{12} = \frac{1}{6}$.

For events A and B to be independent, the condition $P(A \cap B) = P(A) \times P(B)$ must be satisfied.

First, find the intersection of events A and B, $A \cap B$. This is the event where a head appears on the coin AND a 3 appears on the die.

$A \cap B = \{(H, 3)\}$

The number of outcomes in $A \cap B$ is $|A \cap B| = 1$.

The probability of $A \cap B$ is $P(A \cap B) = \frac{|A \cap B|}{|S|} = \frac{1}{12}$.

Now, let's calculate the product $P(A) \times P(B)$:

$P(A) \times P(B) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}$

Comparing $P(A \cap B)$ and $P(A) \times P(B)$:

Since $P(A \cap B) = P(A) \times P(B)$, the events A and B are independent.

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Final Answer:

Yes, the events A and B are independent.

Given:

A die is tossed with faces marked 1, 2, 3 (red) and 4, 5, 6 (green).

Event A: The number is even.

Event B: The number is red.

---

To Check:

Whether events A and B are independent.

---

Solution:

The sample space S when the die is tossed is:

$S = \{1, 2, 3, 4, 5, 6\}$

The total number of outcomes is $|S| = 6$. Each outcome is equally likely.

Event A: The number is even. The even numbers in the sample space are 2, 4, and 6.

$A = \{2, 4, 6\}$

The number of outcomes in A is $|A| = 3$.

The probability of event A is $P(A) = \frac{|A|}{|S|} = \frac{3}{6} = \frac{1}{2}$.

Event B: The number is red. The red numbers in the sample space are 1, 2, and 3.

$B = \{1, 2, 3\}$

The number of outcomes in B is $|B| = 3$.

The probability of event B is $P(B) = \frac{|B|}{|S|} = \frac{3}{6} = \frac{1}{2}$.

For events A and B to be independent, the condition $P(A \cap B) = P(A) \times P(B)$ must be satisfied.

First, find the intersection of events A and B, $A \cap B$. This is the event where the number is even AND the number is red.

$A \cap B = \{2, 4, 6\} \cap \{1, 2, 3\} = \{2\}$

The number of outcomes in $A \cap B$ is $|A \cap B| = 1$.

The probability of $A \cap B$ is $P(A \cap B) = \frac{|A \cap B|}{|S|} = \frac{1}{6}$.

Now, let's calculate the product $P(A) \times P(B)$:

$P(A) \times P(B) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$

Comparing $P(A \cap B)$ and $P(A) \times P(B)$:

Since $P(A \cap B) \neq P(A) \times P(B)$ ($\frac{1}{6} \neq \frac{1}{4}$), the events A and B are not independent.

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Final Answer:

No, the events A and B are not independent.

Given:

$P(E) = \frac{3}{5}$

$P(F) = \frac{3}{10}$

$P(E \cap F) = \frac{1}{5}$

---

To Check:

Whether events E and F are independent.

---

Solution:

For events E and F to be independent, the following condition must be satisfied:

We are given $P(E \cap F) = \frac{1}{5}$.

Now, let's calculate the product $P(E) \times P(F)$ using the given probabilities:

$P(E) \times P(F) = \frac{3}{5} \times \frac{3}{10}$

$P(E) \times P(F) = \frac{3 \times 3}{5 \times 10}$

$P(E) \times P(F) = \frac{9}{50}$

Now, we compare the given value of $P(E \cap F)$ with the calculated value of $P(E) \times P(F)$.

Given: $P(E \cap F) = \frac{1}{5}$

Calculated: $P(E) \times P(F) = \frac{9}{50}$

To compare $\frac{1}{5}$ and $\frac{9}{50}$, we can find a common denominator. The common denominator is 50.

$\frac{1}{5} = \frac{1 \times 10}{5 \times 10} = \frac{10}{50}$

Comparing $\frac{10}{50}$ and $\frac{9}{50}$, we see that $\frac{10}{50} \neq \frac{9}{50}$.

Therefore, $P(E \cap F) \neq P(E) \times P(F)$.

Since the condition for independence is not satisfied, the events E and F are not independent.

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Final Answer:

No, the events E and F are not independent.

Given:

$P(A) = \frac{1}{2}$

$P(A \cup B) = \frac{3}{5}$

$P(B) = p$

To Find:

The value of p.

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Solution (i): Mutually Exclusive Events

If events A and B are mutually exclusive, then $A \cap B = \emptyset$, which implies $P(A \cap B) = 0$.

The general formula for the probability of the union of two events is:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

For mutually exclusive events, the formula simplifies to:

$P(A \cup B) = P(A) + P(B)$

Substitute the given values:

$\frac{3}{5} = \frac{1}{2} + p$

To find p, subtract $P(A)$ from $P(A \cup B)$:

$p = \frac{3}{5} - \frac{1}{2}$

Find a common denominator, which is 10:

$p = \frac{3 \times 2}{5 \times 2} - \frac{1 \times 5}{2 \times 5}$

$p = \frac{6}{10} - \frac{5}{10}$

$p = \frac{6 - 5}{10}$

$p = \frac{1}{10}$

Thus, if A and B are mutually exclusive, $p = \frac{1}{10}$.

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Solution (ii): Independent Events

If events A and B are independent, then the probability of their intersection is the product of their individual probabilities:

$P(A \cap B) = P(A) \times P(B)$

The general formula for the probability of the union of two events is:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

Substitute the condition for independence, $P(A \cap B) = P(A) \times P(B)$:

$P(A \cup B) = P(A) + P(B) - P(A) \times P(B)$

Substitute the given values into this equation:

$\frac{3}{5} = \frac{1}{2} + p - \left(\frac{1}{2}\right) \times p$

$\frac{3}{5} = \frac{1}{2} + p - \frac{p}{2}$

Combine the terms involving p:

$\frac{3}{5} = \frac{1}{2} + \left(p - \frac{p}{2}\right)$

$\frac{3}{5} = \frac{1}{2} + \frac{p}{2}$

Subtract $\frac{1}{2}$ from both sides:

$\frac{3}{5} - \frac{1}{2} = \frac{p}{2}$

Calculate the difference on the left side (common denominator is 10):

$\frac{6}{10} - \frac{5}{10} = \frac{p}{2}$

$\frac{1}{10} = \frac{p}{2}$

Multiply both sides by 2 to solve for p:

$p = \frac{1}{10} \times 2$

$p = \frac{2}{10}$

Simplify the fraction:

$p = \frac{1}{5}$

Thus, if A and B are independent, $p = \frac{1}{5}$.

Given:

Events A and B are independent.

$P(A) = 0.3$

$P(B) = 0.4$

To Find:

(i) $P(A \cap B)$

(ii) $P(A \cup B)$

(iii) $P(A|B)$

(iv) $P(B|A)$

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Solution (i): $P(A \cap B)$

Since A and B are independent events, the probability of their intersection is the product of their individual probabilities.

$P(A \cap B) = P(A) \times P(B)$

Substitute the given values:

$P(A \cap B) = 0.3 \times 0.4$

$P(A \cap B) = 0.12$

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Solution (ii): $P(A \cup B)$

The general formula for the probability of the union of two events is:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

We already know $P(A)$, $P(B)$, and we calculated $P(A \cap B)$ in part (i).

Substitute the values:

$P(A \cup B) = 0.3 + 0.4 - 0.12$

$P(A \cup B) = 0.7 - 0.12$

$P(A \cup B) = 0.58$

Alternatively, for independent events, the formula can be written as:

$P(A \cup B) = P(A) + P(B) - P(A)P(B)$

$P(A \cup B) = 0.3 + 0.4 - (0.3)(0.4)$

$P(A \cup B) = 0.7 - 0.12$

$P(A \cup B) = 0.58$

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Solution (iii): $P(A|B)$

The formula for conditional probability $P(A|B)$ is:

$P(A|B) = \frac{P(A \cap B)}{P(B)}$

Substitute the values $P(A \cap B) = 0.12$ (from part i) and $P(B) = 0.4$:

$P(A|B) = \frac{0.12}{0.4}$

$P(A|B) = \frac{12}{40}$

$P(A|B) = \frac{3}{10}$

$P(A|B) = 0.3$

Alternatively, since A and B are independent events, the occurrence of event B does not affect the probability of event A. Therefore, $P(A|B) = P(A)$.

$P(A|B) = P(A) = 0.3$

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Solution (iv): $P(B|A)$

The formula for conditional probability $P(B|A)$ is:

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

Substitute the values $P(A \cap B) = 0.12$ (from part i) and $P(A) = 0.3$:

$P(B|A) = \frac{0.12}{0.3}$

$P(B|A) = \frac{12}{30}$

$P(B|A) = \frac{2}{5}$

$P(B|A) = 0.4$

Alternatively, since A and B are independent events, the occurrence of event A does not affect the probability of event B. Therefore, $P(B|A) = P(B)$.

$P(B|A) = P(B) = 0.4$

Given:

$P(A) = \frac{1}{4}$

$P(B) = \frac{1}{2}$

$P(A \cap B) = \frac{1}{8}$

To Find:

$P(\text{not A and not B})$, which is $P(A' \cap B')$.

---

Solution:

We need to find $P(A' \cap B')$.

Using De Morgan's Law, we know that $A' \cap B' = (A \cup B)'$.

So, $P(A' \cap B') = P((A \cup B)')$.

The probability of the complement of an event is 1 minus the probability of the event:

$P((A \cup B)') = 1 - P(A \cup B)$.

First, we need to find $P(A \cup B)$ using the formula for the probability of the union of two events:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

Substitute the given values:

$P(A \cup B) = \frac{1}{4} + \frac{1}{2} - \frac{1}{8}$

To add and subtract these fractions, find a common denominator, which is 8.

$P(A \cup B) = \frac{1 \times 2}{4 \times 2} + \frac{1 \times 4}{2 \times 4} - \frac{1}{8}$

$P(A \cup B) = \frac{2}{8} + \frac{4}{8} - \frac{1}{8}$

$P(A \cup B) = \frac{2 + 4 - 1}{8}$

$P(A \cup B) = \frac{5}{8}$

Now, substitute this value into the expression for $P(A' \cap B')$:

$P(A' \cap B') = 1 - P(A \cup B)$

$P(A' \cap B') = 1 - \frac{5}{8}$

$P(A' \cap B') = \frac{8}{8} - \frac{5}{8}$

$P(A' \cap B') = \frac{3}{8}$

Therefore, $P(\text{not A and not B}) = \frac{3}{8}$.

Given:

$P(A) = \frac{1}{2}$

$P(B) = \frac{7}{12}$

$P(\text{not A or not B}) = P(A' \cup B') = \frac{1}{4}$

To Determine:

Whether events A and B are independent.

---

Solution:

Events A and B are independent if and only if $P(A \cap B) = P(A) \times P(B)$.

We are given $P(A' \cup B') = \frac{1}{4}$.

Using De Morgan's Law, we know that $A' \cup B' = (A \cap B)'$.

So, $P(A' \cup B') = P((A \cap B)')$.

The probability of the complement of an event is $1$ minus the probability of the event.

$P((A \cap B)') = 1 - P(A \cap B)$

Substituting the given value:

$\frac{1}{4} = 1 - P(A \cap B)$

Now, we can find $P(A \cap B)$:

$P(A \cap B) = 1 - \frac{1}{4}$

$P(A \cap B) = \frac{4}{4} - \frac{1}{4}$

$P(A \cap B) = \frac{3}{4}$

Next, let's calculate the product of the individual probabilities $P(A) \times P(B)$:

$P(A) \times P(B) = \frac{1}{2} \times \frac{7}{12}$

$P(A) \times P(B) = \frac{1 \times 7}{2 \times 12}$

$P(A) \times P(B) = \frac{7}{24}$

Now, we compare $P(A \cap B)$ with $P(A) \times P(B)$:

We found $P(A \cap B) = \frac{3}{4}$ and $P(A) \times P(B) = \frac{7}{24}$.

Let's express $\frac{3}{4}$ with a denominator of 24 to compare easily:

$\frac{3}{4} = \frac{3 \times 6}{4 \times 6} = \frac{18}{24}$

Comparing $\frac{18}{24}$ and $\frac{7}{24}$:

$\frac{18}{24} \neq \frac{7}{24}$

So, $P(A \cap B) \neq P(A) \times P(B)$.

Since the condition for independence is not met, the events A and B are not independent.

Given:

Events A and B are independent.

$P(A) = 0.3$

$P(B) = 0.6$

To Find:

(i) $P(A \text{ and } B)$ i.e., $P(A \cap B)$

(ii) $P(A \text{ and not } B)$ i.e., $P(A \cap B')$

(iii) $P(A \text{ or } B)$ i.e., $P(A \cup B)$

(iv) $P(\text{neither } A \text{ nor } B)$ i.e., $P(A' \cap B')$

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Solution (i): $P(A \text{ and } B)$

Since events A and B are independent, the probability of their intersection is the product of their individual probabilities.

$P(A \cap B) = P(A) \times P(B)$

Substitute the given values:

$P(A \cap B) = 0.3 \times 0.6$

$P(A \cap B) = 0.18$

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Solution (ii): $P(A \text{ and not } B)$

We need to find $P(A \cap B')$. Since A and B are independent, events A and B' are also independent.

First, find the probability of not B:

$P(B') = 1 - P(B)$

$P(B') = 1 - 0.6$

$P(B') = 0.4$

Now, use the property of independence for A and B':

$P(A \cap B') = P(A) \times P(B')$

Substitute the values:

$P(A \cap B') = 0.3 \times 0.4$

$P(A \cap B') = 0.12$

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Solution (iii): $P(A \text{ or } B)$

We need to find $P(A \cup B)$. The general formula for the union of two events is:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

We know $P(A) = 0.3$, $P(B) = 0.6$, and we found $P(A \cap B) = 0.18$ in part (i).

Substitute these values:

$P(A \cup B) = 0.3 + 0.6 - 0.18$

$P(A \cup B) = 0.9 - 0.18$

$P(A \cup B) = 0.72$

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Solution (iv): $P(\text{neither } A \text{ nor } B)$

We need to find $P(A' \cap B')$. Using De Morgan's Law, we know that $A' \cap B' = (A \cup B)'$.

So, $P(A' \cap B') = P((A \cup B)')$.

The probability of the complement of an event is 1 minus the probability of the event:

$P((A \cup B)') = 1 - P(A \cup B)$

We found $P(A \cup B) = 0.72$ in part (iii).

Substitute this value:

$P(A' \cap B') = 1 - 0.72$

$P(A' \cap B') = 0.28$

Alternatively, since A and B are independent, events A' and B' are also independent.

First, find the probability of not A:

$P(A') = 1 - P(A)$

$P(A') = 1 - 0.3$

$P(A') = 0.7$

We already found $P(B') = 0.4$ in part (ii).

Now, use the property of independence for A' and B':

$P(A' \cap B') = P(A') \times P(B')$

$P(A' \cap B') = 0.7 \times 0.4$

$P(A' \cap B') = 0.28$

Given:

A fair die is tossed thrice.

To Find:

The probability of getting an odd number at least once.

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Solution:

When a fair die is tossed once, the possible outcomes are $\{1, 2, 3, 4, 5, 6\}$.

The total number of outcomes is 6.

The odd numbers are $\{1, 3, 5\}$. The number of odd outcomes is 3.

The probability of getting an odd number in a single toss is $P(\text{Odd}) = \frac{\text{Number of odd outcomes}}{\text{Total number of outcomes}} = \frac{3}{6} = \frac{1}{2}$.

The even numbers are $\{2, 4, 6\}$. The number of even outcomes is 3.

The probability of getting an even number in a single toss is $P(\text{Even}) = \frac{\text{Number of even outcomes}}{\text{Total number of outcomes}} = \frac{3}{6} = \frac{1}{2}$.

The die is tossed thrice, and each toss is an independent event.

Let E be the event of getting an odd number at least once in three tosses.

It is easier to find the probability of the complement event, E', which is the event of getting an odd number zero times (i.e., getting an even number on all three tosses).

$P(E') = P(\text{Even on 1st toss AND Even on 2nd toss AND Even on 3rd toss})$

Since the tosses are independent:

$P(E') = P(\text{Even on 1st}) \times P(\text{Even on 2nd}) \times P(\text{Even on 3rd})$

$P(E') = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}$

$P(E') = \frac{1}{8}$

Now, we can find the probability of event E using the complement rule:

$P(E) = 1 - P(E')$

$P(E) = 1 - \frac{1}{8}$

$P(E) = \frac{8}{8} - \frac{1}{8}$

$P(E) = \frac{7}{8}$

The probability of getting an odd number at least once in three tosses is $\frac{7}{8}$.

Given:

Number of black balls ($N_B$) = 10

Number of red balls ($N_R$) = 8

Total number of balls = $N_B + N_R = 10 + 8 = 18$

Two balls are drawn at random with replacement.

To Find:

(i) Probability that both balls are red.

(ii) Probability that the first ball is black and the second is red.

(iii) Probability that one of them is black and the other is red.

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Solution:

Let R be the event of drawing a red ball in a single draw.

Let B be the event of drawing a black ball in a single draw.

The probability of drawing a red ball in a single draw is:

$P(R) = \frac{\text{Number of red balls}}{\text{Total number of balls}} = \frac{8}{18} = \frac{4}{9}$

The probability of drawing a black ball in a single draw is:

$P(B) = \frac{\text{Number of black balls}}{\text{Total number of balls}} = \frac{10}{18} = \frac{5}{9}$

Since the balls are drawn with replacement, the outcome of the first draw does not affect the outcome of the second draw. Thus, the two draws are independent events.

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Solution (i): Both balls are red

This means the first ball is red AND the second ball is red.

Let $R_1$ be the event that the first ball is red.

Let $R_2$ be the event that the second ball is red.

We need to find $P(R_1 \cap R_2)$. Since $R_1$ and $R_2$ are independent:

$P(R_1 \cap R_2) = P(R_1) \times P(R_2)$

$P(R_1) = P(R) = \frac{4}{9}$

$P(R_2) = P(R) = \frac{4}{9}$ (because of replacement)

$P(\text{both balls are red}) = \frac{4}{9} \times \frac{4}{9} = \frac{16}{81}$

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Solution (ii): First ball is black and second is red

Let $B_1$ be the event that the first ball is black.

Let $R_2$ be the event that the second ball is red.

We need to find $P(B_1 \cap R_2)$. Since $B_1$ and $R_2$ are independent:

$P(B_1 \cap R_2) = P(B_1) \times P(R_2)$

$P(B_1) = P(B) = \frac{5}{9}$

$P(R_2) = P(R) = \frac{4}{9}$ (because of replacement)

$P(\text{first black and second red}) = \frac{5}{9} \times \frac{4}{9} = \frac{20}{81}$

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Solution (iii): One of them is black and other is red

This event can occur in two mutually exclusive ways:

Case 1: First ball is black and second ball is red ($B_1 \cap R_2$).

Case 2: First ball is red and second ball is black ($R_1 \cap B_2$).

We need to find $P((B_1 \cap R_2) \cup (R_1 \cap B_2))$. Since these cases are mutually exclusive, we can add their probabilities:

$P(\text{one black and one red}) = P(B_1 \cap R_2) + P(R_1 \cap B_2)$

From part (ii), we know $P(B_1 \cap R_2) = \frac{20}{81}$.

Now, let's calculate $P(R_1 \cap B_2)$. Since $R_1$ and $B_2$ are independent:

$P(R_1 \cap B_2) = P(R_1) \times P(B_2)$

$P(R_1) = P(R) = \frac{4}{9}$

$P(B_2) = P(B) = \frac{5}{9}$ (because of replacement)

$P(R_1 \cap B_2) = \frac{4}{9} \times \frac{5}{9} = \frac{20}{81}$

Now, add the probabilities of the two cases:

$P(\text{one black and one red}) = \frac{20}{81} + \frac{20}{81} = \frac{40}{81}$

Given:

Probability that A solves the problem, $P(A) = \frac{1}{2}$.

Probability that B solves the problem, $P(B) = \frac{1}{3}$.

Events A and B are independent.

To Find:

(i) Probability that the problem is solved.

(ii) Probability that exactly one of them solves the problem.

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Solution (i): The problem is solved

The problem is solved if A solves it, or B solves it, or both solve it. This corresponds to the event $A \cup B$.

We can find $P(A \cup B)$ using the formula for independent events, or by considering the complement.

Let's use the complement approach. The complement of the problem being solved is that the problem is NOT solved. This means A does NOT solve the problem AND B does NOT solve the problem.

Let $A'$ be the event that A does not solve the problem.

Let $B'$ be the event that B does not solve the problem.

$P(A') = 1 - P(A) = 1 - \frac{1}{2} = \frac{1}{2}$

$P(B') = 1 - P(B) = 1 - \frac{1}{3} = \frac{2}{3}$

Since A and B are independent, $A'$ and $B'$ are also independent.

The probability that the problem is not solved is $P(A' \cap B')$.

$P(A' \cap B') = P(A') \times P(B')$

$P(A' \cap B') = \frac{1}{2} \times \frac{2}{3} = \frac{2}{6} = \frac{1}{3}$

The probability that the problem is solved is $1 - P(\text{problem not solved})$.

$P(\text{problem solved}) = 1 - P(A' \cap B')$

$P(\text{problem solved}) = 1 - \frac{1}{3} = \frac{3}{3} - \frac{1}{3} = \frac{2}{3}$

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Solution (ii): Exactly one of them solves the problem

Exactly one of them solves the problem means either A solves and B does not, OR B solves and A does not.

This corresponds to the event $(A \cap B') \cup (A' \cap B)$.

The events $(A \cap B')$ and $(A' \cap B)$ are mutually exclusive.

So, $P(\text{exactly one solves}) = P(A \cap B') + P(A' \cap B)$.

Since A and B are independent, A and B' are independent, and A' and B are independent.

Calculate $P(A \cap B')$:

$P(A \cap B') = P(A) \times P(B')$

$P(A \cap B') = \frac{1}{2} \times \frac{2}{3} = \frac{2}{6} = \frac{1}{3}$

Calculate $P(A' \cap B)$:

$P(A' \cap B) = P(A') \times P(B)$

$P(A' \cap B) = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6}$

Now, add these probabilities:

$P(\text{exactly one solves}) = P(A \cap B') + P(A' \cap B)$

$P(\text{exactly one solves}) = \frac{1}{3} + \frac{1}{6}$

Find a common denominator, which is 6:

$P(\text{exactly one solves}) = \frac{1 \times 2}{3 \times 2} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6}$

$P(\text{exactly one solves}) = \frac{1}{2}$

Given:

A well-shuffled deck of 52 cards.

One card is drawn at random.

To Determine:

Whether events E and F are independent in each given case.

---

Solution:

Events E and F are independent if and only if $P(E \cap F) = P(E) \times P(F)$.

Total number of outcomes = 52.

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Case (i):

E : ‘the card drawn is a spade’

F : ‘the card drawn is an ace’

Number of spades = 13.

$P(E) = \frac{13}{52} = \frac{1}{4}$

Number of aces = 4.

$P(F) = \frac{4}{52} = \frac{1}{13}$

$E \cap F$: The card is a spade AND an ace (Ace of Spades).

Number of cards in $E \cap F$ = 1.

$P(E \cap F) = \frac{1}{52}$

Check for independence:

$P(E) \times P(F) = \frac{1}{4} \times \frac{1}{13} = \frac{1}{52}$

Since $P(E \cap F) = P(E) \times P(F)$, the events E and F are independent in this case.

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Case (ii):

E : ‘the card drawn is black’

F : ‘the card drawn is a king’

Number of black cards = 26 (13 spades + 13 clubs).

$P(E) = \frac{26}{52} = \frac{1}{2}$

Number of kings = 4 (King of Spades, King of Clubs, King of Hearts, King of Diamonds).

$P(F) = \frac{4}{52} = \frac{1}{13}$

$E \cap F$: The card is black AND a king (King of Spades, King of Clubs).

Number of cards in $E \cap F$ = 2.

$P(E \cap F) = \frac{2}{52} = \frac{1}{26}$

Check for independence:

$P(E) \times P(F) = \frac{1}{2} \times \frac{1}{13} = \frac{1}{26}$

Since $P(E \cap F) = P(E) \times P(F)$, the events E and F are independent in this case.

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Case (iii):

E : ‘the card drawn is a king or queen’

F : ‘the card drawn is a queen or jack’

Event E: The card is a King or a Queen.

Number of kings = 4.

Number of queens = 4.

Number of cards in E = 4 + 4 = 8.

$P(E) = \frac{8}{52} = \frac{2}{13}$

Event F: The card is a Queen or a Jack.

Number of queens = 4.

Number of jacks = 4.

Number of cards in F = 4 + 4 = 8.

$P(F) = \frac{8}{52} = \frac{2}{13}$

$E \cap F$: The card is (King or Queen) AND (Queen or Jack).

The cards common to both conditions are the Queens.

$E \cap F$: The card is a Queen.

Number of cards in $E \cap F$ = 4.

$P(E \cap F) = \frac{4}{52} = \frac{1}{13}$

Check for independence:

$P(E) \times P(F) = \frac{2}{13} \times \frac{2}{13} = \frac{4}{169}$

Since $P(E \cap F) = \frac{1}{13} = \frac{13}{169}$ and $P(E) \times P(F) = \frac{4}{169}$, we see that $P(E \cap F) \neq P(E) \times P(F)$.

Therefore, the events E and F are not independent in this case.

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Conclusion:

Events E and F are independent in case (i) and case (ii), but not in case (iii).

Given:

Let H be the event that a student reads Hindi newspaper.

Let E be the event that a student reads English newspaper.

$P(H) = 60\% = 0.60$

$P(E) = 40\% = 0.40$

$P(H \cap E) = 20\% = 0.20$ (Probability of reading both Hindi and English)

To Find:

(a) $P(\text{neither H nor E})$ i.e., $P(H' \cap E')$

(b) $P(\text{E | H})$ (Probability of reading English given she reads Hindi)

(c) $P(\text{H | E})$ (Probability of reading Hindi given she reads English)

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Solution (a): Probability of reading neither Hindi nor English news papers

We need to find $P(H' \cap E')$.

Using De Morgan's Law, $H' \cap E' = (H \cup E)'$.

So, $P(H' \cap E') = P((H \cup E)')$.

The probability of the complement of an event is $1$ minus the probability of the event:

$P((H \cup E)') = 1 - P(H \cup E)$.

First, find the probability that a student reads at least one of the newspapers, $P(H \cup E)$, using the formula:

$P(H \cup E) = P(H) + P(E) - P(H \cap E)$

Substitute the given probabilities:

$P(H \cup E) = 0.60 + 0.40 - 0.20$

$P(H \cup E) = 1.00 - 0.20$

$P(H \cup E) = 0.80$

Now, find the probability of reading neither newspaper:

$P(H' \cap E') = 1 - P(H \cup E)$

$P(H' \cap E') = 1 - 0.80$

$P(H' \cap E') = 0.20$

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Solution (b): Probability that she reads English news paper if she reads Hindi news paper

This is a conditional probability, $P(E | H)$.

The formula for conditional probability is:

$P(E | H) = \frac{P(E \cap H)}{P(H)}$

We are given $P(H \cap E) = 0.20$ and $P(H) = 0.60$. Note that $P(E \cap H) = P(H \cap E)$.

Substitute the values:

$P(E | H) = \frac{0.20}{0.60}$

$P(E | H) = \frac{20}{60} = \frac{1}{3}$

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Solution (c): Probability that she reads Hindi news paper if she reads English news paper

This is a conditional probability, $P(H | E)$.

The formula for conditional probability is:

$P(H | E) = \frac{P(H \cap E)}{P(E)}$

We are given $P(H \cap E) = 0.20$ and $P(E) = 0.40$.

Substitute the values:

$P(H | E) = \frac{0.20}{0.40}$

$P(H | E) = \frac{20}{40} = \frac{1}{2}$

Given:

A pair of fair dice is rolled.

We are looking for the probability of obtaining an even prime number on each die.

On a standard die, the possible outcomes are $\{1, 2, 3, 4, 5, 6\}$.

The prime numbers on a die are $\{2, 3, 5\}$.

The only even prime number on a die is 2.

So, the desired outcome for a single die is getting a 2.

To Find:

The probability of getting a 2 on the first die AND a 2 on the second die.

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Solution:

For a single die roll, the total number of outcomes is 6.

The number of outcomes that is an even prime number (which is 2) is 1.

The probability of getting an even prime number on a single die is $P(\text{getting a 2}) = \frac{1}{6}$.

When a pair of dice is rolled, the outcome of one die is independent of the outcome of the other die.

Let $E_1$ be the event of getting an even prime number on the first die.

Let $E_2$ be the event of getting an even prime number on the second die.

We want to find $P(E_1 \cap E_2)$. Since $E_1$ and $E_2$ are independent events:

$P(E_1 \cap E_2) = P(E_1) \times P(E_2)$

$P(E_1) = \frac{1}{6}$

$P(E_2) = \frac{1}{6}$

Therefore, the probability of obtaining an even prime number on each die is:

$P(E_1 \cap E_2) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$

This corresponds to option (D).

The correct answer is (D) $\frac{1}{36}$.

To Determine:

The condition under which two events A and B are independent.

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Solution:

Two events A and B are defined as independent if and only if the probability of their intersection is equal to the product of their individual probabilities:

$P(A \cap B) = P(A) \times P(B)$

Let's examine the given options:

(A) A and B are mutually exclusive:

If A and B are mutually exclusive, then $P(A \cap B) = 0$. For independent events, $P(A \cap B) = P(A)P(B)$. Thus, if mutually exclusive events are independent, $P(A)P(B) = 0$, which means $P(A)=0$ or $P(B)=0$. This is a specific case, not a general condition for independence when $P(A) > 0$ and $P(B) > 0$.

(B) $P(A' \cap B') = [1 – P(A)] [1 – P(B)]$:

We know that $P(A') = 1 - P(A)$ and $P(B') = 1 - P(B)$. So the given condition is $P(A' \cap B') = P(A') \times P(B')$. This is the definition of independence for the events A' and B'.

If A and B are independent, then $A'$ and $B'$ are also independent. We can show this:

$P(A' \cap B') = P((A \cup B)')$ (By De Morgan's Law)

$P((A \cup B)') = 1 - P(A \cup B)$ (Using complement rule)

$1 - P(A \cup B) = 1 - (P(A) + P(B) - P(A \cap B))$ (Union formula)

If A and B are independent, $P(A \cap B) = P(A)P(B)$. Substitute this:

$1 - (P(A) + P(B) - P(A)P(B))$

$1 - P(A) - P(B) + P(A)P(B)$

$(1 - P(A))(1 - P(B))$

$P(A') \times P(B')$

So, if A and B are independent, then $P(A' \cap B') = P(A')P(B')$, which matches option (B).

Conversely, if $P(A' \cap B') = (1 - P(A))(1 - P(B))$, by reversing the steps, we get $P(A \cap B) = P(A)P(B)$. Thus, this condition is equivalent to the independence of A and B.

(C) P(A) = P(B):

This condition does not imply independence. For example, consider drawing a card from a deck. Let A be drawing a King ($P(A)=4/52$) and B be drawing a Queen ($P(B)=4/52$). $P(A) = P(B)$, but $P(A \cap B) = P(\text{drawing a King and a Queen}) = 0$. $P(A)P(B) = (4/52)(4/52) \neq 0$. Since $P(A \cap B) \neq P(A)P(B)$, A and B are not independent.

(D) P(A) + P(B) = 1:

This condition also does not imply independence. For example, let A be drawing a red card ($P(A)=26/52=1/2$) and B be drawing a black card ($P(B)=26/52=1/2$). $P(A) + P(B) = 1/2 + 1/2 = 1$. However, A and B are mutually exclusive ($P(A \cap B) = 0$), and $P(A)P(B) = (1/2)(1/2) = 1/4 \neq 0$. Since $P(A \cap B) \neq P(A)P(B)$, A and B are not independent.

Therefore, the only condition among the options that guarantees A and B are independent is option (B).

The correct answer is (B) P(A′B′) = [1 – P(A)] [1 – P(B)].

Given:

Let S be the event that there is a strike.

Let S' be the event that there is no strike.

Let C be the event that the construction job will be completed on time.

We are given the following probabilities:

$P(S) = 0.65$

$P(C|S') = 0.80$ (Probability of completion on time given no strike)

$P(C|S) = 0.32$ (Probability of completion on time given strike)

To Find:

The probability that the construction job will be completed on time, $P(C)$.

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Solution:

The events S (strike) and S' (no strike) are complementary events. This means that either S occurs or S' occurs, and they cannot occur simultaneously. Thus, $\{S, S'\}$ forms a partition of the sample space.

The probability of no strike is:

$P(S') = 1 - P(S)$

$P(S') = 1 - 0.65$

$P(S') = 0.35$

To find the probability that the construction job will be completed on time, $P(C)$, we can use the Law of Total Probability.

The Law of Total Probability states:

$P(C) = P(C \cap S) + P(C \cap S')$

Using the definition of conditional probability, $P(A \cap B) = P(A|B)P(B)$:

$P(C \cap S) = P(C|S)P(S)$

$P(C \cap S') = P(C|S')P(S')$

Substituting these into the Law of Total Probability formula:

$P(C) = P(C|S)P(S) + P(C|S')P(S')$

Substitute the given values:

$P(C) = (0.32)(0.65) + (0.80)(0.35)$

Calculate the products:

$0.32 \times 0.65 = 0.2080$

$0.80 \times 0.35 = 0.2800$

Add the results:

$P(C) = 0.2080 + 0.2800$

$P(C) = 0.4880$

The probability that the construction job will be completed on time is 0.4880.

Given:

Bag I contains: 3 Red (R) + 4 Black (B) = 7 balls.

Bag II contains: 5 Red (R) + 6 Black (B) = 11 balls.

A ball is drawn at random from one of the bags.

The ball drawn is found to be red.

To Find:

The probability that the red ball was drawn from Bag II.

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Solution:

Let $B_1$ be the event that Bag I is selected.

Let $B_2$ be the event that Bag II is selected.

Since a bag is selected at random, the probability of selecting either bag is equal:

$P(B_1) = \frac{1}{2}$

$P(B_2) = \frac{1}{2}$

Let R be the event that the ball drawn is red.

We need to find the conditional probabilities of drawing a red ball given that a specific bag was selected:

Probability of drawing a red ball given Bag I was selected:

$P(R|B_1) = \frac{\text{Number of red balls in Bag I}}{\text{Total balls in Bag I}} = \frac{3}{7}$

Probability of drawing a red ball given Bag II was selected:

$P(R|B_2) = \frac{\text{Number of red balls in Bag II}}{\text{Total balls in Bag II}} = \frac{5}{11}$

We are given that a red ball has been drawn (event R occurred), and we need to find the probability that it was drawn from Bag II, i.e., $P(B_2 | R)$.

We can use Bayes' Theorem for this:

$P(B_2 | R) = \frac{P(R | B_2) P(B_2)}{P(R)}$

To use this formula, we first need to find the overall probability of drawing a red ball, $P(R)$. We can find $P(R)$ using the Law of Total Probability:

$P(R) = P(R|B_1) P(B_1) + P(R|B_2) P(B_2)$

Substitute the values we have:

$P(R) = \left(\frac{3}{7}\right) \left(\frac{1}{2}\right) + \left(\frac{5}{11}\right) \left(\frac{1}{2}\right)$

$P(R) = \frac{3}{14} + \frac{5}{22}$

To add these fractions, find a common denominator for 14 and 22. The least common multiple of 14 and 22 is 154.

$P(R) = \frac{3 \times 11}{14 \times 11} + \frac{5 \times 7}{22 \times 7}$

$P(R) = \frac{33}{154} + \frac{35}{154}$

$P(R) = \frac{33 + 35}{154} = \frac{68}{154}$

Simplify the fraction by dividing numerator and denominator by 2:

$P(R) = \frac{\cancel{68}^{34}}{\cancel{154}_{77}} = \frac{34}{77}$

Now, substitute the values into Bayes' Theorem formula for $P(B_2 | R)$:

$P(B_2 | R) = \frac{P(R | B_2) P(B_2)}{P(R)}$

$P(B_2 | R) = \frac{\left(\frac{5}{11}\right) \left(\frac{1}{2}\right)}{\frac{34}{77}}$

$P(B_2 | R) = \frac{\frac{5}{22}}{\frac{34}{77}}$

To divide by a fraction, multiply by its reciprocal:

$P(B_2 | R) = \frac{5}{22} \times \frac{77}{34}$

Cancel the common factor 11 between 22 and 77:

$P(B_2 | R) = \frac{5}{\cancel{22}_{2}} \times \frac{\cancel{77}^{7}}{34}$

$P(B_2 | R) = \frac{5 \times 7}{2 \times 34}$

$P(B_2 | R) = \frac{35}{68}$

The probability that the red ball was drawn from Bag II is $\frac{35}{68}$.

Given:

Box I: 2 Gold coins (GG)

Box II: 2 Silver coins (SS)

Box III: 1 Gold coin and 1 Silver coin (GS)

There are three identical boxes, and one is chosen at random.

A coin is drawn from the chosen box and it is found to be gold.

To Find:

The probability that the other coin in the box is also gold, given that the first coin drawn is gold.

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Solution:

Let $B_1$ be the event that Box I is chosen.

Let $B_2$ be the event that Box II is chosen.

Let $B_3$ be the event that Box III is chosen.

Since the boxes are identical and chosen at random:

$P(B_1) = P(B_2) = P(B_3) = \frac{1}{3}$

Let G be the event that the coin drawn is gold.

We need to find the probability of drawing a gold coin given that a specific box was chosen:

$P(G|B_1) = \frac{\text{Number of gold coins in Box I}}{\text{Total coins in Box I}} = \frac{2}{2} = 1$

$P(G|B_2) = \frac{\text{Number of gold coins in Box II}}{\text{Total coins in Box II}} = \frac{0}{2} = 0$

$P(G|B_3) = \frac{\text{Number of gold coins in Box III}}{\text{Total coins in Box III}} = \frac{1}{2}$

We are given that the drawn coin is gold (event G has occurred). We need to find the probability that the other coin in the box is also gold. The other coin is also gold only if the chosen box was Box I (GG). So, we need to find $P(B_1 | G)$.

We can use Bayes' Theorem to find $P(B_1 | G)$:

$P(B_1 | G) = \frac{P(G | B_1) P(B_1)}{P(G)}$

First, we need to calculate the overall probability of drawing a gold coin, $P(G)$, using the Law of Total Probability:

$P(G) = P(G|B_1) P(B_1) + P(G|B_2) P(B_2) + P(G|B_3) P(B_3)$

Substitute the known probabilities:

$P(G) = (1) \left(\frac{1}{3}\right) + (0) \left(\frac{1}{3}\right) + \left(\frac{1}{2}\right) \left(\frac{1}{3}\right)$

$P(G) = \frac{1}{3} + 0 + \frac{1}{6}$

$P(G) = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}$

Now, substitute the values into the Bayes' Theorem formula for $P(B_1 | G)$:

$P(B_1 | G) = \frac{P(G | B_1) P(B_1)}{P(G)}$

$P(B_1 | G) = \frac{(1) \left(\frac{1}{3}\right)}{\frac{1}{2}}$

$P(B_1 | G) = \frac{\frac{1}{3}}{\frac{1}{2}}$

$P(B_1 | G) = \frac{1}{3} \times \frac{2}{1}$

$P(B_1 | G) = \frac{2}{3}$

Given that a gold coin was drawn, the probability that the other coin in the box is also gold (meaning the box chosen was Box I) is $\frac{2}{3}$.

Given:

Let H be the event that a person has HIV.

Let H' be the event that a person does not have HIV.

Let T+ be the event that the HIV test is positive.

Let T- be the event that the HIV test is negative.

We are given the following probabilities:

$P(H) = 0.1\% = \frac{0.1}{100} = 0.001$

The probability of not having HIV is $P(H') = 1 - P(H)$.

$P(H') = 1 - 0.001 = 0.999$

The reliability of the test is given by conditional probabilities:

$P(T+|H) = 90\% = 0.90$ (True Positive Rate - Test detects HIV given the person has it)

$P(T-|H) = 10\% = 0.10$ (False Negative Rate - Test undetected given the person has HIV)

$P(T-|H') = 99\% = 0.99$ (True Negative Rate - Test is negative given the person does not have HIV)

$P(T+|H') = 1\% = 0.01$ (False Positive Rate - Test is positive given the person does not have HIV)

We are given that a randomly selected person's test is reported as HIV+ive (event T+ has occurred).

To Find:

The probability that the person actually has HIV given that the test is positive, i.e., $P(H | T+)$.

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Solution:

We need to find the conditional probability $P(H | T+)$. We can use Bayes' Theorem for this.

Bayes' Theorem states:

$P(H | T+) = \frac{P(T+ | H) P(H)}{P(T+)}$

To use this formula, we first need to calculate the overall probability of getting a positive test result, $P(T+)$. We can do this using the Law of Total Probability. The events H and H' form a partition of the sample space.

The Law of Total Probability gives:

$P(T+) = P(T+ | H) P(H) + P(T+ | H') P(H')$

Substitute the known probabilities into this formula:

$P(T+) = (0.90)(0.001) + (0.01)(0.999)$

$P(T+) = 0.0009 + 0.00999$

$P(T+) = 0.01089$

Now, substitute the values of $P(T+|H)$, $P(H)$, and $P(T+)$ into the Bayes' Theorem formula for $P(H | T+)$:

$P(H | T+) = \frac{P(T+ | H) P(H)}{P(T+)}$

$P(H | T+) = \frac{(0.90)(0.001)}{0.01089}$

$P(H | T+) = \frac{0.0009}{0.01089}$

To simplify the fraction, we can multiply the numerator and denominator by 100000:

$P(H | T+) = \frac{0.0009 \times 100000}{0.01089 \times 100000} = \frac{90}{1089}$

Both 90 and 1089 are divisible by 9:

$P(H | T+) = \frac{\cancel{90}^{10}}{\cancel{1089}_{121}}$

$P(H | T+) = \frac{10}{121}$

Thus, the probability that the person actually has HIV given a positive test result is $\frac{10}{121}$.

Given:

Let A, B, and C be the events that a bolt is manufactured by machine A, B, and C respectively.

Let D be the event that a bolt is defective.

We are given the following probabilities:

$P(A) = 25\% = 0.25$

$P(B) = 35\% = 0.35$

$P(C) = 40\% = 0.40$

Note that $P(A) + P(B) + P(C) = 0.25 + 0.35 + 0.40 = 1.00$, so these events form a partition of the sample space of bolts.

We are also given the conditional probabilities of a bolt being defective, given which machine manufactured it:

$P(D|A) = 5\% = 0.05$

$P(D|B) = 4\% = 0.04$

$P(D|C) = 2\% = 0.02$

We are given that a bolt is drawn at random and is found to be defective (event D has occurred).

To Find:

The probability that the defective bolt was manufactured by machine B, i.e., $P(B|D)$.

---

Solution:

We need to find the conditional probability $P(B|D)$. We can use Bayes' Theorem for this.

Bayes' Theorem states:

$P(B | D) = \frac{P(D | B) P(B)}{P(D)}$

First, we need to calculate the overall probability of drawing a defective bolt, $P(D)$. We can do this using the Law of Total Probability. Since events A, B, and C form a partition of the sample space:

$P(D) = P(D|A) P(A) + P(D|B) P(B) + P(D|C) P(C)$

Substitute the known probabilities into this formula:

$P(D) = (0.05)(0.25) + (0.04)(0.35) + (0.02)(0.40)$

Calculate the products:

$0.05 \times 0.25 = 0.0125$

$0.04 \times 0.35 = 0.0140$

$0.02 \times 0.40 = 0.0080$

Add the results to find $P(D)$:

$P(D) = 0.0125 + 0.0140 + 0.0080$

$P(D) = 0.0345$

Now, substitute the values of $P(D|B)$, $P(B)$, and $P(D)$ into the Bayes' Theorem formula for $P(B|D)$:

$P(B | D) = \frac{P(D | B) P(B)}{P(D)}$

$P(B | D) = \frac{(0.04)(0.35)}{0.0345}$

$P(B | D) = \frac{0.0140}{0.0345}$

To simplify the fraction, we can multiply the numerator and denominator by 10000:

$P(B | D) = \frac{0.0140 \times 10000}{0.0345 \times 10000} = \frac{140}{345}$

Both 140 and 345 are divisible by 5:

$P(B | D) = \frac{\cancel{140}^{28}}{\cancel{345}_{69}}$

$P(B | D) = \frac{28}{69}$

Thus, the probability that the defective bolt was manufactured by machine B is $\frac{28}{69}$.

Given:

Let T be the event that the doctor comes by train.

Let B be the event that the doctor comes by bus.

Let S be the event that the doctor comes by scooter.

Let O be the event that the doctor comes by other means of transport.

Let L be the event that the doctor is late.

We are given the following probabilities for the modes of transport:

$P(T) = \frac{3}{10}$

$P(B) = \frac{1}{5} = \frac{2}{10}$

$P(S) = \frac{1}{10}$

$P(O) = \frac{2}{5} = \frac{4}{10}$

The events T, B, S, and O form a partition of the sample space, as their probabilities sum to 1 ($0.3 + 0.2 + 0.1 + 0.4 = 1$).

We are given the following conditional probabilities of being late given the mode of transport:

$P(L|T) = \frac{1}{4}$

$P(L|B) = \frac{1}{3}$

$P(L|S) = \frac{1}{12}$

$P(L|O) = 0$ (The doctor is not late if he comes by other means)

We are given that the doctor arrived late (event L occurred).

To Find:

The probability that the doctor came by train given that he was late, i.e., $P(T|L)$.

---

Solution:

We need to find the conditional probability $P(T|L)$. Using Bayes' Theorem:

$P(T | L) = \frac{P(L | T) P(T)}{P(L)}$

To use Bayes' Theorem, we first need to calculate the overall probability of the doctor being late, $P(L)$. We can use the Law of Total Probability, since T, B, S, and O form a partition of the sample space:

$P(L) = P(L|T) P(T) + P(L|B) P(B) + P(L|S) P(S) + P(L|O) P(O)$

Substitute the given probabilities into the formula for $P(L)$:

$P(L) = \left(\frac{1}{4}\right) \left(\frac{3}{10}\right) + \left(\frac{1}{3}\right) \left(\frac{1}{5}\right) + \left(\frac{1}{12}\right) \left(\frac{1}{10}\right) + (0) \left(\frac{2}{5}\right)$

$P(L) = \frac{3}{40} + \frac{1}{15} + \frac{1}{120} + 0$

To add these fractions, find the least common multiple (LCM) of the denominators 40, 15, and 120. The LCM is 120.

$P(L) = \frac{3 \times 3}{40 \times 3} + \frac{1 \times 8}{15 \times 8} + \frac{1 \times 1}{120 \times 1}$

$P(L) = \frac{9}{120} + \frac{8}{120} + \frac{1}{120}$

$P(L) = \frac{9 + 8 + 1}{120}$

$P(L) = \frac{18}{120}$

Simplify the fraction for $P(L)$:

$P(L) = \frac{\cancel{18}^{3}}{\cancel{120}_{20}} = \frac{3}{20}$

Now, substitute the values $P(L|T) = \frac{1}{4}$, $P(T) = \frac{3}{10}$, and $P(L) = \frac{3}{20}$ into Bayes' Theorem formula for $P(T | L)$:

$P(T | L) = \frac{\left(\frac{1}{4}\right) \left(\frac{3}{10}\right)}{\frac{3}{20}}$

$P(T | L) = \frac{\frac{3}{40}}{\frac{3}{20}}$

To divide by a fraction, multiply by its reciprocal:

$P(T | L) = \frac{3}{40} \times \frac{20}{3}$

Cancel the common factors:

$P(T | L) = \frac{\cancel{3}}{40} \times \frac{20}{\cancel{3}}$

$P(T | L) = \frac{20}{40}$

$P(T | L) = \frac{\cancel{20}^{1}}{\cancel{40}_{2}} = \frac{1}{2}$

Thus, the probability that the doctor comes by train, given that he is late, is $\frac{1}{2}$.

Given:

Probability that the man speaks the truth, $P(\text{Truth}) = \frac{3}{4}$.

Probability that the man lies, $P(\text{Lie}) = 1 - P(\text{Truth}) = 1 - \frac{3}{4} = \frac{1}{4}$.

A fair die is thrown.

The man reports that the outcome is a six.

To Find:

The probability that the outcome is actually a six, given that the man reports it is a six.

---

Solution:

Let A be the event that the die actually shows a six.

Let A' be the event that the die does not show a six.

Let E be the event that the man reports that the outcome is a six.

When a fair die is thrown, the probability of getting a six is:

$P(A) = \frac{1}{6}$

The probability of not getting a six is:

$P(A') = 1 - P(A) = 1 - \frac{1}{6} = \frac{5}{6}$

We are given information about the man's truthfulness when reporting a specific outcome (like "it is a six").

The probability that he reports a six, given that it is actually a six ($P(E|A)$), is the probability that he tells the truth:

$P(E|A) = P(\text{Truth}) = \frac{3}{4}$

The probability that he reports a six, given that it is not actually a six ($P(E|A')$), is the probability that he lies and reports a six when the outcome was something else. Based on the standard interpretation of such problems, this is the probability that he lies:

$P(E|A') = P(\text{Lie}) = \frac{1}{4}$

We want to find the probability that the outcome is actually a six, given that he reports it is a six. This is the conditional probability $P(A|E)$. We can use Bayes' Theorem:

$P(A | E) = \frac{P(E | A) P(A)}{P(E)}$

First, we need to find the overall probability that the man reports that the outcome is a six, $P(E)$. We can use the Law of Total Probability. The events A and A' form a partition of the sample space (the outcome is either a six or not a six).

$P(E) = P(E | A) P(A) + P(E | A') P(A')$

Substitute the known probabilities into the formula for $P(E)$:

$P(E) = \left(\frac{3}{4}\right) \left(\frac{1}{6}\right) + \left(\frac{1}{4}\right) \left(\frac{5}{6}\right)$

$P(E) = \frac{3}{24} + \frac{5}{24}$

$P(E) = \frac{3 + 5}{24} = \frac{8}{24} = \frac{1}{3}$

Now, substitute the values $P(E|A) = \frac{3}{4}$, $P(A) = \frac{1}{6}$, and $P(E) = \frac{1}{3}$ into Bayes' Theorem formula for $P(A | E)$:

$P(A | E) = \frac{\left(\frac{3}{4}\right) \left(\frac{1}{6}\right)}{\frac{1}{3}}$

$P(A | E) = \frac{\frac{3}{24}}{\frac{1}{3}}$

$P(A | E) = \frac{\frac{1}{8}}{\frac{1}{3}}$

To divide by a fraction, multiply by its reciprocal:

$P(A | E) = \frac{1}{8} \times \frac{3}{1}$

$P(A | E) = \frac{3}{8}$

Thus, the probability that the outcome was actually a six, given that the man reported it was a six, is $\frac{3}{8}$.

Given:

Initial number of red balls in the urn = 5

Initial number of black balls in the urn = 5

Total initial number of balls = $5 + 5 = 10$

A ball is drawn at random, its colour is noted, and it is returned to the urn.

2 additional balls of the colour drawn are put in the urn.

Then, a second ball is drawn at random from the modified urn.

To Find:

The probability that the second ball drawn is red.

---

Solution:

Let $R_1$ be the event that the first ball drawn is red.

Let $B_1$ be the event that the first ball drawn is black.

Let $R_2$ be the event that the second ball drawn is red.

Let $B_2$ be the event that the second ball drawn is black.

The probability of drawing a red ball in the first draw from the initial urn is:

$P(R_1) = \frac{\text{Number of red balls}}{\text{Total number of balls}} = \frac{5}{10} = \frac{1}{2}$

The probability of drawing a black ball in the first draw from the initial urn is:

$P(B_1) = \frac{\text{Number of black balls}}{\text{Total number of balls}} = \frac{5}{10} = \frac{1}{2}$

After the first draw, the ball is returned to the urn. Then, 2 additional balls of the drawn colour are added.

We need to find the probability that the second ball drawn is red, $P(R_2)$. We can use the Law of Total Probability, considering the outcomes of the first draw ($R_1$ and $B_1$) as the partition events.

$P(R_2) = P(R_2 | R_1) P(R_1) + P(R_2 | B_1) P(B_1)$

Let's calculate the conditional probabilities $P(R_2 | R_1)$ and $P(R_2 | B_1)$.

Case 1: The first ball drawn was Red ($R_1$).

The red ball was returned, and 2 additional red balls were added.

The number of red balls in the urn becomes $5 + 2 = 7$.

The number of black balls remains 5.

The total number of balls in the urn becomes $7 + 5 = 12$.

The probability of drawing a red ball in the second draw, given the first was red, is:

$P(R_2 | R_1) = \frac{\text{Number of red balls}}{\text{Total number of balls}} = \frac{7}{12}$

Case 2: The first ball drawn was Black ($B_1$).

The black ball was returned, and 2 additional black balls were added.

The number of red balls remains 5.

The number of black balls becomes $5 + 2 = 7$.

The total number of balls in the urn becomes $5 + 7 = 12$.

The probability of drawing a red ball in the second draw, given the first was black, is:

$P(R_2 | B_1) = \frac{\text{Number of red balls}}{\text{Total number of balls}} = \frac{5}{12}$

Now, substitute the probabilities into the Law of Total Probability formula:

$P(R_2) = P(R_2 | R_1) P(R_1) + P(R_2 | B_1) P(B_1)$

$P(R_2) = \left(\frac{7}{12}\right) \left(\frac{1}{2}\right) + \left(\frac{5}{12}\right) \left(\frac{1}{2}\right)$

$P(R_2) = \frac{7}{24} + \frac{5}{24}$

$P(R_2) = \frac{7 + 5}{24}$

$P(R_2) = \frac{12}{24}$

$P(R_2) = \frac{1}{2}$

The probability that the second ball drawn is red is $\frac{1}{2}$.

Given:

Bag I contains: 4 Red (R) + 4 Black (B) = 8 balls.

Bag II contains: 2 Red (R) + 6 Black (B) = 8 balls.

One of the two bags is selected at random.

A ball is drawn from the selected bag and is found to be red.

To Find:

The probability that the red ball was drawn from the first bag (Bag I).

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Solution:

Let $B_1$ be the event that Bag I is selected.

Let $B_2$ be the event that Bag II is selected.

Since one of the two bags is selected at random, the probability of selecting either bag is equal:

$P(B_1) = \frac{1}{2}$

$P(B_2) = \frac{1}{2}$

Let R be the event that the ball drawn is red.

We need to find the conditional probabilities of drawing a red ball given that a specific bag was selected:

Probability of drawing a red ball given Bag I was selected:

$P(R|B_1) = \frac{\text{Number of red balls in Bag I}}{\text{Total balls in Bag I}} = \frac{4}{8} = \frac{1}{2}$

Probability of drawing a red ball given Bag II was selected:

$P(R|B_2) = \frac{\text{Number of red balls in Bag II}}{\text{Total balls in Bag II}} = \frac{2}{8} = \frac{1}{4}$

We are given that a red ball has been drawn (event R occurred), and we need to find the probability that it was drawn from Bag I, i.e., $P(B_1 | R)$.

We can use Bayes' Theorem for this:

$P(B_1 | R) = \frac{P(R | B_1) P(B_1)}{P(R)}$

To use this formula, we first need to find the overall probability of drawing a red ball, $P(R)$. We can find $P(R)$ using the Law of Total Probability. The events $B_1$ and $B_2$ form a partition of the sample space.

$P(R) = P(R|B_1) P(B_1) + P(R|B_2) P(B_2)$

Substitute the values we have:

$P(R) = \left(\frac{1}{2}\right) \left(\frac{1}{2}\right) + \left(\frac{1}{4}\right) \left(\frac{1}{2}\right)$

$P(R) = \frac{1}{4} + \frac{1}{8}$

To add these fractions, find a common denominator, which is 8:

$P(R) = \frac{1 \times 2}{4 \times 2} + \frac{1}{8}$

$P(R) = \frac{2}{8} + \frac{1}{8} = \frac{3}{8}$

Now, substitute the values into Bayes' Theorem formula for $P(B_1 | R)$:

$P(B_1 | R) = \frac{P(R | B_1) P(B_1)}{P(R)}$

$P(B_1 | R) = \frac{\left(\frac{1}{2}\right) \left(\frac{1}{2}\right)}{\frac{3}{8}}$

$P(B_1 | R) = \frac{\frac{1}{4}}{\frac{3}{8}}$

To divide by a fraction, multiply by its reciprocal:

$P(B_1 | R) = \frac{1}{4} \times \frac{8}{3}$

Cancel the common factor 4:

$P(B_1 | R) = \frac{1}{\cancel{4}_{1}} \times \frac{\cancel{8}^{2}}{3}$

$P(B_1 | R) = \frac{1 \times 2}{1 \times 3}$

$P(B_1 | R) = \frac{2}{3}$

The probability that the red ball was drawn from the first bag is $\frac{2}{3}$.

Given:

Let H be the event that a student resides in the hostel.

Let D be the event that a student is a day scholar.

Let A be the event that a student attains A grade.

We are given the following probabilities:

$P(H) = 60\% = 0.60$

$P(D) = 40\% = 0.40$

Note that H and D are mutually exclusive and exhaustive events, so they form a partition of the sample space ($P(H) + P(D) = 0.60 + 0.40 = 1.00$).

We are also given the conditional probabilities of attaining an A grade given the student's residency status:

$P(A|H) = 30\% = 0.30$ (Probability of A grade given the student is a hostlier)

$P(A|D) = 20\% = 0.20$ (Probability of A grade given the student is a day scholar)

A student is chosen at random and has an A grade (event A has occurred).

---

To Find:

The probability that the student is a hostlier, given that they attained an A grade, i.e., $P(H|A)$.

---

Solution:

We need to find the conditional probability $P(H|A)$. We can use Bayes' Theorem for this.

Bayes' Theorem states:

$P(H | A) = \frac{P(A | H) P(H)}{P(A)}$

To use this formula, we first need to calculate the overall probability of a student attaining an A grade, $P(A)$. We can do this using the Law of Total Probability. Since events H and D form a partition of the student population:

$P(A) = P(A | H) P(H) + P(A | D) P(D)$

Substitute the known probabilities into the formula for $P(A)$:

$P(A) = (0.30)(0.60) + (0.20)(0.40)$

$P(A) = 0.18 + 0.08$

$P(A) = 0.26$

Now, substitute the values of $P(A|H)$, $P(H)$, and $P(A)$ into the Bayes' Theorem formula for $P(H | A)$:

$P(H | A) = \frac{(0.30)(0.60)}{0.26}$

$P(H | A) = \frac{0.18}{0.26}$

To simplify the fraction, multiply the numerator and denominator by 100:

$P(H | A) = \frac{18}{26}$

Divide the numerator and denominator by their greatest common divisor, which is 2:

$P(H | A) = \frac{\cancel{18}^{9}}{\cancel{26}_{13}}$

$P(H | A) = \frac{9}{13}$

Thus, the probability that the student is a hostlier, given they attained an A grade, is $\frac{9}{13}$.

Given:

Let K be the event that the student knows the answer.

Let G be the event that the student guesses the answer.

Let C be the event that the student answers the question correctly.

We are given the following probabilities:

$P(K) = \frac{3}{4}$

$P(G) = \frac{1}{4}$

Note that the events K and G are mutually exclusive and exhaustive, forming a partition of the sample space ($P(K) + P(G) = \frac{3}{4} + \frac{1}{4} = 1$).

We are given the probability of answering correctly if the student guesses:

$P(C|G) = \frac{1}{4}$

We assume that if a student knows the answer, they answer it correctly with certainty.

$P(C|K) = 1$

The student answered the question correctly (event C has occurred).

To Find:

The probability that the student knows the answer, given that they answered it correctly, i.e., $P(K|C)$.

---

Solution:

We need to find the conditional probability $P(K|C)$. We can use Bayes' Theorem for this.

Bayes' Theorem states:

$P(K | C) = \frac{P(C | K) P(K)}{P(C)}$

To use this formula, we first need to calculate the overall probability of the student answering correctly, $P(C)$. We can do this using the Law of Total Probability. Since events K and G form a partition:

$P(C) = P(C | K) P(K) + P(C | G) P(G)$

Substitute the known probabilities into the formula for $P(C)$:

$P(C) = (1) \times \left(\frac{3}{4}\right) + \left(\frac{1}{4}\right) \times \left(\frac{1}{4}\right)$

$P(C) = \frac{3}{4} + \frac{1}{16}$

To add these fractions, find a common denominator, which is 16:

$P(C) = \frac{3 \times 4}{4 \times 4} + \frac{1}{16}$

$P(C) = \frac{12}{16} + \frac{1}{16} = \frac{12 + 1}{16} = \frac{13}{16}$

Now, substitute the values $P(C|K) = 1$, $P(K) = \frac{3}{4}$, and $P(C) = \frac{13}{16}$ into Bayes' Theorem formula for $P(K | C)$:

$P(K | C) = \frac{(1) \times \left(\frac{3}{4}\right)}{\frac{13}{16}}$

$P(K | C) = \frac{\frac{3}{4}}{\frac{13}{16}}$

To divide by a fraction, multiply by its reciprocal:

$P(K | C) = \frac{3}{4} \times \frac{16}{13}$

Cancel the common factor 4 between 4 and 16:

$P(K | C) = \frac{3}{\cancel{4}_{1}} \times \frac{\cancel{16}^{4}}{13}$

$P(K | C) = \frac{3 \times 4}{1 \times 13} = \frac{12}{13}$

Thus, the probability that the student knows the answer, given that they answered it correctly, is $\frac{12}{13}$.

Given:

Let D be the event that a person has the disease.

Let D' be the event that a person does not have the disease (is healthy).

Let T+ be the event that the test result is positive.

Let T- be the event that the test result is negative.

We are given the following probabilities:

Probability of having the disease in the population, $P(D) = 0.1\% = \frac{0.1}{100} = 0.001$.

Probability of not having the disease, $P(D') = 1 - P(D) = 1 - 0.001 = 0.999$.

Probability of a positive test result given the person has the disease (True Positive Rate), $P(T+|D) = 99\% = 0.99$.

Probability of a positive test result given the person does not have the disease (False Positive Rate), $P(T+|D') = 0.5\% = \frac{0.5}{100} = 0.005$.

We are given that a randomly selected person's test result is positive (event T+ has occurred).

---

To Find:

The probability that the person actually has the disease given that the test result is positive, i.e., $P(D | T+)$.

---

Solution:

We need to find the conditional probability $P(D | T+)$. We can use Bayes' Theorem for this.

Bayes' Theorem states:

$P(D | T+) = \frac{P(T+ | D) P(D)}{P(T+)}$

To use this formula, we first need to calculate the overall probability of getting a positive test result, $P(T+)$. We can do this using the Law of Total Probability. The events D and D' form a partition of the sample space (a person either has the disease or does not).

$P(T+) = P(T+ | D) P(D) + P(T+ | D') P(D')$

Substitute the known probabilities into the formula for $P(T+)$:

$P(T+) = (0.99)(0.001) + (0.005)(0.999)$

$P(T+) = 0.00099 + 0.004995$

$P(T+) = 0.005985$

Now, substitute the values of $P(T+|D)$, $P(D)$, and $P(T+)$ into the Bayes' Theorem formula for $P(D | T+)$:

$P(D | T+) = \frac{(0.99)(0.001)}{0.005985}$

$P(D | T+) = \frac{0.00099}{0.005985}$

To simplify the fraction, we can multiply the numerator and denominator by 1,000,000:

$P(D | T+) = \frac{0.00099 \times 1000000}{0.005985 \times 1000000} = \frac{990}{5985}$

We can simplify this fraction by dividing the numerator and denominator by common factors. Both are divisible by 5:

$\frac{\cancel{990}^{198}}{\cancel{5985}_{1197}}$

The fraction is now $\frac{198}{1197}$. Both are divisible by 9:

$\frac{\cancel{198}^{22}}{\cancel{1197}_{133}}$

The simplified fraction is $\frac{22}{133}$.

Thus, the probability that a person has the disease, given that their test result is positive, is $\frac{22}{133}$.

Given:

There are three coins:

Coin 1 ($C_1$): Two-headed coin (Head on both faces).

Coin 2 ($C_2$): Biased coin (Heads 75% of the time).

Coin 3 ($C_3$): Unbiased coin (Heads 50% of the time).

One of the three coins is chosen at random.

The chosen coin is tossed and shows heads.

Let $C_1$ be the event that Coin 1 is chosen.

Let $C_2$ be the event that Coin 2 is chosen.

Let $C_3$ be the event that Coin 3 is chosen.

Since one of the three identical coins is chosen at random:

$P(C_1) = \frac{1}{3}$

$P(C_2) = \frac{1}{3}$

$P(C_3) = \frac{1}{3}$

These events form a partition of the sample space (choosing a coin).

Let H be the event that the tossed coin shows heads.

We are given the probability of getting heads for each coin:

$P(H|C_1) = 1$ (A two-headed coin always shows heads)

$P(H|C_2) = 75\% = \frac{75}{100} = \frac{3}{4}$ (Biased coin)

$P(H|C_3) = 50\% = \frac{50}{100} = \frac{1}{2}$ (Unbiased coin)

We are given that the tossed coin shows heads (event H has occurred).

---

To Find:

The probability that the chosen coin was the two-headed coin (Coin 1), given that it shows heads, i.e., $P(C_1 | H)$.

---

Solution:

We need to find the conditional probability $P(C_1 | H)$. We can use Bayes' Theorem for this.

Bayes' Theorem states:

$P(C_1 | H) = \frac{P(H | C_1) P(C_1)}{P(H)}$

To use this formula, we first need to calculate the overall probability of getting heads, $P(H)$. We can do this using the Law of Total Probability. Since events $C_1$, $C_2$, and $C_3$ form a partition:

$P(H) = P(H | C_1) P(C_1) + P(H | C_2) P(C_2) + P(H | C_3) P(C_3)$

Substitute the known probabilities into the formula for $P(H)$:

$P(H) = (1) \left(\frac{1}{3}\right) + \left(\frac{3}{4}\right) \left(\frac{1}{3}\right) + \left(\frac{1}{2}\right) \left(\frac{1}{3}\right)$

$P(H) = \frac{1}{3} + \frac{3}{12} + \frac{1}{6}$

To add these fractions, find the least common multiple (LCM) of the denominators 3, 12, and 6. The LCM is 12.

$P(H) = \frac{1 \times 4}{3 \times 4} + \frac{3}{12} + \frac{1 \times 2}{6 \times 2}$

$P(H) = \frac{4}{12} + \frac{3}{12} + \frac{2}{12}$

$P(H) = \frac{4 + 3 + 2}{12}$

$P(H) = \frac{9}{12}$

Simplify the fraction for $P(H)$:

$P(H) = \frac{\cancel{9}^{3}}{\cancel{12}_{4}} = \frac{3}{4}$

Now, substitute the values $P(H|C_1) = 1$, $P(C_1) = \frac{1}{3}$, and $P(H) = \frac{3}{4}$ into Bayes' Theorem formula for $P(C_1 | H)$:

$P(C_1 | H) = \frac{(1) \times \left(\frac{1}{3}\right)}{\frac{3}{4}}$

$P(C_1 | H) = \frac{\frac{1}{3}}{\frac{3}{4}}$

To divide by a fraction, multiply by its reciprocal:

$P(C_1 | H) = \frac{1}{3} \times \frac{4}{3}$

$P(C_1 | H) = \frac{1 \times 4}{3 \times 3}$

$P(C_1 | H) = \frac{4}{9}$

Thus, the probability that the chosen coin was the two-headed coin, given that it shows heads, is $\frac{4}{9}$.

Given:

Number of insured scooter drivers ($N_S$) = 2000

Number of insured car drivers ($N_C$) = 4000

Number of insured truck drivers ($N_T$) = 6000

Total number of insured persons = $2000 + 4000 + 6000 = 12000$

Let S be the event that the insured person is a scooter driver.

Let C be the event that the insured person is a car driver.

Let T be the event that the insured person is a truck driver.

The probabilities of an insured person being of each type are:

$P(S) = \frac{N_S}{\text{Total}} = \frac{2000}{12000} = \frac{2}{12} = \frac{1}{6}$

$P(C) = \frac{N_C}{\text{Total}} = \frac{4000}{12000} = \frac{4}{12} = \frac{1}{3}$

$P(T) = \frac{N_T}{\text{Total}} = \frac{6000}{12000} = \frac{6}{12} = \frac{1}{2}$

Note that these events form a partition of the sample space ($1/6 + 1/3 + 1/2 = 1/6 + 2/6 + 3/6 = 6/6 = 1$).

Let A be the event that an insured person meets with an accident.

We are given the conditional probabilities of having an accident given the type of driver:

$P(A|S) = 0.01$

$P(A|C) = 0.03$

$P(A|T) = 0.15$

One of the insured persons meets with an accident (event A has occurred).

---

To Find:

The probability that the person is a scooter driver, given that they met with an accident, i.e., $P(S|A)$.

---

Solution:

We need to find the conditional probability $P(S|A)$. We can use Bayes' Theorem for this.

Bayes' Theorem states:

$P(S | A) = \frac{P(A | S) P(S)}{P(A)}$

To use this formula, we first need to calculate the overall probability of an insured person meeting with an accident, $P(A)$. We can do this using the Law of Total Probability. Since events S, C, and T form a partition:

$P(A) = P(A | S) P(S) + P(A | C) P(C) + P(A | T) P(T)$

Substitute the known probabilities into the formula for $P(A)$:

$P(A) = (0.01) \left(\frac{1}{6}\right) + (0.03) \left(\frac{1}{3}\right) + (0.15) \left(\frac{1}{2}\right)$

$P(A) = \frac{0.01}{6} + \frac{0.03}{3} + \frac{0.15}{2}$

$P(A) = \frac{0.01}{6} + 0.01 + 0.075$

Let's work with decimals or fractions. Using fractions:

$P(A) = \frac{1}{100} \times \frac{1}{6} + \frac{3}{100} \times \frac{1}{3} + \frac{15}{100} \times \frac{1}{2}$

$P(A) = \frac{1}{600} + \frac{\cancel{3}^{1}}{\cancel{300}_{100}} + \frac{15}{200}$

$P(A) = \frac{1}{600} + \frac{1}{100} + \frac{15}{200}$

Find the LCM of 600, 100, and 200. The LCM is 600.

$P(A) = \frac{1}{600} + \frac{1 \times 6}{100 \times 6} + \frac{15 \times 3}{200 \times 3}$

$P(A) = \frac{1}{600} + \frac{6}{600} + \frac{45}{600}$

$P(A) = \frac{1 + 6 + 45}{600} = \frac{52}{600}$

Simplify the fraction for $P(A)$ by dividing by 4:

$P(A) = \frac{\cancel{52}^{13}}{\cancel{600}_{150}} = \frac{13}{150}$

Now, substitute the values $P(A|S) = 0.01 = \frac{1}{100}$, $P(S) = \frac{1}{6}$, and $P(A) = \frac{13}{150}$ into Bayes' Theorem formula for $P(S | A)$:

$P(S | A) = \frac{P(A | S) P(S)}{P(A)}$

$P(S | A) = \frac{\left(\frac{1}{100}\right) \left(\frac{1}{6}\right)}{\frac{13}{150}}$

$P(S | A) = \frac{\frac{1}{600}}{\frac{13}{150}}$

To divide by a fraction, multiply by its reciprocal:

$P(S | A) = \frac{1}{600} \times \frac{150}{13}$

Cancel the common factor 150:

$P(S | A) = \frac{1}{\cancel{600}_{4}} \times \frac{\cancel{150}^{1}}{13}$

$P(S | A) = \frac{1 \times 1}{4 \times 13}$

$P(S | A) = \frac{1}{52}$

Thus, the probability that the insured person who met with an accident is a scooter driver is $\frac{1}{52}$.

Given:

Let A be the event that an item is produced by machine A.

Let B be the event that an item is produced by machine B.

Let D be the event that an item is defective.

We are given the following probabilities:

$P(A) = 60\% = 0.60$

$P(B) = 40\% = 0.40$

Note that events A and B form a partition of the sample space of all items produced ($P(A) + P(B) = 0.60 + 0.40 = 1.00$).

We are given the conditional probabilities of an item being defective, given which machine produced it:

$P(D|A) = 2\% = 0.02$

$P(D|B) = 1\% = 0.01$

An item is chosen at random from the stockpile and is found to be defective (event D has occurred).

---

To Find:

The probability that the defective item was produced by machine B, i.e., $P(B|D)$.

---

Solution:

We need to find the conditional probability $P(B|D)$. We can use Bayes' Theorem for this.

Bayes' Theorem states:

$P(B | D) = \frac{P(D | B) P(B)}{P(D)}$

To use this formula, we first need to calculate the overall probability of an item being defective, $P(D)$. We can do this using the Law of Total Probability. Since events A and B form a partition:

$P(D) = P(D | A) P(A) + P(D | B) P(B)$

Substitute the known probabilities into the formula for $P(D)$:

$P(D) = (0.02)(0.60) + (0.01)(0.40)$

$P(D) = 0.0120 + 0.0040$

$P(D) = 0.0160$

Now, substitute the values of $P(D|B)$, $P(B)$, and $P(D)$ into the Bayes' Theorem formula for $P(B|D)$:

$P(B | D) = \frac{(0.01)(0.40)}{0.0160}$

$P(B | D) = \frac{0.0040}{0.0160}$

To simplify the fraction, we can multiply the numerator and denominator by 10000:

$P(B | D) = \frac{0.0040 \times 10000}{0.0160 \times 10000} = \frac{40}{160}$

Simplify the fraction by dividing by 40:

$P(B | D) = \frac{\cancel{40}^{1}}{\cancel{160}_{4}}$

$P(B | D) = \frac{1}{4}$

Thus, the probability that the defective item was produced by machine B is $\frac{1}{4}$ or 0.25.

Given:

Let $G_1$ be the event that the first group wins.

Let $G_2$ be the event that the second group wins.

Let N be the event that a new product is introduced.

We are given the following probabilities:

$P(G_1) = 0.6$

$P(G_2) = 0.4$

Note that $P(G_1) + P(G_2) = 0.6 + 0.4 = 1.0$, and the events $G_1$ and $G_2$ are mutually exclusive, so they form a partition of the sample space.

We are given the conditional probabilities of introducing a new product:

$P(N|G_1) = 0.7$ (Probability of new product given the first group wins)

$P(N|G_2) = 0.3$ (Probability of new product given the second group wins)

A new product is introduced (event N has occurred).

---

To Find:

The probability that the new product was introduced by the second group, given that a new product was introduced, i.e., $P(G_2 | N)$.

---

Solution:

We need to find the conditional probability $P(G_2 | N)$. We can use Bayes' Theorem for this.

Bayes' Theorem states:

$P(G_2 | N) = \frac{P(N | G_2) P(G_2)}{P(N)}$

To use this formula, we first need to calculate the overall probability of introducing a new product, $P(N)$. We can do this using the Law of Total Probability. Since events $G_1$ and $G_2$ form a partition:

$P(N) = P(N | G_1) P(G_1) + P(N | G_2) P(G_2)$

Substitute the known probabilities into the formula for $P(N)$:

$P(N) = (0.7)(0.6) + (0.3)(0.4)$

$P(N) = 0.42 + 0.12$

$P(N) = 0.54$

Now, substitute the values of $P(N|G_2)$, $P(G_2)$, and $P(N)$ into the Bayes' Theorem formula for $P(G_2 | N)$:

$P(G_2 | N) = \frac{(0.3)(0.4)}{0.54}$

$P(G_2 | N) = \frac{0.12}{0.54}$

To simplify the fraction, multiply the numerator and denominator by 100:

$P(G_2 | N) = \frac{12}{54}$

Divide the numerator and denominator by their greatest common divisor, which is 6:

$P(G_2 | N) = \frac{\cancel{12}^{2}}{\cancel{54}_{9}}$

$P(G_2 | N) = \frac{2}{9}$

Thus, the probability that the new product was introduced by the second group, given that a new product was introduced, is $\frac{2}{9}$.

Given:

A girl throws a fair die.

Let $D_1$ be the event that she gets a 1, 2, 3, or 4 on the die.

Let $D_2$ be the event that she gets a 5 or 6 on the die.

The possible outcomes when throwing a die are {1, 2, 3, 4, 5, 6}.

$P(D_1) = \frac{\text{Number of outcomes in } D_1}{\text{Total outcomes}} = \frac{4}{6} = \frac{2}{3}$

$P(D_2) = \frac{\text{Number of outcomes in } D_2}{\text{Total outcomes}} = \frac{2}{6} = \frac{1}{3}$

Note that $D_1$ and $D_2$ are mutually exclusive and exhaustive events, forming a partition ($P(D_1) + P(D_2) = \frac{2}{3} + \frac{1}{3} = 1$).

If she gets a 5 or 6 ($D_2$), she tosses a coin three times. The possible outcomes are HHH, HHT, HTH, THH, HTT, THT, TTH, TTT (8 outcomes).

If she gets 1, 2, 3, or 4 ($D_1$), she tosses a coin once. The possible outcomes are H, T (2 outcomes).

Let E be the event that she obtains exactly one head.

We need to find the conditional probability of obtaining exactly one head given which die outcome occurred:

If she threw 1, 2, 3, or 4 ($D_1$), she tosses the coin once.

$P(E | D_1) = P(\text{exactly one head in 1 toss}) = P(\text{H}) = \frac{1}{2}$

If she threw 5 or 6 ($D_2$), she tosses the coin three times.

The outcomes with exactly one head are HTT, THT, TTH.

Number of outcomes with exactly one head = 3.

Total number of outcomes from three coin tosses = $2^3 = 8$.

$P(E | D_2) = P(\text{exactly one head in 3 tosses}) = \frac{3}{8}$

We are given that she obtained exactly one head (event E has occurred).

---

To Find:

The probability that she threw 1, 2, 3, or 4 with the die, given that she obtained exactly one head, i.e., $P(D_1 | E)$.

---

Solution:

We need to find the conditional probability $P(D_1 | E)$. We can use Bayes' Theorem for this.

Bayes' Theorem states:

$P(D_1 | E) = \frac{P(E | D_1) P(D_1)}{P(E)}$

To use this formula, we first need to calculate the overall probability of obtaining exactly one head, $P(E)$. We can do this using the Law of Total Probability. Since events $D_1$ and $D_2$ form a partition:

$P(E) = P(E | D_1) P(D_1) + P(E | D_2) P(D_2)$

Substitute the known probabilities into the formula for $P(E)$:

$P(E) = \left(\frac{1}{2}\right) \left(\frac{2}{3}\right) + \left(\frac{3}{8}\right) \left(\frac{1}{3}\right)$

$P(E) = \frac{2}{6} + \frac{3}{24}$

$P(E) = \frac{1}{3} + \frac{3}{24}$

Find a common denominator for 3 and 24, which is 24.

$P(E) = \frac{1 \times 8}{3 \times 8} + \frac{3}{24}$

$P(E) = \frac{8}{24} + \frac{3}{24} = \frac{8 + 3}{24} = \frac{11}{24}$

Now, substitute the values $P(E|D_1) = \frac{1}{2}$, $P(D_1) = \frac{2}{3}$, and $P(E) = \frac{11}{24}$ into Bayes' Theorem formula for $P(D_1 | E)$:

$P(D_1 | E) = \frac{\left(\frac{1}{2}\right) \left(\frac{2}{3}\right)}{\frac{11}{24}}$

$P(D_1 | E) = \frac{\frac{2}{6}}{\frac{11}{24}}$

$P(D_1 | E) = \frac{\frac{1}{3}}{\frac{11}{24}}$

To divide by a fraction, multiply by its reciprocal:

$P(D_1 | E) = \frac{1}{3} \times \frac{24}{11}$

Cancel the common factor 3 between 3 and 24:

$P(D_1 | E) = \frac{1}{\cancel{3}_{1}} \times \frac{\cancel{24}^{8}}{11}$

$P(D_1 | E) = \frac{1 \times 8}{1 \times 11} = \frac{8}{11}$

Thus, the probability that she threw 1, 2, 3, or 4 with the die, given that she obtained exactly one head, is $\frac{8}{11}$.

Given:

Let A, B, and C be the events that an item is produced by operator A, B, and C respectively.

Let D be the event that an item is defective.

The proportion of time each operator is on the job represents the probability that a randomly selected item was produced by that operator:

$P(A) = 50\% = 0.50$

$P(B) = 30\% = 0.30$

$P(C) = 20\% = 0.20$

Note that $P(A) + P(B) + P(C) = 0.50 + 0.30 + 0.20 = 1.00$, so these events form a partition of the sample space of all items produced.

We are given the conditional probabilities of an item being defective, given which operator produced it:

$P(D|A) = 1\% = 0.01$

$P(D|B) = 5\% = 0.05$

$P(D|C) = 7\% = 0.07$

A defective item is produced (event D has occurred).

---

To Find:

The probability that the defective item was produced by operator A, given that it is defective, i.e., $P(A|D)$.

---

Solution:

We need to find the conditional probability $P(A|D)$. We can use Bayes' Theorem for this.

Bayes' Theorem states:

$P(A | D) = \frac{P(D | A) P(A)}{P(D)}$

To use this formula, we first need to calculate the overall probability of an item being defective, $P(D)$. We can do this using the Law of Total Probability. Since events A, B, and C form a partition:

$P(D) = P(D | A) P(A) + P(D | B) P(B) + P(D | C) P(C)$

Substitute the known probabilities into the formula for $P(D)$:

$P(D) = (0.01)(0.50) + (0.05)(0.30) + (0.07)(0.20)$

$P(D) = 0.0050 + 0.0150 + 0.0140$

$P(D) = 0.0340$

Now, substitute the values of $P(D|A)$, $P(A)$, and $P(D)$ into the Bayes' Theorem formula for $P(A|D)$:

$P(A | D) = \frac{(0.01)(0.50)}{0.0340}$

$P(A | D) = \frac{0.0050}{0.0340}$

To simplify the fraction, we can multiply the numerator and denominator by 10000:

$P(A | D) = \frac{0.0050 \times 10000}{0.0340 \times 10000} = \frac{50}{340}$

Simplify the fraction by dividing by 10:

$P(A | D) = \frac{\cancel{50}^{5}}{\cancel{340}_{34}}$

$P(A | D) = \frac{5}{34}$

Thus, the probability that the defective item was produced by operator A is $\frac{5}{34}$.

Given:

Initial deck size = 52 cards (13 diamonds, 39 non-diamonds).

One card is lost.

From the remaining 51 cards, two cards are drawn and found to be both diamonds.

To Find:

The probability that the lost card was a diamond, given that the two drawn cards are diamonds.

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Solution:

Let $L_D$ be the event that the lost card is a diamond.

Let $L_{ND}$ be the event that the lost card is not a diamond.

Let E be the event that two cards drawn from the remaining 51 cards are both diamonds.

We want to find $P(L_D | E)$. We can use Bayes' Theorem:

$P(L_D | E) = \frac{P(E | L_D) P(L_D)}{P(E)}$

First, determine the prior probabilities of the lost card's type:

$P(L_D) = \frac{\text{Number of diamonds initially}}{\text{Total number of cards initially}} = \frac{13}{52} = \frac{1}{4}$

$P(L_{ND}) = 1 - P(L_D) = 1 - \frac{1}{4} = \frac{3}{4}$

Next, determine the conditional probabilities of drawing two diamonds (Event E) given the type of the lost card:

Case 1: The lost card was a diamond ($L_D$).

The remaining pack has $52 - 1 = 51$ cards.

Number of diamonds remaining = $13 - 1 = 12$.

Number of non-diamonds remaining = 39.

The probability of drawing two diamonds from these 51 cards is:

$P(E | L_D) = \frac{\binom{12}{2}}{\binom{51}{2}}$

$\binom{12}{2} = \frac{12 \times 11}{2} = 66$

$\binom{51}{2} = \frac{51 \times 50}{2} = 51 \times 25 = 1275$

$P(E | L_D) = \frac{66}{1275} = \frac{\cancel{66}^{22}}{\cancel{1275}_{425}} = \frac{22}{425}$

Case 2: The lost card was not a diamond ($L_{ND}$).

The remaining pack has $52 - 1 = 51$ cards.

Number of diamonds remaining = 13.

Number of non-diamonds remaining = $39 - 1 = 38$.

The probability of drawing two diamonds from these 51 cards is:

$P(E | L_{ND}) = \frac{\binom{13}{2}}{\binom{51}{2}}$

$\binom{13}{2} = \frac{13 \times 12}{2} = 78$

$\binom{51}{2} = 1275$

$P(E | L_{ND}) = \frac{78}{1275} = \frac{\cancel{78}^{26}}{\cancel{1275}_{425}} = \frac{26}{425}$

Now, calculate the overall probability of event E occurring, $P(E)$, using the Law of Total Probability:

$P(E) = P(E | L_D) P(L_D) + P(E | L_{ND}) P(L_{ND})$

$P(E) = \left(\frac{22}{425}\right) \left(\frac{1}{4}\right) + \left(\frac{26}{425}\right) \left(\frac{3}{4}\right)$

$P(E) = \frac{22}{1700} + \frac{78}{1700}$

$P(E) = \frac{22 + 78}{1700} = \frac{100}{1700} = \frac{\cancel{100}^{1}}{\cancel{1700}_{17}} = \frac{1}{17}$

Finally, apply Bayes' Theorem to find $P(L_D | E)$:

$P(L_D | E) = \frac{P(E | L_D) P(L_D)}{P(E)}$

$P(L_D | E) = \frac{\left(\frac{22}{425}\right) \left(\frac{1}{4}\right)}{\frac{1}{17}}$

$P(L_D | E) = \frac{\frac{22}{1700}}{\frac{1}{17}}$

$P(L_D | E) = \frac{22}{1700} \times \frac{17}{1}$

$P(L_D | E) = \frac{22}{\cancel{1700}_{100}} \times \cancel{17}^{1}$

$P(L_D | E) = \frac{22}{100}$

$P(L_D | E) = \frac{\cancel{22}^{11}}{\cancel{100}_{50}} = \frac{11}{50}$

The probability that the lost card was a diamond, given that two diamonds were drawn from the remaining cards, is $\frac{11}{50}$.

Given:

Probability that A speaks the truth, $P(\text{Truth}) = \frac{4}{5}$.

Probability that A lies, $P(\text{Lie}) = 1 - P(\text{Truth}) = 1 - \frac{4}{5} = \frac{1}{5}$.

A fair coin is tossed.

A reports that a head appears.

To Find:

The probability that the coin actually showed a head, given that A reports it is a head.

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Solution:

Let H be the event that the coin actually shows a head.

Let H' be the event that the coin actually shows a tail.

Let R be the event that A reports that a head appears.

For a fair coin toss:

$P(H) = \frac{1}{2}$

$P(H') = 1 - P(H) = 1 - \frac{1}{2} = \frac{1}{2}$

We need the conditional probabilities of the report given the actual outcome:

The probability that A reports a head, given that it is actually a head ($P(R|H)$), is the probability that A speaks the truth:

$P(R|H) = P(\text{Truth}) = \frac{4}{5}$

The probability that A reports a head, given that it is actually a tail ($P(R|H')$), is the probability that A lies:

$P(R|H') = P(\text{Lie}) = \frac{1}{5}$

We want to find the probability that it was actually a head, given that A reports it is a head, which is $P(H|R)$. We use Bayes' Theorem:

$P(H | R) = \frac{P(R | H) P(H)}{P(R)}$

First, calculate the overall probability of the man reporting a head, $P(R)$, using the Law of Total Probability. The events H and H' form a partition.

$P(R) = P(R | H) P(H) + P(R | H') P(H')$

Substitute the known probabilities into the formula for $P(R)$:

$P(R) = \left(\frac{4}{5}\right) \left(\frac{1}{2}\right) + \left(\frac{1}{5}\right) \left(\frac{1}{2}\right)$

$P(R) = \frac{4}{10} + \frac{1}{10}$

$P(R) = \frac{4 + 1}{10} = \frac{5}{10} = \frac{1}{2}$

Now, substitute the values $P(R|H) = \frac{4}{5}$, $P(H) = \frac{1}{2}$, and $P(R) = \frac{1}{2}$ into Bayes' Theorem formula for $P(H | R)$:

$P(H | R) = \frac{\left(\frac{4}{5}\right) \left(\frac{1}{2}\right)}{\frac{1}{2}}$

$P(H | R) = \frac{\frac{4}{10}}{\frac{1}{2}}$

$P(H | R) = \frac{\frac{2}{5}}{\frac{1}{2}}$

To divide by a fraction, multiply by its reciprocal:

$P(H | R) = \frac{2}{5} \times \frac{2}{1}$

$P(H | R) = \frac{4}{5}$

The probability that the coin actually showed a head, given that A reports it is a head, is $\frac{4}{5}$.

This corresponds to option (A).

The correct answer is (A) $\frac{4}{5}$.

Given:

Events A and B such that $A \subset B$ and $P(B) \neq 0$.

To Determine:

Which of the given statements about $P(A|B)$ is correct.

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Solution:

The definition of conditional probability is $P(A|B) = \frac{P(A \cap B)}{P(B)}$, provided $P(B) \neq 0$.

We are given that $A \subset B$. This means that event A is a subset of event B. If event A occurs, event B must also occur. Therefore, the intersection of A and B, $A \cap B$, is equal to A.

$A \cap B = A$

Substituting this into the formula for conditional probability:

$P(A|B) = \frac{P(A)}{P(B)}$

Since $A \subset B$, the probability of A occurring must be less than or equal to the probability of B occurring. That is, $P(A) \leq P(B)$.

Since $P(B) \neq 0$, and probability is non-negative, we have $0 < P(B) \leq 1$.

From $P(A) \leq P(B)$, since $P(B) > 0$, we can divide by $P(B)$:

$\frac{P(A)}{P(B)} \leq \frac{P(B)}{P(B)}$

$\frac{P(A)}{P(B)} \leq 1$

This inequality $P(A|B) \leq 1$ is always true for any probability.

Let's compare $P(A|B)$ with $P(A)$. We have $P(A|B) = \frac{P(A)}{P(B)}$.

Since $P(B) \leq 1$ and $P(B) > 0$, the reciprocal $\frac{1}{P(B)} \geq 1$.

So, $P(A|B) = P(A) \times \frac{1}{P(B)}$.

Multiplying $P(A)$ by a factor $\frac{1}{P(B)}$ which is greater than or equal to 1 means that $P(A|B)$ must be greater than or equal to $P(A)$.

$P(A|B) \geq P(A)$

Equality holds if $P(B) = 1$. If $P(B) < 1$ and $P(A) > 0$, then $P(A|B) > P(A)$. If $P(A) = 0$, then $P(A|B) = 0/P(B) = 0 = P(A)$, so equality holds.

Thus, in all cases where $A \subset B$ and $P(B) \neq 0$, we have $P(A|B) \geq P(A)$.

Comparing this with the given options, option (C) is correct.

The correct answer is (C) P(A|B) ≥ P(A).

Given:

A person tosses a coin thrice.

Gain for each head = $\textsf{₹} 2$.

Loss for each tail = $\textsf{₹} 1.50$.

X denotes the amount gained or lost by the person.

To Show:

X is a random variable and exhibit it as a function on the sample space.

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Solution:

The experiment is tossing a coin thrice. The sample space S consists of all possible sequences of three coin tosses.

The possible outcomes are:

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

The random variable X is defined as the amount gained or lost by the person. This amount depends on the number of heads and tails obtained in the three tosses.

Let $n_H$ be the number of heads and $n_T$ be the number of tails in a sequence of three tosses. Note that $n_H + n_T = 3$.

The amount gained is $2 \times n_H$.

The amount lost is $1.50 \times n_T$.

The amount gained or lost (X) is given by the gain minus the loss:

$X = 2n_H - 1.50n_T$

Since $n_T = 3 - n_H$, we can express X in terms of $n_H$ only:

$X = 2n_H - 1.50(3 - n_H)$

$X = 2n_H - 4.50 + 1.50n_H$

$X = 3.50n_H - 4.50$

Now, let's evaluate the value of X for each outcome in the sample space S:

For HHH: $n_H = 3$. $X = 3.50(3) - 4.50 = 10.50 - 4.50 = 6.00$. Amount gained = $\textsf{₹} 6.00$.

For HHT: $n_H = 2$. $X = 3.50(2) - 4.50 = 7.00 - 4.50 = 2.50$. Amount gained = $\textsf{₹} 2.50$.

For HTH: $n_H = 2$. $X = 3.50(2) - 4.50 = 7.00 - 4.50 = 2.50$. Amount gained = $\textsf{₹} 2.50$.

For THH: $n_H = 2$. $X = 3.50(2) - 4.50 = 7.00 - 4.50 = 2.50$. Amount gained = $\textsf{₹} 2.50$.

For HTT: $n_H = 1$. $X = 3.50(1) - 4.50 = 3.50 - 4.50 = -1.00$. Amount lost = $\textsf{₹} 1.00$.

For THT: $n_H = 1$. $X = 3.50(1) - 4.50 = 3.50 - 4.50 = -1.00$. Amount lost = $\textsf{₹} 1.00$.

For TTH: $n_H = 1$. $X = 3.50(1) - 4.50 = 3.50 - 4.50 = -1.00$. Amount lost = $\textsf{₹} 1.00$.

For TTT: $n_H = 0$. $X = 3.50(0) - 4.50 = 0 - 4.50 = -4.50$. Amount lost = $\textsf{₹} 4.50$.

A random variable is a real-valued function whose domain is the sample space of a random experiment.

In this case, X assigns a real number (the amount gained or lost) to each outcome in the sample space S. Therefore, X is a random variable.

We can exhibit X as a function on the sample space:

$X(\text{HHH}) = 6.00$

$X(\text{HHT}) = 2.50$

$X(\text{HTH}) = 2.50$

$X(\text{THH}) = 2.50$

$X(\text{HTT}) = -1.00$

$X(\text{THT}) = -1.00$

$X(\text{TTH}) = -1.00$

$X(\text{TTT}) = -4.50$

The possible values that the random variable X can take are $\{6.00, 2.50, -1.00, -4.50\}$.

Given:

A bag contains 2 white (W) and 1 red (R) ball.

Total balls = $2 + 1 = 3$.

A ball is drawn at random, its colour is noted, and it is put back in the bag (drawing with replacement).

The process is repeated (two draws in total).

X denotes the number of red balls recorded in the two draws.

To Describe:

The random variable X.

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Solution:

The experiment consists of two draws with replacement from the bag. The possible outcomes of a single draw are either a Red (R) ball or a White (W) ball.

The probability of drawing a red ball in a single draw is $P(R) = \frac{\text{Number of red balls}}{\text{Total balls}} = \frac{1}{3}$.

The probability of drawing a white ball in a single draw is $P(W) = \frac{\text{Number of white balls}}{\text{Total balls}} = \frac{2}{3}$.

Since the drawing is done with replacement, the two draws are independent.

The sample space S for the two draws is the set of all possible ordered pairs of outcomes:

S = {(R, R), (R, W), (W, R), (W, W)}

The random variable X is defined as the number of red balls in the two draws. We can find the value of X for each outcome in the sample space:

For (R, R): Number of red balls = 2. So, X = 2.

For (R, W): Number of red balls = 1. So, X = 1.

For (W, R): Number of red balls = 1. So, X = 1.

For (W, W): Number of red balls = 0. So, X = 0.

The possible values that the random variable X can take are 0, 1, and 2.

We can also describe the probability distribution of X by finding the probability for each possible value of X.

$P(X=0)$ = Probability of getting 0 red balls (W, W)

$P(X=0) = P(\text{W in 1st}) \times P(\text{W in 2nd})$ (Independent draws)

$P(X=0) = \frac{2}{3} \times \frac{2}{3} = \frac{4}{9}$

$P(X=1)$ = Probability of getting exactly 1 red ball (R, W) or (W, R)

$P(X=1) = P(\text{R in 1st and W in 2nd}) + P(\text{W in 1st and R in 2nd})$ (Mutually exclusive outcomes)

$P(X=1) = [P(\text{R in 1st}) \times P(\text{W in 2nd})] + [P(\text{W in 1st}) \times P(\text{R in 2nd})]$ (Independent draws)

$P(X=1) = \left(\frac{1}{3} \times \frac{2}{3}\right) + \left(\frac{2}{3} \times \frac{1}{3}\right) = \frac{2}{9} + \frac{2}{9} = \frac{4}{9}$

$P(X=2)$ = Probability of getting 2 red balls (R, R)

$P(X=2) = P(\text{R in 1st}) \times P(\text{R in 2nd})$ (Independent draws)

$P(X=2) = \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$

Check that the probabilities sum to 1: $P(X=0) + P(X=1) + P(X=2) = \frac{4}{9} + \frac{4}{9} + \frac{1}{9} = \frac{9}{9} = 1$.

The random variable X takes values in the set $\{0, 1, 2\}$ with the following probability distribution:

$P(X=0) = \frac{4}{9}$

$P(X=1) = \frac{4}{9}$

$P(X=2) = \frac{1}{9}$

Given:

A well-shuffled deck of 52 cards.

Two cards are drawn successively with replacement.

X denotes the number of aces drawn.

To Find:

The probability distribution of X.

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Solution:

A standard deck has 52 cards.

Number of aces in the deck = 4.

Number of non-aces in the deck = $52 - 4 = 48$.

A card is drawn at random, its outcome is noted, and it is returned to the deck. The process is repeated for the second draw. Since the draws are with replacement, the two draws are independent events.

Let A be the event that an ace is drawn in a single draw.

Let A' be the event that a non-ace is drawn in a single draw.

The probability of drawing an ace in a single draw is:

$P(A) = \frac{\text{Number of aces}}{\text{Total number of cards}} = \frac{4}{52} = \frac{1}{13}$

The probability of drawing a non-ace in a single draw is:

$P(A') = \frac{\text{Number of non-aces}}{\text{Total number of cards}} = \frac{48}{52} = \frac{12}{13}$

The random variable X represents the number of aces in two draws. Since there are two draws, the possible number of aces can be 0, 1, or 2.

The possible values of X are $x = 0, 1, 2$.

We need to find the probability for each of these values.

$P(X=0)$: Getting 0 aces in two draws.

This means the first card is not an ace AND the second card is not an ace. Since the draws are independent:

$P(X=0) = P(\text{Not Ace on 1st draw}) \times P(\text{Not Ace on 2nd draw})$

$P(X=0) = P(A') \times P(A')$

$P(X=0) = \frac{12}{13} \times \frac{12}{13} = \frac{144}{169}$

$P(X=1)$: Getting exactly 1 ace in two draws.

This can happen in two mutually exclusive ways: (Ace on 1st AND Not Ace on 2nd) OR (Not Ace on 1st AND Ace on 2nd). Since the draws are independent:

$P(X=1) = [P(\text{Ace on 1st}) \times P(\text{Not Ace on 2nd})] + [P(\text{Not Ace on 1st}) \times P(\text{Ace on 2nd})]$

$P(X=1) = [P(A) \times P(A')] + [P(A') \times P(A)]$

$P(X=1) = \left(\frac{1}{13} \times \frac{12}{13}\right) + \left(\frac{12}{13} \times \frac{1}{13}\right)$

$P(X=1) = \frac{12}{169} + \frac{12}{169} = \frac{12 + 12}{169} = \frac{24}{169}$

$P(X=2)$: Getting 2 aces in two draws.

This means the first card is an ace AND the second card is an ace. Since the draws are independent:

$P(X=2) = P(\text{Ace on 1st draw}) \times P(\text{Ace on 2nd draw})$

$P(X=2) = P(A) \times P(A)$

$P(X=2) = \frac{1}{13} \times \frac{1}{13} = \frac{1}{169}$

The probability distribution of X is a table showing the possible values of X and their corresponding probabilities.

X (Number of Aces)	P(X)
0	$\frac{144}{169}$
1	$\frac{24}{169}$
2	$\frac{1}{169}$

Check: $\frac{144}{169} + \frac{24}{169} + \frac{1}{169} = \frac{144 + 24 + 1}{169} = \frac{169}{169} = 1$. The probabilities sum to 1.

Given:

A pair of fair dice is thrown three times.

A doublet is an outcome where both dice show the same number (e.g., (1, 1), (2, 2), etc.).

X denotes the number of doublets in three throws.

To Find:

The probability distribution of X.

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Solution:

Consider a single throw of a pair of dice.

The total number of possible outcomes is $6 \times 6 = 36$.

The possible doublets are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6).

The number of doublet outcomes is 6.

Let D be the event of getting a doublet in a single throw of a pair of dice.

The probability of getting a doublet in a single throw is:

$P(D) = \frac{\text{Number of doublets}}{\text{Total number of outcomes}} = \frac{6}{36} = \frac{1}{6}$

Let D' be the event of not getting a doublet in a single throw.

The probability of not getting a doublet in a single throw is:

$P(D') = 1 - P(D) = 1 - \frac{1}{6} = \frac{5}{6}$

The experiment consists of three throws of a pair of dice. Each throw is independent of the others.

The random variable X represents the number of doublets in these three throws. The possible number of doublets can be 0, 1, 2, or 3.

The possible values of X are $x = 0, 1, 2, 3$.

This is a binomial distribution scenario, where a "success" is getting a doublet (with probability $p = \frac{1}{6}$) and a "failure" is not getting a doublet (with probability $q = \frac{5}{6}$). The number of trials is $n=3$.

The probability mass function for a binomial distribution is given by $P(X=k) = \binom{n}{k} p^k q^{n-k}$, where $k$ is the number of successes.

In this case, $n=3$, $p = \frac{1}{6}$, and $q = \frac{5}{6}$. We need to find $P(X=k)$ for $k=0, 1, 2, 3$.

$P(X=0)$: Getting 0 doublets in 3 throws.

$P(X=0) = \binom{3}{0} \left(\frac{1}{6}\right)^0 \left(\frac{5}{6}\right)^{3-0}$

$P(X=0) = 1 \times 1 \times \left(\frac{5}{6}\right)^3 = \frac{5^3}{6^3} = \frac{125}{216}$

$P(X=1)$: Getting exactly 1 doublet in 3 throws.

$P(X=1) = \binom{3}{1} \left(\frac{1}{6}\right)^1 \left(\frac{5}{6}\right)^{3-1}$

$P(X=1) = 3 \times \left(\frac{1}{6}\right) \times \left(\frac{5}{6}\right)^2 = 3 \times \frac{1}{6} \times \frac{25}{36} = \frac{3 \times 1 \times 25}{6 \times 36} = \frac{75}{216}$

Simplify the fraction:

$P(X=1) = \frac{\cancel{75}^{25}}{\cancel{216}^{72}} = \frac{25}{72}$

$P(X=2)$: Getting exactly 2 doublets in 3 throws.

$P(X=2) = \binom{3}{2} \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^{3-2}$

$P(X=2) = 3 \times \left(\frac{1}{6}\right)^2 \times \left(\frac{5}{6}\right)^1 = 3 \times \frac{1}{36} \times \frac{5}{6} = \frac{3 \times 1 \times 5}{36 \times 6} = \frac{15}{216}$

Simplify the fraction:

$P(X=2) = \frac{\cancel{15}^{5}}{\cancel{216}^{72}} = \frac{5}{72}$

$P(X=3)$: Getting exactly 3 doublets in 3 throws.

$P(X=3) = \binom{3}{3} \left(\frac{1}{6}\right)^3 \left(\frac{5}{6}\right)^{3-3}$

$P(X=3) = 1 \times \left(\frac{1}{6}\right)^3 \times \left(\frac{5}{6}\right)^0 = 1 \times \frac{1}{216} \times 1 = \frac{1}{216}$

The probability distribution of X is:

X (Number of Doublets)	P(X)
0	$\frac{125}{216}$
1	$\frac{75}{216}$ or $\frac{25}{72}$
2	$\frac{15}{216}$ or $\frac{5}{72}$
3	$\frac{1}{216}$

Check: $\frac{125}{216} + \frac{75}{216} + \frac{15}{216} + \frac{1}{216} = \frac{125 + 75 + 15 + 1}{216} = \frac{216}{216} = 1$. The probabilities sum to 1.

Given:

The probability distribution function of the random variable X (number of hours studied) is given by:

$P(X = x) = \begin{cases} 0.1, & \text{if } x = 0 \\ kx, & \text{if } x = 1 \text{ or } 2 \\ k(5 − x), & \text{if } x = 3 \text{ or } 4 \\ 0, & \text{otherwise} \end{cases}$

where k is an unknown constant.

The possible values for X with non-zero probability are $x = 0, 1, 2, 3, 4$.

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To Find:

(a) The value of k.

(b) $P(X \geq 2)$, $P(X = 2)$, and $P(X \leq 2)$.

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Solution (a): Finding the value of k

For any probability distribution, the sum of the probabilities for all possible values of the random variable must equal 1.

So, we have:

$\sum P(X=x) = 1$

$P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) = 1$

Substitute the given probability function for each value of x:

$P(X=0) = 0.1$

$P(X=1) = k \times 1 = k$

$P(X=2) = k \times 2 = 2k$

$P(X=3) = k(5 - 3) = 2k$

$P(X=4) = k(5 - 4) = k$

Now, sum these probabilities and set the sum equal to 1:

$0.1 + k + 2k + 2k + k = 1$

Combine the terms involving k:

$0.1 + 6k = 1$

Subtract 0.1 from both sides:

$6k = 1 - 0.1$

$6k = 0.9$

Divide by 6 to find k:

$k = \frac{0.9}{6}$

$k = 0.15$

The value of k is 0.15.

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Solution (b): Calculating Probabilities

Now that we have the value of k, we can write out the complete probability distribution:

$P(X=0) = 0.1$

$P(X=1) = k \times 1 = 0.15 \times 1 = 0.15$

$P(X=2) = k \times 2 = 0.15 \times 2 = 0.30$

$P(X=3) = k(5 - 3) = 0.15 \times 2 = 0.30$

$P(X=4) = k(5 - 4) = 0.15 \times 1 = 0.15$

Sum = $0.1 + 0.15 + 0.30 + 0.30 + 0.15 = 1.00$. This confirms our value of k is correct.

Probability of studying at least two hours ($P(X \geq 2)$):

$P(X \geq 2) = P(X=2) + P(X=3) + P(X=4)$

$P(X \geq 2) = 0.30 + 0.30 + 0.15$

$P(X \geq 2) = 0.75$

Probability of studying exactly two hours ($P(X = 2)$):

This is directly from our calculated probabilities:

$P(X = 2) = 0.30$

Probability of studying at most two hours ($P(X \leq 2)$):

$P(X \leq 2) = P(X=0) + P(X=1) + P(X=2)$

$P(X \leq 2) = 0.1 + 0.15 + 0.30$

$P(X \leq 2) = 0.55$

Given:

A pair of fair dice is thrown.

X is the random variable representing the sum of the numbers that appear on the two dice.

To Find:

The mean or expectation of X, $E(X)$.

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Solution:

When a pair of fair dice is thrown, the total number of possible outcomes is $6 \times 6 = 36$. Each outcome is equally likely with probability $\frac{1}{36}$.

The random variable X represents the sum of the numbers on the two dice. The possible values of X are the integers from $1+1=2$ to $6+6=12$.

The possible values of X are $\{2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\}$.

We list the outcomes corresponding to each sum and their probabilities:

• $X=2$: (1,1) - 1 outcome, $P(X=2) = \frac{1}{36}$
• $X=3$: (1,2), (2,1) - 2 outcomes, $P(X=3) = \frac{2}{36}$
• $X=4$: (1,3), (2,2), (3,1) - 3 outcomes, $P(X=4) = \frac{3}{36}$
• $X=5$: (1,4), (2,3), (3,2), (4,1) - 4 outcomes, $P(X=5) = \frac{4}{36}$
• $X=6$: (1,5), (2,4), (3,3), (4,2), (5,1) - 5 outcomes, $P(X=6) = \frac{5}{36}$
• $X=7$: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) - 6 outcomes, $P(X=7) = \frac{6}{36}$
• $X=8$: (2,6), (3,5), (4,4), (5,3), (6,2) - 5 outcomes, $P(X=8) = \frac{5}{36}$
• $X=9$: (3,6), (4,5), (5,4), (6,3) - 4 outcomes, $P(X=9) = \frac{4}{36}$
• $X=10$: (4,6), (5,5), (6,4) - 3 outcomes, $P(X=10) = \frac{3}{36}$
• $X=11$: (5,6), (6,5) - 2 outcomes, $P(X=11) = \frac{2}{36}$
• $X=12$: (6,6) - 1 outcome, $P(X=12) = \frac{1}{36}$

The probability distribution of X is:

X (Sum)	2	3	4	5	6	7	8	9	10	11	12
P(X)	$\frac{1}{36}$	$\frac{2}{36}$	$\frac{3}{36}$	$\frac{4}{36}$	$\frac{5}{36}$	$\frac{6}{36}$	$\frac{5}{36}$	$\frac{4}{36}$	$\frac{3}{36}$	$\frac{2}{36}$	$\frac{1}{36}$

The mean or expectation of a random variable X is given by the formula:

$E(X) = \sum\limits_{i} x_i P(X=x_i)$

Summing the products of each value of X and its probability:

$E(X) = 2 \left(\frac{1}{36}\right) + 3 \left(\frac{2}{36}\right) + 4 \left(\frac{3}{36}\right) + 5 \left(\frac{4}{36}\right) + 6 \left(\frac{5}{36}\right) + 7 \left(\frac{6}{36}\right) + 8 \left(\frac{5}{36}\right) + 9 \left(\frac{4}{36}\right) + 10 \left(\frac{3}{36}\right) + 11 \left(\frac{2}{36}\right) + 12 \left(\frac{1}{36}\right)$

$E(X) = \frac{1}{36} (2 \times 1 + 3 \times 2 + 4 \times 3 + 5 \times 4 + 6 \times 5 + 7 \times 6 + 8 \times 5 + 9 \times 4 + 10 \times 3 + 11 \times 2 + 12 \times 1)$

$E(X) = \frac{1}{36} (2 + 6 + 12 + 20 + 30 + 42 + 40 + 36 + 30 + 22 + 12)$

$E(X) = \frac{252}{36}$

$E(X) = 7$

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Alternate Solution:

Let $X_1$ be the random variable representing the number on the first die, and $X_2$ be the random variable representing the number on the second die.

Then $X = X_1 + X_2$.

The expectation of a sum of random variables is the sum of their expectations:

$E(X) = E(X_1 + X_2) = E(X_1) + E(X_2)$

For a single fair die, the possible values are $\{1, 2, 3, 4, 5, 6\}$, each with probability $\frac{1}{6}$. The expectation of a single die roll is:

$E(X_1) = \sum\limits_{i=1}^6 i P(X_1=i) = 1\left(\frac{1}{6}\right) + 2\left(\frac{1}{6}\right) + 3\left(\frac{1}{6}\right) + 4\left(\frac{1}{6}\right) + 5\left(\frac{1}{6}\right) + 6\left(\frac{1}{6}\right)$

$E(X_1) = \frac{1}{6}(1+2+3+4+5+6) = \frac{21}{6} = \frac{7}{2}$

Similarly, for the second die:

$E(X_2) = \frac{7}{2}$

Therefore, the expectation of the sum is:

$E(X) = E(X_1) + E(X_2) = \frac{7}{2} + \frac{7}{2} = \frac{14}{2} = 7$

The mean or expectation of X is 7.

Given:

An unbiased die is thrown.

X is the random variable representing the number obtained on the die.

To Find:

The variance of X, $Var(X)$.

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Solution:

When an unbiased die is thrown, the possible outcomes are $\{1, 2, 3, 4, 5, 6\}$. Each outcome has a probability of $\frac{1}{6}$.

The possible values of the random variable X are $x = 1, 2, 3, 4, 5, 6$, and the probability of each value is $P(X=x) = \frac{1}{6}$.

The variance of a random variable X is given by the formula:

$Var(X) = E(X^2) - [E(X)]^2$

First, calculate the expected value of X, $E(X)$:

$E(X) = \sum\limits_{x} x P(X=x)$

$E(X) = 1\left(\frac{1}{6}\right) + 2\left(\frac{1}{6}\right) + 3\left(\frac{1}{6}\right) + 4\left(\frac{1}{6}\right) + 5\left(\frac{1}{6}\right) + 6\left(\frac{1}{6}\right)$

$E(X) = \frac{1}{6}(1 + 2 + 3 + 4 + 5 + 6)$

$E(X) = \frac{1}{6}(21) = \frac{21}{6} = \frac{7}{2}$

Next, calculate the expected value of $X^2$, $E(X^2)$:

$E(X^2) = \sum\limits_{x} x^2 P(X=x)$

$E(X^2) = 1^2\left(\frac{1}{6}\right) + 2^2\left(\frac{1}{6}\right) + 3^2\left(\frac{1}{6}\right) + 4^2\left(\frac{1}{6}\right) + 5^2\left(\frac{1}{6}\right) + 6^2\left(\frac{1}{6}\right)$

$E(X^2) = 1\left(\frac{1}{6}\right) + 4\left(\frac{1}{6}\right) + 9\left(\frac{1}{6}\right) + 16\left(\frac{1}{6}\right) + 25\left(\frac{1}{6}\right) + 36\left(\frac{1}{6}\right)$

$E(X^2) = \frac{1}{6}(1 + 4 + 9 + 16 + 25 + 36)$

$E(X^2) = \frac{1}{6}(91) = \frac{91}{6}$

Now, substitute the values of $E(X^2)$ and $E(X)$ into the variance formula:

$Var(X) = E(X^2) - [E(X)]^2$

$Var(X) = \frac{91}{6} - \left(\frac{7}{2}\right)^2$

$Var(X) = \frac{91}{6} - \frac{49}{4}$

Find a common denominator for 6 and 4, which is 12:

$Var(X) = \frac{91 \times 2}{6 \times 2} - \frac{49 \times 3}{4 \times 3}$

$Var(X) = \frac{182}{12} - \frac{147}{12}$

$Var(X) = \frac{182 - 147}{12}$

$Var(X) = \frac{35}{12}$

The variance of the number obtained on a throw of an unbiased die is $\frac{35}{12}$.

Given:

A well-shuffled deck of 52 cards.

Two cards are drawn simultaneously (or successively without replacement).

X is the random variable representing the number of kings drawn.

To Find:

The mean, variance, and standard deviation of X.

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Solution:

A standard deck has 52 cards.

Number of kings in the deck = 4.

Number of non-kings in the deck = $52 - 4 = 48$.

Two cards are drawn without replacement. The total number of ways to draw 2 cards from 52 is $\binom{52}{2}$.

$\binom{52}{2} = \frac{52 \times 51}{2 \times 1} = 26 \times 51 = 1326$

The random variable X represents the number of kings in the two draws. The possible number of kings can be 0, 1, or 2.

The possible values of X are $x = 0, 1, 2$.

We find the probability for each value of X:

$P(X=0)$: Getting 0 kings and 2 non-kings.

Number of ways to choose 0 kings from 4 is $\binom{4}{0} = 1$.

Number of ways to choose 2 non-kings from 48 is $\binom{48}{2}$.

$\binom{48}{2} = \frac{48 \times 47}{2 \times 1} = 24 \times 47 = 1128$

$P(X=0) = \frac{\binom{4}{0} \times \binom{48}{2}}{\binom{52}{2}} = \frac{1 \times 1128}{1326} = \frac{1128}{1326}$

Simplify the fraction by dividing numerator and denominator by 6:

$P(X=0) = \frac{\cancel{1128}^{188}}{\cancel{1326}_{221}}$

Both are divisible by 13:

$P(X=0) = \frac{\cancel{188}^{12}}{\cancel{221}_{13 \times 17}} = \frac{188}{221}$ (Correction: 188 is not divisible by 13. $188/221$ is already simplified).

Let's keep the denominator as 1326 for now and simplify later if needed.

$P(X=0) = \frac{1128}{1326}$

$P(X=1)$: Getting exactly 1 king and 1 non-king.

Number of ways to choose 1 king from 4 is $\binom{4}{1} = 4$.

Number of ways to choose 1 non-king from 48 is $\binom{48}{1} = 48$.

$P(X=1) = \frac{\binom{4}{1} \times \binom{48}{1}}{\binom{52}{2}} = \frac{4 \times 48}{1326} = \frac{192}{1326}$

$P(X=2)$: Getting exactly 2 kings and 0 non-kings.

Number of ways to choose 2 kings from 4 is $\binom{4}{2} = \frac{4 \times 3}{2} = 6$.

Number of ways to choose 0 non-kings from 48 is $\binom{48}{0} = 1$.

$P(X=2) = \frac{\binom{4}{2} \times \binom{48}{0}}{\binom{52}{2}} = \frac{6 \times 1}{1326} = \frac{6}{1326}$

The probability distribution of X is:

X (Number of Kings)	0	1	2
P(X)	$\frac{1128}{1326}$	$\frac{192}{1326}$	$\frac{6}{1326}$

Check: $\frac{1128}{1326} + \frac{192}{1326} + \frac{6}{1326} = \frac{1128 + 192 + 6}{1326} = \frac{1326}{1326} = 1$. The probabilities sum to 1.

Mean $E(X)$:

$E(X) = \sum\limits_{x} x P(X=x)$

$E(X) = 0 \times P(X=0) + 1 \times P(X=1) + 2 \times P(X=2)$

$E(X) = 0 \times \frac{1128}{1326} + 1 \times \frac{192}{1326} + 2 \times \frac{6}{1326}$

$E(X) = 0 + \frac{192}{1326} + \frac{12}{1326}$

$E(X) = \frac{192 + 12}{1326} = \frac{204}{1326}$

Simplify the fraction $\frac{204}{1326}$. Both are divisible by 6:

$E(X) = \frac{\cancel{204}^{34}}{\cancel{1326}_{221}}$

$E(X) = \frac{34}{221}$

Both are divisible by 17:

$E(X) = \frac{\cancel{34}^{2}}{\cancel{221}_{13}}$

$E(X) = \frac{2}{13}$

The mean number of kings is $\frac{2}{13}$.

Variance $Var(X)$:

$Var(X) = E(X^2) - [E(X)]^2$

First, calculate $E(X^2) = \sum\limits_{x} x^2 P(X=x)$:

$E(X^2) = 0^2 \times P(X=0) + 1^2 \times P(X=1) + 2^2 \times P(X=2)$

$E(X^2) = 0 \times \frac{1128}{1326} + 1 \times \frac{192}{1326} + 4 \times \frac{6}{1326}$

$E(X^2) = 0 + \frac{192}{1326} + \frac{24}{1326}$

$E(X^2) = \frac{192 + 24}{1326} = \frac{216}{1326}$

Simplify the fraction $\frac{216}{1326}$. Both are divisible by 6:

$E(X^2) = \frac{\cancel{216}^{36}}{\cancel{1326}_{221}}$

$E(X^2) = \frac{36}{221}$

Now, substitute $E(X^2) = \frac{36}{221}$ and $E(X) = \frac{2}{13}$ into the variance formula:

$Var(X) = \frac{36}{221} - \left(\frac{2}{13}\right)^2$

$Var(X) = \frac{36}{221} - \frac{4}{169}$

Find a common denominator for 221 and 169. Note that $221 = 13 \times 17$ and $169 = 13^2$. The LCM is $13^2 \times 17 = 169 \times 17 = 2873$.

$Var(X) = \frac{36 \times 13}{221 \times 13} - \frac{4 \times 17}{169 \times 17}$

$36 \times 13 = 468$

$4 \times 17 = 68$

$221 \times 13 = 2873$

$169 \times 17 = 2873$

$Var(X) = \frac{468}{2873} - \frac{68}{2873}$

$Var(X) = \frac{468 - 68}{2873} = \frac{400}{2873}$

The variance of the number of kings is $\frac{400}{2873}$.

Standard Deviation $\sigma_X$:

The standard deviation is the square root of the variance.

$\sigma_X = \sqrt{Var(X)} = \sqrt{\frac{400}{2873}}$

$\sigma_X = \frac{\sqrt{400}}{\sqrt{2873}} = \frac{20}{\sqrt{2873}}$

The standard deviation of the number of kings is $\frac{20}{\sqrt{2873}}$.

Conditions for a Valid Probability Distribution:

A function $P(X=x)$ is a probability distribution of a random variable X if it satisfies the following conditions:

1. $0 \leq P(X=x) \leq 1$ for all possible values of x.

2. $\sum P(X=x) = 1$, where the sum is taken over all possible values of x.

---

Let's check each given case:

Case (i):

X	0	1	2
P(X)	0.4	0.4	0.2

1. All probabilities (0.4, 0.4, 0.2) are between 0 and 1.

2. Sum of probabilities = $0.4 + 0.4 + 0.2 = 1.0$.

Both conditions are satisfied.

Therefore, this is a valid probability distribution.

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Case (ii):

X	0	1	2	3	4
P(X)	0.1	0.5	0.2	– 0.1	0.3

1. One of the probabilities, $P(X=3) = -0.1$, is negative.

This violates the condition that $0 \leq P(X=x) \leq 1$.

Therefore, this is not a valid probability distribution.

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Case (iii):

Y	–1	0	1
P(Y)	0.6	0.1	0.2

1. All probabilities (0.6, 0.1, 0.2) are between 0 and 1.

2. Sum of probabilities = $0.6 + 0.1 + 0.2 = 0.9$.

This violates the condition that $\sum P(Y=y) = 1$.

Therefore, this is not a valid probability distribution.

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Case (iv):

Z	3	2	1	0	–1
P(Z)	0.3	0.2	0.4	0.1	0.05

1. All probabilities (0.3, 0.2, 0.4, 0.1, 0.05) are between 0 and 1.

2. Sum of probabilities = $0.3 + 0.2 + 0.4 + 0.1 + 0.05 = 1.05$.

This violates the condition that $\sum P(Z=z) = 1$.

Therefore, this is not a valid probability distribution.

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In summary, the probability distributions that are not valid are (ii), (iii), and (iv) for the reasons stated above.

Given:

An urn contains 5 red balls and 2 black balls.

Total number of balls = $5 + 2 = 7$.

Two balls are drawn randomly from the urn. (Assuming without replacement, as is standard unless stated otherwise).

X represents the number of black balls drawn.

To Find:

The possible values of X.

Whether X is a random variable.

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Solution:

The random variable X represents the number of black balls when two balls are drawn from the urn.

When drawing two balls without replacement from a total of 7 balls (5 Red, 2 Black), we can have the following outcomes regarding the number of black balls:

1. We can draw 0 black balls. This occurs if both drawn balls are red.

The number of red balls is 5. We need to choose 2 red balls.

2. We can draw 1 black ball. This occurs if one drawn ball is black and the other is red.

The number of black balls is 2, and the number of red balls is 5. We need to choose 1 black and 1 red ball.

3. We can draw 2 black balls. This occurs if both drawn balls are black.

The number of black balls is 2. We need to choose 2 black balls. We cannot draw more than 2 black balls since there are only 2 black balls in the urn.

Thus, the possible number of black balls drawn is 0, 1, or 2.

The possible values of the random variable X are $\{0, 1, 2\}$.

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A random variable is a real-valued function defined on the sample space of a random experiment.

In this experiment, the sample space consists of all possible pairs of two balls drawn from the urn (without replacement). Each outcome in the sample space is a pair of balls.

The random variable X assigns a numerical value (the count of black balls) to each outcome in the sample space.

For example:

If the outcome is {Red, Red}, X assigns the value 0.

If the outcome is {Red, Black}, X assigns the value 1.

If the outcome is {Black, Red}, X assigns the value 1.

If the outcome is {Black, Black}, X assigns the value 2.

Since X is a function that maps the outcomes of the random experiment (drawing two balls) to real numbers (0, 1, or 2), X is indeed a random variable.

Given:

A coin is tossed 6 times.

X represents the difference between the number of heads and the number of tails obtained.

To Find:

The possible values of the random variable X.

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Solution:

Let $n_H$ be the number of heads obtained in 6 tosses.

Let $n_T$ be the number of tails obtained in 6 tosses.

The total number of tosses is 6, so the sum of the number of heads and tails is always 6:

$n_H + n_T = 6$

The random variable X is defined as the difference between the number of heads and the number of tails:

$X = n_H - n_T$

From the first equation, we can express $n_T$ in terms of $n_H$: $n_T = 6 - n_H$.

Substitute this into the expression for X:

$X = n_H - (6 - n_H)$

$X = n_H - 6 + n_H$

$X = 2n_H - 6$

The number of heads, $n_H$, in 6 tosses of a coin can be any integer from 0 (no heads, all tails) to 6 (all heads, no tails).

The possible values for $n_H$ are $\{0, 1, 2, 3, 4, 5, 6\}$.

Now we find the corresponding values of X for each possible value of $n_H$:

• If $n_H = 0$, $X = 2(0) - 6 = -6$. ($n_T = 6$)
• If $n_H = 1$, $X = 2(1) - 6 = 2 - 6 = -4$. ($n_T = 5$)
• If $n_H = 2$, $X = 2(2) - 6 = 4 - 6 = -2$. ($n_T = 4$)
• If $n_H = 3$, $X = 2(3) - 6 = 6 - 6 = 0$. ($n_T = 3$)
• If $n_H = 4$, $X = 2(4) - 6 = 8 - 6 = 2$. ($n_T = 2$)
• If $n_H = 5$, $X = 2(5) - 6 = 10 - 6 = 4$. ($n_T = 1$)
• If $n_H = 6$, $X = 2(6) - 6 = 12 - 6 = 6$. ($n_T = 0$)

The possible values of X are the distinct values obtained in the list above.

The possible values of X are $\{-6, -4, -2, 0, 2, 4, 6\}$.

Solution (i): Number of heads in two tosses of a coin

Let X be the random variable representing the number of heads in two tosses of a fair coin.

The sample space for tossing a coin twice is S = {HH, HT, TH, TT}.

The possible values of X are 0, 1, or 2.

$P(\text{Head}) = \frac{1}{2}$

$P(\text{Tail}) = \frac{1}{2}$

Since the tosses are independent:

$P(X=0)$: No heads (TT) = $P(T) \times P(T) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$

$P(X=1)$: Exactly one head (HT or TH) = $P(\text{HT}) + P(\text{TH}) = (P(H) \times P(T)) + (P(T) \times P(H)) = (\frac{1}{2} \times \frac{1}{2}) + (\frac{1}{2} \times \frac{1}{2}) = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$

$P(X=2)$: Two heads (HH) = $P(H) \times P(H) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$

The probability distribution of X is:

X	0	1	2
P(X)	$\frac{1}{4}$	$\frac{1}{2}$	$\frac{1}{4}$

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Solution (ii): Number of tails in the simultaneous tosses of three coins

Let X be the random variable representing the number of tails in the simultaneous tosses of three fair coins.

The sample space for tossing three coins is S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}. The total number of outcomes is 8.

The possible values of X (number of tails) are 0, 1, 2, or 3.

Assuming fair coins, each outcome in S has a probability of $\frac{1}{8}$.

$P(X=0)$: 0 tails (HHH). Number of outcomes = 1. $P(X=0) = \frac{1}{8}$.

$P(X=1)$: 1 tail (HHT, HTH, THH). Number of outcomes = 3. $P(X=1) = \frac{3}{8}$.

$P(X=2)$: 2 tails (HTT, THT, TTH). Number of outcomes = 3. $P(X=2) = \frac{3}{8}$.

$P(X=3)$: 3 tails (TTT). Number of outcomes = 1. $P(X=3) = \frac{1}{8}$.

The probability distribution of X is:

X	0	1	2	3
P(X)	$\frac{1}{8}$	$\frac{3}{8}$	$\frac{3}{8}$	$\frac{1}{8}$

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Solution (iii): Number of heads in four tosses of a coin

Let X be the random variable representing the number of heads in four tosses of a fair coin.

The experiment is tossing a fair coin 4 times independently. This is a binomial experiment with $n=4$ trials and probability of success (getting a head) $p = \frac{1}{2}$.

The number of heads X can take values $k = 0, 1, 2, 3, 4$.

The probability of getting exactly k heads in 4 tosses is given by the binomial probability formula:

$P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$

$P(X=k) = \binom{4}{k} \left(\frac{1}{2}\right)^k \left(\frac{1}{2}\right)^{4-k} = \binom{4}{k} \left(\frac{1}{2}\right)^4 = \binom{4}{k} \frac{1}{16}$

Calculate the probabilities for each possible value of k:

$P(X=0) = \binom{4}{0} \frac{1}{16} = 1 \times \frac{1}{16} = \frac{1}{16}$

$P(X=1) = \binom{4}{1} \frac{1}{16} = 4 \times \frac{1}{16} = \frac{4}{16} = \frac{1}{4}$

$P(X=2) = \binom{4}{2} \frac{1}{16} = 6 \times \frac{1}{16} = \frac{6}{16} = \frac{3}{8}$

$P(X=3) = \binom{4}{3} \frac{1}{16} = 4 \times \frac{1}{16} = \frac{4}{16} = \frac{1}{4}$

$P(X=4) = \binom{4}{4} \frac{1}{16} = 1 \times \frac{1}{16} = \frac{1}{16}$

The probability distribution of X is:

X	0	1	2	3	4
P(X)	$\frac{1}{16}$	$\frac{4}{16}$ or $\frac{1}{4}$	$\frac{6}{16}$ or $\frac{3}{8}$	$\frac{4}{16}$ or $\frac{1}{4}$	$\frac{1}{16}$

The experiment consists of tossing a single die two times. Each toss is an independent trial.

The possible outcomes for a single toss of a fair die are $\{1, 2, 3, 4, 5, 6\}$. The total number of outcomes is 6, and each outcome is equally likely with probability $\frac{1}{6}$.

Let X be the random variable representing the number of successes in two tosses. The number of trials is $n=2$. The possible number of successes is $x = 0, 1, 2$.

We can use the binomial probability formula $P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$, where $p$ is the probability of success in a single trial.

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Solution (i): Success is 'number greater than 4'

In a single toss of a die, a success is getting a number greater than 4. The outcomes greater than 4 are $\{5, 6\}$.

The probability of success in a single toss is $p = P(\text{number > 4}) = \frac{\text{Number of outcomes > 4}}{\text{Total outcomes}} = \frac{2}{6} = \frac{1}{3}$.

The probability of failure in a single toss is $q = 1 - p = 1 - \frac{1}{3} = \frac{2}{3}$.

The number of trials is $n=2$. The possible values of X are $x = 0, 1, 2$.

We calculate the probability for each value of X:

$P(X=0)$: 0 successes in 2 tosses.

$P(X=0) = \binom{2}{0} \left(\frac{1}{3}\right)^0 \left(\frac{2}{3}\right)^{2-0} = 1 \times 1 \times \left(\frac{2}{3}\right)^2 = \frac{4}{9}$

$P(X=1)$: 1 success in 2 tosses.

$P(X=1) = \binom{2}{1} \left(\frac{1}{3}\right)^1 \left(\frac{2}{3}\right)^{2-1} = 2 \times \left(\frac{1}{3}\right) \times \left(\frac{2}{3}\right) = \frac{4}{9}$

$P(X=2)$: 2 successes in 2 tosses.

$P(X=2) = \binom{2}{2} \left(\frac{1}{3}\right)^2 \left(\frac{2}{3}\right)^{2-2} = 1 \times \left(\frac{1}{3}\right)^2 \times 1 = \frac{1}{9}$

The probability distribution of the number of successes in this case is:

X (Number of Successes)	0	1	2
P(X)	$\frac{4}{9}$	$\frac{4}{9}$	$\frac{1}{9}$

Check: $\frac{4}{9} + \frac{4}{9} + \frac{1}{9} = \frac{9}{9} = 1$.

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Solution (ii): Success is 'six appears on at least one die'

Given the phrasing "in two tosses of a die", we interpret this as two sequential tosses of a single die. A "success" is defined as getting a six on a single toss.

In a single toss of a die, a success is getting a six ({6}).

The probability of success in a single toss is $p = P(\text{getting a six}) = \frac{\text{Number of outcomes = 6}}{\text{Total outcomes}} = \frac{1}{6}$.

The probability of failure in a single toss is $q = 1 - p = 1 - \frac{1}{6} = \frac{5}{6}$.

The number of trials is $n=2$. The possible values of X (number of successes, i.e., number of sixes, in two tosses) are $x = 0, 1, 2$.

We calculate the probability for each value of X:

$P(X=0)$: 0 successes (no sixes) in 2 tosses.

$P(X=0) = \binom{2}{0} \left(\frac{1}{6}\right)^0 \left(\frac{5}{6}\right)^{2-0} = 1 \times 1 \times \left(\frac{5}{6}\right)^2 = \frac{25}{36}$

$P(X=1)$: 1 success (exactly one six) in 2 tosses.

$P(X=1) = \binom{2}{1} \left(\frac{1}{6}\right)^1 \left(\frac{5}{6}\right)^{2-1} = 2 \times \left(\frac{1}{6}\right) \times \left(\frac{5}{6}\right) = \frac{10}{36} = \frac{5}{18}$

$P(X=2)$: 2 successes (two sixes) in 2 tosses.

$P(X=2) = \binom{2}{2} \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^{2-2} = 1 \times \left(\frac{1}{6}\right)^2 \times 1 = \frac{1}{36}$

The probability distribution of the number of successes in this case is:

X (Number of Successes)	0	1	2
P(X)	$\frac{25}{36}$	$\frac{10}{36}$ or $\frac{5}{18}$	$\frac{1}{36}$

Check: $\frac{25}{36} + \frac{10}{36} + \frac{1}{36} = \frac{36}{36} = 1$.

Note: If the phrasing was intended to mean "two throws of a pair of dice", where a success in a single throw of a pair is getting at least one six, the probability of success would be $p = P(\text{at least one 6}) = 1 - P(\text{no 6s}) = 1 - (\frac{5}{6} \times \frac{5}{6}) = 1 - \frac{25}{36} = \frac{11}{36}$. Then the probabilities for $X=0, 1, 2$ would be calculated using $n=2$ and $p=\frac{11}{36}$. However, the phrase "two tosses of a die" implies a single die being tossed twice, so the solution above follows that interpretation.

Given:

Total number of bulbs in the lot = 30.

Number of defective bulbs = 6.

Number of non-defective bulbs = $30 - 6 = 24$.

A sample of 4 bulbs is drawn at random with replacement.

Let X be the random variable representing the number of defective bulbs in the sample.

To Find:

The probability distribution of X.

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Solution:

Since the bulbs are drawn with replacement, each draw is an independent trial.

Consider a single draw. A "success" is defined as drawing a defective bulb.

The probability of drawing a defective bulb in a single draw is:

$p = P(\text{Defective}) = \frac{\text{Number of defective bulbs}}{\text{Total number of bulbs}} = \frac{6}{30} = \frac{1}{5}$

The probability of drawing a non-defective bulb in a single draw is:

$q = P(\text{Non-defective}) = 1 - p = 1 - \frac{1}{5} = \frac{4}{5}$

We are drawing a sample of 4 bulbs with replacement, so the number of trials is $n=4$.

The random variable X, representing the number of defective bulbs in 4 draws, follows a binomial distribution with parameters $n=4$ and $p = \frac{1}{5}$.

The possible values of X are $x = 0, 1, 2, 3, 4$.

The probability of getting exactly x defective bulbs in 4 draws is given by the binomial probability formula:

$P(X=x) = \binom{n}{x} p^x q^{n-x}$

$P(X=x) = \binom{4}{x} \left(\frac{1}{5}\right)^x \left(\frac{4}{5}\right)^{4-x}$, for $x = 0, 1, 2, 3, 4$.

We calculate the probability for each possible value of X:

$P(X=0) = \binom{4}{0} \left(\frac{1}{5}\right)^0 \left(\frac{4}{5}\right)^{4-0} = 1 \times 1 \times \left(\frac{4}{5}\right)^4 = \frac{4^4}{5^4} = \frac{256}{625}$

$P(X=1) = \binom{4}{1} \left(\frac{1}{5}\right)^1 \left(\frac{4}{5}\right)^{4-1} = 4 \times \frac{1}{5} \times \left(\frac{4}{5}\right)^3 = 4 \times \frac{1}{5} \times \frac{64}{125} = \frac{4 \times 1 \times 64}{5 \times 125} = \frac{256}{625}$

$P(X=2) = \binom{4}{2} \left(\frac{1}{5}\right)^2 \left(\frac{4}{5}\right)^{4-2} = 6 \times \left(\frac{1}{5}\right)^2 \times \left(\frac{4}{5}\right)^2 = 6 \times \frac{1}{25} \times \frac{16}{25} = \frac{6 \times 1 \times 16}{25 \times 25} = \frac{96}{625}$

$P(X=3) = \binom{4}{3} \left(\frac{1}{5}\right)^3 \left(\frac{4}{5}\right)^{4-3} = 4 \times \left(\frac{1}{5}\right)^3 \times \left(\frac{4}{5}\right)^1 = 4 \times \frac{1}{125} \times \frac{4}{5} = \frac{4 \times 1 \times 4}{125 \times 5} = \frac{16}{625}$

$P(X=4) = \binom{4}{4} \left(\frac{1}{5}\right)^4 \left(\frac{4}{5}\right)^{4-4} = 1 \times \left(\frac{1}{5}\right)^4 \times \left(\frac{4}{5}\right)^0 = 1 \times \frac{1}{625} \times 1 = \frac{1}{625}$

The probability distribution of the number of defective bulbs is:

X (Number of Defective Bulbs)	0	1	2	3	4
P(X)	$\frac{256}{625}$	$\frac{256}{625}$	$\frac{96}{625}$	$\frac{16}{625}$	$\frac{1}{625}$

Check: $\frac{256+256+96+16+1}{625} = \frac{625}{625} = 1$.

Given:

A biased coin where Head (H) is 3 times as likely as Tail (T).

Let $P(T) = p$. Then $P(H) = 3p$.

The sum of probabilities must be 1:

$P(H) + P(T) = 1$

$3p + p = 1$

$4p = 1$

$p = \frac{1}{4}$

So, the probability of getting a tail in a single toss is $P(T) = \frac{1}{4}$.

The probability of getting a head in a single toss is $P(H) = 3 \times \frac{1}{4} = \frac{3}{4}$.

The coin is tossed twice.

Let X be the random variable representing the number of tails in two tosses.

To Find:

The probability distribution of X.

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Solution:

The experiment consists of two independent tosses of the biased coin.

The number of tails X can take values $x = 0, 1, 2$.

We can find the probabilities for each value of X using the probabilities of H and T for a single toss.

$P(X=0)$: Getting 0 tails (HH).

$P(X=0) = P(H \text{ on 1st}) \times P(H \text{ on 2nd})$ (Independent tosses)

$P(X=0) = P(H) \times P(H) = \frac{3}{4} \times \frac{3}{4} = \frac{9}{16}$

$P(X=1)$: Getting exactly 1 tail (HT or TH).

$P(X=1) = P(H \text{ on 1st and T on 2nd}) + P(T \text{ on 1st and H on 2nd})$ (Mutually exclusive outcomes)

$P(X=1) = [P(H) \times P(T)] + [P(T) \times P(H)]$ (Independent tosses)

$P(X=1) = \left(\frac{3}{4} \times \frac{1}{4}\right) + \left(\frac{1}{4} \times \frac{3}{4}\right) = \frac{3}{16} + \frac{3}{16} = \frac{6}{16} = \frac{3}{8}$

$P(X=2)$: Getting 2 tails (TT).

$P(X=2) = P(T \text{ on 1st}) \times P(T \text{ on 2nd})$ (Independent tosses)

$P(X=2) = P(T) \times P(T) = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16}$

The probability distribution of the number of tails (X) is:

X (Number of Tails)	0	1	2
P(X)	$\frac{9}{16}$	$\frac{6}{16}$ or $\frac{3}{8}$	$\frac{1}{16}$

Check: $\frac{9}{16} + \frac{6}{16} + \frac{1}{16} = \frac{16}{16} = 1$.

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Alternate Solution (using Binomial Distribution):

This is a binomial experiment with $n=2$ trials.

A "success" is defined as getting a tail in a single toss. The probability of success is $p = P(T) = \frac{1}{4}$.

The number of tails X follows a binomial distribution with parameters $n=2$ and $p = \frac{1}{4}$.

$P(X=x) = \binom{n}{x} p^x (1-p)^{n-x} = \binom{2}{x} \left(\frac{1}{4}\right)^x \left(\frac{3}{4}\right)^{2-x}$, for $x = 0, 1, 2$.

$P(X=0) = \binom{2}{0} \left(\frac{1}{4}\right)^0 \left(\frac{3}{4}\right)^2 = 1 \times 1 \times \frac{9}{16} = \frac{9}{16}$

$P(X=1) = \binom{2}{1} \left(\frac{1}{4}\right)^1 \left(\frac{3}{4}\right)^1 = 2 \times \frac{1}{4} \times \frac{3}{4} = \frac{6}{16} = \frac{3}{8}$

$P(X=2) = \binom{2}{2} \left(\frac{1}{4}\right)^2 \left(\frac{3}{4}\right)^0 = 1 \times \frac{1}{16} \times 1 = \frac{1}{16}$

This gives the same probability distribution as obtained earlier.

Given:

The probability distribution of the random variable X is given by:

X	0	1	2	3	4	5	6	7
P(X)	0	k	2k	2k	3k	$k^2$	$2k^2$	$7k^2+k$

To Determine:

(i) The value of k.

(ii) $P(X < 3)$.

(iii) $P(X > 6)$.

(iv) $P(0 < X < 3)$.

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Solution (i): Finding the value of k

For a valid probability distribution, two conditions must be met:

1. $P(X=x) \geq 0$ for all values of x.

2. $\sum P(X=x) = 1$ over all possible values of x.

Let's use the second condition first:

$P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6) + P(X=7) = 1$

$0 + k + 2k + 2k + 3k + k^2 + 2k^2 + (7k^2+k) = 1$

Combine like terms:

$(k + 2k + 2k + 3k + k) + (k^2 + 2k^2 + 7k^2) = 1$

$9k + 10k^2 = 1$

Rearrange the equation into a quadratic form:

$10k^2 + 9k - 1 = 0$

We can solve this quadratic equation using factorization or the quadratic formula. Let's factorize:

We need two numbers that multiply to $10 \times (-1) = -10$ and add up to 9. These numbers are 10 and -1.

$10k^2 + 10k - k - 1 = 0$

$10k(k + 1) - 1(k + 1) = 0$

$(10k - 1)(k + 1) = 0$

This gives two possible values for k:

$10k - 1 = 0 \implies 10k = 1 \implies k = \frac{1}{10} = 0.1$

$k + 1 = 0 \implies k = -1$

Now, we must check the first condition: $P(X=x) \geq 0$ for all x.

If $k = -1$, then $P(X=1) = k = -1$, which is a negative probability. Thus, $k = -1$ is not a valid value for k.

If $k = 0.1$, let's check the probabilities:

$P(X=0) = 0 \geq 0$

$P(X=1) = k = 0.1 \geq 0$

$P(X=2) = 2k = 2(0.1) = 0.2 \geq 0$

$P(X=3) = 2k = 2(0.1) = 0.2 \geq 0$

$P(X=4) = 3k = 3(0.1) = 0.3 \geq 0$

$P(X=5) = k^2 = (0.1)^2 = 0.01 \geq 0$

$P(X=6) = 2k^2 = 2(0.1)^2 = 2(0.01) = 0.02 \geq 0$

$P(X=7) = 7k^2 + k = 7(0.1)^2 + 0.1 = 7(0.01) + 0.1 = 0.07 + 0.1 = 0.17 \geq 0$

All probabilities are non-negative when $k = 0.1$.

Therefore, the correct value of k is 0.1.

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Solution (ii): $P(X < 3)$

$P(X < 3)$ means the probability that X takes values less than 3. The possible values less than 3 are $X=0, 1, 2$.

$P(X < 3) = P(X=0) + P(X=1) + P(X=2)$

Substitute the probabilities using $k = 0.1$:

$P(X < 3) = 0 + k + 2k$

$P(X < 3) = 0 + 0.1 + 2(0.1)$

$P(X < 3) = 0.1 + 0.2 = 0.3$

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Solution (iii): $P(X > 6)$

$P(X > 6)$ means the probability that X takes values greater than 6. The only possible value greater than 6 is $X=7$.

$P(X > 6) = P(X=7)$

Substitute the probability using $k = 0.1$:

$P(X > 6) = 7k^2 + k$

$P(X > 6) = 7(0.1)^2 + 0.1$

$P(X > 6) = 7(0.01) + 0.1$

$P(X > 6) = 0.07 + 0.1 = 0.17$

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Solution (iv): $P(0 < X < 3)$

$P(0 < X < 3)$ means the probability that X takes values strictly between 0 and 3. The possible integer values are $X=1, 2$.

$P(0 < X < 3) = P(X=1) + P(X=2)$

Substitute the probabilities using $k = 0.1$:

$P(0 < X < 3) = k + 2k$

$P(0 < X < 3) = 0.1 + 2(0.1)$

$P(0 < X < 3) = 0.1 + 0.2 = 0.3$

Given:

The probability distribution of the random variable X is given by:

$P(X = x) = \begin{cases} k, & \text{if } x = 0 \\ 2k, & \text{if } x = 1 \\ 3k, & \text{if } x = 2 \\ 0, & \text{otherwise} \end{cases}$

The possible values for X with non-zero probability are $x = 0, 1, 2$.

To Determine:

(a) The value of k.

(b) $P(X < 2)$, $P(X \leq 2)$, and $P(X \geq 2)$.

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Solution (a): Finding the value of k

For a valid probability distribution, the sum of the probabilities for all possible values of the random variable must equal 1.

So, we have:

$\sum\limits_{x} P(X=x) = 1$

$P(X=0) + P(X=1) + P(X=2) = 1$

Substitute the given probability function for each value of x:

$k + 2k + 3k = 1$

Combine the terms involving k:

$6k = 1$

Divide by 6 to find k:

$k = \frac{1}{6}$

We also need to check if the probabilities are non-negative for this value of k.

$P(X=0) = k = \frac{1}{6} \geq 0$

$P(X=1) = 2k = 2 \times \frac{1}{6} = \frac{2}{6} = \frac{1}{3} \geq 0$

$P(X=2) = 3k = 3 \times \frac{1}{6} = \frac{3}{6} = \frac{1}{2} \geq 0$

All probabilities are non-negative when $k = \frac{1}{6}$.

Therefore, the value of k is $\frac{1}{6}$.

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Solution (b): Calculating Probabilities

Now that we have the value of k, we can find the probabilities for each possible value of X:

$P(X=0) = k = \frac{1}{6}$

$P(X=1) = 2k = 2 \times \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$

$P(X=2) = 3k = 3 \times \frac{1}{6} = \frac{3}{6} = \frac{1}{2}$

Probability $P(X < 2)$:

$P(X < 2)$ means the probability that X takes values less than 2. The possible values less than 2 are $X=0$ and $X=1$.

$P(X < 2) = P(X=0) + P(X=1)$

$P(X < 2) = \frac{1}{6} + \frac{1}{3}$

To add these fractions, find a common denominator, which is 6:

$P(X < 2) = \frac{1}{6} + \frac{1 \times 2}{3 \times 2} = \frac{1}{6} + \frac{2}{6} = \frac{1 + 2}{6} = \frac{3}{6} = \frac{1}{2}$

So, $P(X < 2) = \frac{1}{2}$.

Probability $P(X \leq 2)$:

$P(X \leq 2)$ means the probability that X takes values less than or equal to 2. The possible values are $X=0, 1, 2$.

$P(X \leq 2) = P(X=0) + P(X=1) + P(X=2)$

$P(X \leq 2) = \frac{1}{6} + \frac{1}{3} + \frac{1}{2}$

Find a common denominator, which is 6:

$P(X \leq 2) = \frac{1}{6} + \frac{2}{6} + \frac{3}{6} = \frac{1 + 2 + 3}{6} = \frac{6}{6} = 1$

So, $P(X \leq 2) = 1$.

Probability $P(X \geq 2)$:

$P(X \geq 2)$ means the probability that X takes values greater than or equal to 2. The only possible value is $X=2$.

$P(X \geq 2) = P(X=2)$

$P(X \geq 2) = \frac{1}{2}$

Alternatively, using the complement rule:

$P(X \geq 2) = 1 - P(X < 2)$

$P(X \geq 2) = 1 - \frac{1}{2} = \frac{1}{2}$

So, $P(X \geq 2) = \frac{1}{2}$.

Given:

A fair coin is tossed three times.

X is the random variable representing the number of heads obtained.

To Find:

The mean (expectation) of X, $E(X)$.

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Solution:

When a fair coin is tossed, the probability of getting a head is $P(H) = \frac{1}{2}$, and the probability of getting a tail is $P(T) = \frac{1}{2}$.

The experiment consists of three independent tosses. The random variable X represents the number of heads in these three tosses. The possible values of X are 0, 1, 2, or 3.

We can find the probability distribution of X:

• $P(X=0)$: 0 heads (TTT) $= P(T) \times P(T) \times P(T) = \left(\frac{1}{2}\right)^3 = \frac{1}{8}$
• $P(X=1)$: 1 head (HTT, THT, TTH) $= P(H)P(T)P(T) + P(T)P(H)P(T) + P(T)P(T)P(H) = 3 \times \left(\frac{1}{2}\right)^3 = \frac{3}{8}$
• $P(X=2)$: 2 heads (HHT, HTH, THH) $= P(H)P(H)P(T) + P(H)P(T)P(H) + P(T)P(H)P(H) = 3 \times \left(\frac{1}{2}\right)^3 = \frac{3}{8}$
• $P(X=3)$: 3 heads (HHH) $= P(H) \times P(H) \times P(H) = \left(\frac{1}{2}\right)^3 = \frac{1}{8}$

The probability distribution of X is:

X (Number of Heads)	0	1	2	3
P(X)	$\frac{1}{8}$	$\frac{3}{8}$	$\frac{3}{8}$	$\frac{1}{8}$

The mean or expectation of X is given by the formula $E(X) = \sum\limits_{x} x P(X=x)$.

$E(X) = 0 \times P(X=0) + 1 \times P(X=1) + 2 \times P(X=2) + 3 \times P(X=3)$

$E(X) = 0 \times \frac{1}{8} + 1 \times \frac{3}{8} + 2 \times \frac{3}{8} + 3 \times \frac{1}{8}$

$E(X) = 0 + \frac{3}{8} + \frac{6}{8} + \frac{3}{8}$

$E(X) = \frac{3 + 6 + 3}{8} = \frac{12}{8} = \frac{3}{2}$

The mean number of heads is $\frac{3}{2}$ or 1.5.

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Alternate Solution (using Binomial Distribution properties):

The number of heads in three tosses of a fair coin follows a binomial distribution with $n=3$ (number of trials) and $p = P(\text{Head}) = \frac{1}{2}$ (probability of success in a single trial).

For a binomial distribution $B(n, p)$, the mean (expectation) is given by the formula:

$E(X) = np$

Substitute the values $n=3$ and $p = \frac{1}{2}$:

$E(X) = 3 \times \frac{1}{2}$

$E(X) = \frac{3}{2}$

The mean number of heads is $\frac{3}{2}$ or 1.5.

Given:

Two fair dice are thrown simultaneously.

X denotes the number of sixes obtained.

To Find:

The expectation of X, $E(X)$.

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Solution:

When a single fair die is thrown, the possible outcomes are $\{1, 2, 3, 4, 5, 6\}$. The probability of getting a six is $P(\text{getting a 6}) = \frac{1}{6}$. The probability of not getting a six is $P(\text{not getting a 6}) = \frac{5}{6}$.

The experiment consists of two independent trials (the two dice throws). The random variable X represents the number of sixes obtained. The possible values of X are 0, 1, or 2.

We can find the probability distribution of X:

• $P(X=0)$: 0 sixes. This means the first die is not a six AND the second die is not a six.

  $P(X=0) = P(\text{not 6 on 1st}) \times P(\text{not 6 on 2nd})$ (Independent throws)

  $P(X=0) = \frac{5}{6} \times \frac{5}{6} = \frac{25}{36}$

• $P(X=1)$: Exactly 1 six. This means (6 on 1st AND not 6 on 2nd) OR (not 6 on 1st AND 6 on 2nd).

  $P(X=1) = P(6 \cap \text{not 6}') + P(\text{not 6}' \cap 6)$ (Mutually exclusive outcomes)

  $P(X=1) = [P(6) \times P(\text{not 6}')] + [P(\text{not 6}') \times P(6)]$ (Independent throws)

  $P(X=1) = \left(\frac{1}{6} \times \frac{5}{6}\right) + \left(\frac{5}{6} \times \frac{1}{6}\right) = \frac{5}{36} + \frac{5}{36} = \frac{10}{36} = \frac{5}{18}$

• $P(X=2)$: 2 sixes. This means the first die is a six AND the second die is a six.

  $P(X=2) = P(6 \text{ on 1st}) \times P(6 \text{ on 2nd})$ (Independent throws)

  $P(X=2) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$

The probability distribution of X is:

X (Number of Sixes)	0	1	2
P(X)	$\frac{25}{36}$	$\frac{10}{36}$ or $\frac{5}{18}$	$\frac{1}{36}$

The expectation of X is given by the formula $E(X) = \sum\limits_{x} x P(X=x)$.

$E(X) = 0 \times P(X=0) + 1 \times P(X=1) + 2 \times P(X=2)$

$E(X) = 0 \times \frac{25}{36} + 1 \times \frac{10}{36} + 2 \times \frac{1}{36}$

$E(X) = 0 + \frac{10}{36} + \frac{2}{36}$

$E(X) = \frac{10 + 2}{36} = \frac{12}{36} = \frac{1}{3}$

The expectation of the number of sixes is $\frac{1}{3}$.

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Alternate Solution (using linearity of expectation):

Let $I_1$ be an indicator random variable that takes value 1 if the first die is a six, and 0 otherwise.

Let $I_2$ be an indicator random variable that takes value 1 if the second die is a six, and 0 otherwise.

The number of sixes X is the sum of these indicator variables: $X = I_1 + I_2$.

The expectation of an indicator variable is the probability of the event it indicates:

$E(I_1) = P(\text{getting a 6 on 1st die}) = \frac{1}{6}$

$E(I_2) = P(\text{getting a 6 on 2nd die}) = \frac{1}{6}$

Using the linearity of expectation, $E(X) = E(I_1 + I_2) = E(I_1) + E(I_2)$.

$E(X) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$

The expectation of the number of sixes is $\frac{1}{3}$.

This problem can also be viewed as a binomial distribution scenario with $n=2$ trials (two dice), where success is getting a six ($p = \frac{1}{6}$). For a binomial distribution $B(n, p)$, the mean is $E(X) = np = 2 \times \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$.

Given:

Two numbers are selected at random without replacement from the first six positive integers: $\{1, 2, 3, 4, 5, 6\}$.

X is the random variable representing the larger of the two numbers obtained.

To Find:

The expectation of X, $E(X)$.

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Solution:

The total number of ways to select two numbers from the set $\{1, 2, 3, 4, 5, 6\}$ without replacement is the number of combinations of 6 items taken 2 at a time, denoted by $\binom{6}{2}$.

Total number of outcomes = $\binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15$

Let the two selected numbers be $n_1$ and $n_2$. The random variable X is the larger of the two, i.e., $X = \max(n_1, n_2)$.

The possible values that X can take are the integers from the second smallest possible value (when {1, 2} are chosen, X=2) up to the largest possible value (when {5, 6} or {any number, 6} are chosen, X=6).

The possible values of X are $\{2, 3, 4, 5, 6\}$.

We need to find the probability $P(X=x)$ for each possible value x.

For X to take the value x, one of the selected numbers must be x, and the other number must be less than x, chosen from the set $\{1, 2, ..., x-1\}$.

The number of ways to select a pair of numbers where the larger number is x is the number of ways to choose x (1 way) times the number of ways to choose one number from the $x-1$ numbers smaller than x ($\binom{x-1}{1}$).

Number of pairs where $\max(n_1, n_2) = x$ is $\binom{1}{1} \times \binom{x-1}{1} = 1 \times (x-1) = x-1$.

The probability $P(X=x)$ is the number of such pairs divided by the total number of possible pairs (15).

$P(X=x) = \frac{x-1}{15}$ for $x = 2, 3, 4, 5, 6$.

Let's calculate the probabilities for each possible value of X:

$P(X=2) = \frac{2-1}{15} = \frac{1}{15}$

$P(X=3) = \frac{3-1}{15} = \frac{2}{15}$

$P(X=4) = \frac{4-1}{15} = \frac{3}{15}$

$P(X=5) = \frac{5-1}{15} = \frac{4}{15}$

$P(X=6) = \frac{6-1}{15} = \frac{5}{15}$

The probability distribution of X is:

X (Larger Number)	2	3	4	5	6
P(X)	$\frac{1}{15}$	$\frac{2}{15}$	$\frac{3}{15}$	$\frac{4}{15}$	$\frac{5}{15}$

Check: $\frac{1}{15} + \frac{2}{15} + \frac{3}{15} + \frac{4}{15} + \frac{5}{15} = \frac{1+2+3+4+5}{15} = \frac{15}{15} = 1$.

The expectation of X, $E(X)$, is given by the formula:

$E(X) = \sum\limits_{x} x P(X=x)$

$E(X) = 2 \times P(X=2) + 3 \times P(X=3) + 4 \times P(X=4) + 5 \times P(X=5) + 6 \times P(X=6)$

$E(X) = 2 \times \frac{1}{15} + 3 \times \frac{2}{15} + 4 \times \frac{3}{15} + 5 \times \frac{4}{15} + 6 \times \frac{5}{15}$

$E(X) = \frac{2}{15} + \frac{6}{15} + \frac{12}{15} + \frac{20}{15} + \frac{30}{15}$

$E(X) = \frac{2 + 6 + 12 + 20 + 30}{15}$

$E(X) = \frac{70}{15}$

Simplify the fraction by dividing the numerator and denominator by 5:

$E(X) = \frac{\cancel{70}^{14}}{\cancel{15}_{3}}$

$E(X) = \frac{14}{3}$

The expectation of X is $\frac{14}{3}$.

Given:

Two fair dice are rolled.

X is the random variable representing the sum of the numbers obtained on the two dice.

To Find:

The variance ($Var(X)$) and standard deviation ($\sigma_X$) of X.

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Solution:

When a pair of fair dice is rolled, the total number of possible outcomes is $6 \times 6 = 36$. Each outcome is equally likely with probability $\frac{1}{36}$.

The random variable X represents the sum of the numbers on the two dice. The possible values of X are the integers from $1+1=2$ to $6+6=12$.

The possible values of X are $\{2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\}$.

The probability distribution of X is determined by the number of ways each sum can be obtained:

X (Sum)	Outcomes	Number of Outcomes	P(X)
2	(1,1)	1	$\frac{1}{36}$
3	(1,2), (2,1)	2	$\frac{2}{36}$
4	(1,3), (2,2), (3,1)	3	$\frac{3}{36}$
5	(1,4), (2,3), (3,2), (4,1)	4	$\frac{4}{36}$
6	(1,5), (2,4), (3,3), (4,2), (5,1)	5	$\frac{5}{36}$
7	(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)	6	$\frac{6}{36}$
8	(2,6), (3,5), (4,4), (5,3), (6,2)	5	$\frac{5}{36}$
9	(3,6), (4,5), (5,4), (6,3)	4	$\frac{4}{36}$
10	(4,6), (5,5), (6,4)	3	$\frac{3}{36}$
11	(5,6), (6,5)	2	$\frac{2}{36}$
12	(6,6)	1	$\frac{1}{36}$

First, calculate the mean (expectation) of X, $E(X)$. As calculated in Example 27:

$E(X) = \sum\limits_{x} x P(X=x)$

$E(X) = 2\left(\frac{1}{36}\right) + 3\left(\frac{2}{36}\right) + 4\left(\frac{3}{36}\right) + 5\left(\frac{4}{36}\right) + 6\left(\frac{5}{36}\right) + 7\left(\frac{6}{36}\right) + 8\left(\frac{5}{36}\right) + 9\left(\frac{4}{36}\right) + 10\left(\frac{3}{36}\right) + 11\left(\frac{2}{36}\right) + 12\left(\frac{1}{36}\right)$

$E(X) = \frac{1}{36}(2 + 6 + 12 + 20 + 30 + 42 + 40 + 36 + 30 + 22 + 12) = \frac{252}{36} = 7$

The mean is $E(X) = 7$.

The variance of X is given by the formula:

$Var(X) = E(X^2) - [E(X)]^2$

Calculate $E(X^2) = \sum\limits_{x} x^2 P(X=x)$:

$E(X^2) = 2^2\left(\frac{1}{36}\right) + 3^2\left(\frac{2}{36}\right) + 4^2\left(\frac{3}{36}\right) + 5^2\left(\frac{4}{36}\right) + 6^2\left(\frac{5}{36}\right) + 7^2\left(\frac{6}{36}\right) + 8^2\left(\frac{5}{36}\right) + 9^2\left(\frac{4}{36}\right) + 10^2\left(\frac{3}{36}\right) + 11^2\left(\frac{2}{36}\right) + 12^2\left(\frac{1}{36}\right)$

$E(X^2) = \frac{1}{36}(4 \times 1 + 9 \times 2 + 16 \times 3 + 25 \times 4 + 36 \times 5 + 49 \times 6 + 64 \times 5 + 81 \times 4 + 100 \times 3 + 121 \times 2 + 144 \times 1)$

$E(X^2) = \frac{1}{36}(4 + 18 + 48 + 100 + 180 + 294 + 320 + 324 + 300 + 242 + 144)$

$E(X^2) = \frac{1974}{36}$

Simplify $\frac{1974}{36}$ by dividing by 6:

$E(X^2) = \frac{\cancel{1974}^{329}}{\cancel{36}_{6}} = \frac{329}{6}$

Now, calculate the variance:

$Var(X) = E(X^2) - [E(X)]^2$

$Var(X) = \frac{329}{6} - (7)^2$

$Var(X) = \frac{329}{6} - 49$

$Var(X) = \frac{329}{6} - \frac{49 \times 6}{6} = \frac{329}{6} - \frac{294}{6}$

$Var(X) = \frac{35}{6}$

The variance is $Var(X) = \frac{35}{6}$.

The standard deviation is the square root of the variance:

$\sigma_X = \sqrt{Var(X)} = \sqrt{\frac{35}{6}}$

The standard deviation is $\sigma_X = \sqrt{\frac{35}{6}}$.

Given:

The ages of 15 students are: 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19, 20.

Total number of students = 15.

One student is selected at random, and X is the age of the selected student.

Each student has the same chance of being chosen, so the probability of selecting any particular student is $\frac{1}{15}$.

To Find:

The probability distribution of X.

The mean, variance, and standard deviation of X.

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Solution:

First, let's identify the distinct ages and their frequencies in the class:

• Age 14: Appears 2 times.
• Age 15: Appears 1 time.
• Age 16: Appears 2 times.
• Age 17: Appears 3 times.
• Age 18: Appears 1 time.
• Age 19: Appears 2 times.
• Age 20: Appears 3 times.
• Age 21: Appears 1 time.

Check the total frequency: $2 + 1 + 2 + 3 + 1 + 2 + 3 + 1 = 15$. This matches the total number of students.

The possible values of the random variable X (the age) are the distinct ages in the class: $\{14, 15, 16, 17, 18, 19, 20, 21\}$.

The probability of selecting a student of a particular age is the frequency of that age divided by the total number of students (15).

Probability Distribution of X:

$P(X=14) = \frac{2}{15}$

$P(X=15) = \frac{1}{15}$

$P(X=16) = \frac{2}{15}$

$P(X=17) = \frac{3}{15} = \frac{1}{5}$

$P(X=18) = \frac{1}{15}$

$P(X=19) = \frac{2}{15}$

$P(X=20) = \frac{3}{15} = \frac{1}{5}$

$P(X=21) = \frac{1}{15}$

The probability distribution is:

X (Age)	14	15	16	17	18	19	20	21
P(X)	$\frac{2}{15}$	$\frac{1}{15}$	$\frac{2}{15}$	$\frac{3}{15}$	$\frac{1}{15}$	$\frac{2}{15}$	$\frac{3}{15}$	$\frac{1}{15}$

Check sum of probabilities: $\frac{2+1+2+3+1+2+3+1}{15} = \frac{15}{15} = 1$.

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Mean $E(X)$:

$E(X) = \sum\limits_{x} x P(X=x)$

$E(X) = 14\left(\frac{2}{15}\right) + 15\left(\frac{1}{15}\right) + 16\left(\frac{2}{15}\right) + 17\left(\frac{3}{15}\right) + 18\left(\frac{1}{15}\right) + 19\left(\frac{2}{15}\right) + 20\left(\frac{3}{15}\right) + 21\left(\frac{1}{15}\right)$

$E(X) = \frac{1}{15} (14 \times 2 + 15 \times 1 + 16 \times 2 + 17 \times 3 + 18 \times 1 + 19 \times 2 + 20 \times 3 + 21 \times 1)$

$E(X) = \frac{1}{15} (28 + 15 + 32 + 51 + 18 + 38 + 60 + 21)$

$E(X) = \frac{1}{15} (263)$

The mean age is $E(X) = \frac{263}{15} \approx 17.533$.

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Variance $Var(X)$:

$Var(X) = E(X^2) - [E(X)]^2$

First, calculate $E(X^2) = \sum\limits_{x} x^2 P(X=x)$:

$E(X^2) = 14^2\left(\frac{2}{15}\right) + 15^2\left(\frac{1}{15}\right) + 16^2\left(\frac{2}{15}\right) + 17^2\left(\frac{3}{15}\right) + 18^2\left(\frac{1}{15}\right) + 19^2\left(\frac{2}{15}\right) + 20^2\left(\frac{3}{15}\right) + 21^2\left(\frac{1}{15}\right)$

$E(X^2) = \frac{1}{15} (196 \times 2 + 225 \times 1 + 256 \times 2 + 289 \times 3 + 324 \times 1 + 361 \times 2 + 400 \times 3 + 441 \times 1)$

$E(X^2) = \frac{1}{15} (392 + 225 + 512 + 867 + 324 + 722 + 1200 + 441)$

$E(X^2) = \frac{4683}{15}$

Simplify $\frac{4683}{15}$ by dividing by 3:

$E(X^2) = \frac{\cancel{4683}^{1561}}{\cancel{15}_{5}} = \frac{1561}{5} = 312.2$

Now, calculate the variance:

$Var(X) = E(X^2) - [E(X)]^2$

$Var(X) = \frac{1561}{5} - \left(\frac{263}{15}\right)^2$

$Var(X) = \frac{1561}{5} - \frac{263^2}{15^2}$

$263^2 = 69169$

$15^2 = 225$

$Var(X) = \frac{1561}{5} - \frac{69169}{225}$

Find a common denominator, which is 225:

$Var(X) = \frac{1561 \times 45}{5 \times 45} - \frac{69169}{225}$

$1561 \times 45 = 70245$

$Var(X) = \frac{70245}{225} - \frac{69169}{225}$

$Var(X) = \frac{70245 - 69169}{225} = \frac{1076}{225}$

The variance is $Var(X) = \frac{1076}{225} \approx 4.7822$.

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Standard Deviation $\sigma_X$:

The standard deviation is the square root of the variance.

$\sigma_X = \sqrt{Var(X)} = \sqrt{\frac{1076}{225}}$

$\sigma_X = \frac{\sqrt{1076}}{\sqrt{225}} = \frac{\sqrt{1076}}{15}$

The standard deviation is $\sigma_X = \frac{\sqrt{1076}}{15}$. (Note: $\sqrt{1076} \approx 32.80$)

$\sigma_X \approx \frac{32.80}{15} \approx 2.187$

Given:

70% of members favour a proposal.

30% of members oppose a proposal.

A member is selected at random.

X is a random variable such that:

X = 0 if the selected member opposed.

X = 1 if the selected member is in favour.

To Find:

The expectation (mean) of X, $E(X)$.

The variance of X, $Var(X)$.

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Solution:

The random variable X can take two possible values: 0 and 1.

The probability that X takes the value 0 is the probability that the selected member opposed the proposal:

$P(X=0) = P(\text{Opposed}) = 30\% = 0.30$

The probability that X takes the value 1 is the probability that the selected member is in favour of the proposal:

$P(X=1) = P(\text{In favour}) = 70\% = 0.70$

The probability distribution of X is:

X	0	1
P(X)	0.30	0.70

Check: $0.30 + 0.70 = 1.00$.

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Expectation $E(X)$:

$E(X) = \sum\limits_{x} x P(X=x)$

$E(X) = 0 \times P(X=0) + 1 \times P(X=1)$

$E(X) = 0 \times 0.30 + 1 \times 0.70$

$E(X) = 0 + 0.70$

$E(X) = 0.70$

The mean of X is 0.70.

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Variance $Var(X)$:

$Var(X) = E(X^2) - [E(X)]^2$

First, calculate $E(X^2) = \sum\limits_{x} x^2 P(X=x)$:

$E(X^2) = 0^2 \times P(X=0) + 1^2 \times P(X=1)$

$E(X^2) = 0 \times 0.30 + 1 \times 0.70$

$E(X^2) = 0 + 0.70$

$E(X^2) = 0.70$

Now, calculate the variance:

$Var(X) = E(X^2) - [E(X)]^2$

$Var(X) = 0.70 - (0.70)^2$

$Var(X) = 0.70 - 0.49$

$Var(X) = 0.21$

The variance of X is 0.21.

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Note: A random variable that takes values 0 and 1, with $P(X=1) = p$ and $P(X=0) = 1-p$, follows a Bernoulli distribution with parameter p. For a Bernoulli distribution, the mean is $E(X) = p$ and the variance is $Var(X) = p(1-p)$. In this case, $p = 0.70$.

$E(X) = 0.70$

$Var(X) = 0.70(1 - 0.70) = 0.70(0.30) = 0.21$

This confirms our results.

Given:

A die with 6 faces.

Number 1 is written on 3 faces.

Number 2 is written on 2 faces.

Number 5 is written on 1 face.

Total number of faces = $3 + 2 + 1 = 6$.

Let X be the random variable representing the number obtained on throwing this die.

To Find:

The mean (expectation) of X, $E(X)$.

---

Solution:

The possible values that X can take are the numbers written on the faces: {1, 2, 5}.

The probability of obtaining each number is the number of faces with that number divided by the total number of faces.

$P(X=1) = \frac{\text{Number of faces with 1}}{\text{Total number of faces}} = \frac{3}{6} = \frac{1}{2}$

$P(X=2) = \frac{\text{Number of faces with 2}}{\text{Total number of faces}} = \frac{2}{6} = \frac{1}{3}$

$P(X=5) = \frac{\text{Number of faces with 5}}{\text{Total number of faces}} = \frac{1}{6}$

The probability distribution of X is:

X	1	2	5
P(X)	$\frac{1}{2}$	$\frac{1}{3}$	$\frac{1}{6}$

Check: $\frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3}{6} + \frac{2}{6} + \frac{1}{6} = \frac{6}{6} = 1$.

The mean (expectation) of X is given by the formula $E(X) = \sum\limits_{x} x P(X=x)$.

$E(X) = 1 \times P(X=1) + 2 \times P(X=2) + 5 \times P(X=5)$

$E(X) = 1 \times \frac{1}{2} + 2 \times \frac{1}{3} + 5 \times \frac{1}{6}$

$E(X) = \frac{1}{2} + \frac{2}{3} + \frac{5}{6}$

Find a common denominator for 2, 3, and 6, which is 6.

$E(X) = \frac{1 \times 3}{2 \times 3} + \frac{2 \times 2}{3 \times 2} + \frac{5}{6}$

$E(X) = \frac{3}{6} + \frac{4}{6} + \frac{5}{6}$

$E(X) = \frac{3 + 4 + 5}{6} = \frac{12}{6} = 2$

The mean of the numbers obtained on throwing the die is 2.

This corresponds to option (B).

The correct answer is (B) 2.

Given:

A well-shuffled deck of 52 cards.

Two cards are drawn at random. This implies drawing without replacement.

X is the random variable representing the number of aces obtained.

To Find:

The expectation of X, $E(X)$.

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Solution:

A standard deck has 52 cards.

Number of aces = 4.

Number of non-aces = $52 - 4 = 48$.

Two cards are drawn without replacement. The total number of ways to draw 2 cards from 52 is $\binom{52}{2}$.

Total number of outcomes = $\binom{52}{2} = \frac{52 \times 51}{2 \times 1} = 26 \times 51 = 1326$

The random variable X represents the number of aces in the two draws. The possible number of aces can be 0, 1, or 2.

The possible values of X are $x = 0, 1, 2$.

We find the probability for each value of X:

$P(X=0)$: Getting 0 aces and 2 non-aces.

Number of ways to choose 0 aces from 4 is $\binom{4}{0} = 1$.

Number of ways to choose 2 non-aces from 48 is $\binom{48}{2} = \frac{48 \times 47}{2} = 1128$.

$P(X=0) = \frac{\binom{4}{0} \times \binom{48}{2}}{\binom{52}{2}} = \frac{1 \times 1128}{1326} = \frac{1128}{1326}$

$P(X=1)$: Getting exactly 1 ace and 1 non-ace.

Number of ways to choose 1 king from 4 is $\binom{4}{1} = 4$.

Number of ways to choose 1 non-king from 48 is $\binom{48}{1} = 48$.

$P(X=1) = \frac{\binom{4}{1} \times \binom{48}{1}}{\binom{52}{2}} = \frac{4 \times 48}{1326} = \frac{192}{1326}$

$P(X=2)$: Getting exactly 2 aces and 0 non-aces.

Number of ways to choose 2 aces from 4 is $\binom{4}{2} = \frac{4 \times 3}{2} = 6$.

Number of ways to choose 0 non-aces from 48 is $\binom{48}{0} = 1$.

$P(X=2) = \frac{\binom{4}{2} \times \binom{48}{0}}{\binom{52}{2}} = \frac{6 \times 1}{1326} = \frac{6}{1326}$

The probability distribution of X is:

X (Number of Aces)	0	1	2
P(X)	$\frac{1128}{1326}$	$\frac{192}{1326}$	$\frac{6}{1326}$

Check sum: $\frac{1128+192+6}{1326} = \frac{1326}{1326} = 1$.

The expectation of X is given by the formula $E(X) = \sum\limits_{x} x P(X=x)$.

$E(X) = 0 \times P(X=0) + 1 \times P(X=1) + 2 \times P(X=2)$

$E(X) = 0 \times \frac{1128}{1326} + 1 \times \frac{192}{1326} + 2 \times \frac{6}{1326}$

$E(X) = 0 + \frac{192}{1326} + \frac{12}{1326}$

$E(X) = \frac{192 + 12}{1326} = \frac{204}{1326}$

Simplify the fraction $\frac{204}{1326}$:

$E(X) = \frac{\cancel{204}^{34}}{\cancel{1326}_{221}}$ (Dividing by 6)

$E(X) = \frac{\cancel{34}^{2}}{\cancel{221}_{13}}$ (Dividing by 17, $221 = 13 \times 17$)

$E(X) = \frac{2}{13}$

The expectation of the number of aces is $\frac{2}{13}$.

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Alternate Solution (using Linearity of Expectation):

Let $I_1$ be an indicator random variable such that $I_1 = 1$ if the first card drawn is an ace, and $I_1 = 0$ otherwise.

Let $I_2$ be an indicator random variable such that $I_2 = 1$ if the second card drawn is an ace, and $I_2 = 0$ otherwise.

The total number of aces drawn is $X = I_1 + I_2$.

The expectation of $I_1$ is the probability that the first card is an ace:

$E(I_1) = P(\text{1st card is Ace}) = \frac{\text{Number of aces}}{\text{Total number of cards}} = \frac{4}{52} = \frac{1}{13}$

The expectation of $I_2$ is the probability that the second card is an ace. By symmetry (or by using Law of Total Probability), the probability that the second card drawn is an ace is the same as the probability that the first card drawn is an ace:

$E(I_2) = P(\text{2nd card is Ace}) = \frac{4}{52} = \frac{1}{13}$

By the linearity of expectation, the expectation of the sum is the sum of the expectations:

$E(X) = E(I_1 + I_2) = E(I_1) + E(I_2)$

$E(X) = \frac{1}{13} + \frac{1}{13} = \frac{2}{13}$

The expectation of the number of aces is $\frac{2}{13}$.

This matches option (D).

The correct answer is (D) $\frac{2}{13}$.

Solution:

For a sequence of independent trials to be considered Bernoulli trials, each trial must satisfy the following conditions:

1. Each trial must have exactly two possible outcomes, conventionally called 'success' and 'failure'.

2. The probability of success must remain constant for each trial.

3. The trials must be independent of each other.

In this problem, drawing a ball from the urn is a trial. We can define 'success' as drawing a red ball and 'failure' as drawing a black ball. This satisfies the first condition as there are only two outcomes for each draw.

The urn contains 7 red balls and 9 black balls. The total number of balls is $7 + 9 = 16$.

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(i) When after each draw the ball drawn is replaced.

When the ball drawn is replaced after each draw, the composition of the urn remains unchanged for every subsequent draw. There are always 7 red balls and 9 black balls, totaling 16 balls.

The probability of drawing a red ball (success) in any trial is given by:

$P(\text{Red}) = \frac{\text{Number of red balls}}{\text{Total number of balls}} = \frac{7}{16}$.

This probability remains constant for each of the six draws because the ball is replaced.

Since the ball is replaced, the outcome of any one draw does not influence the outcome of any other draw. Therefore, the trials are independent.

All three conditions for Bernoulli trials are satisfied.

Thus, the trials of drawing balls successively with replacement are Bernoulli trials.

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(ii) When after each draw the ball drawn is not replaced in the urn.

When the ball drawn is not replaced after each draw, the total number of balls in the urn and the number of balls of a specific color change depending on the outcome of the previous draws.

For the first draw, the probability of drawing a red ball is $P(\text{Red}) = \frac{7}{16}$.

However, if a red ball is drawn in the first trial and not replaced, the urn for the second trial will contain $7-1 = 6$ red balls and 9 black balls, for a total of $16-1 = 15$ balls. The probability of drawing a red ball in the second trial would then be $\frac{6}{15}$.

If a black ball is drawn in the first trial and not replaced, the urn for the second trial will contain 7 red balls and $9-1 = 8$ black balls, for a total of $16-1 = 15$ balls. The probability of drawing a red ball in the second trial would then be $\frac{7}{15}$.

Since the probability of success (drawing a red ball) changes from trial to trial depending on the outcomes of previous trials, the probability of success is not constant.

Also, the outcome of one trial affects the composition of the urn for subsequent trials, so the trials are not independent.

Therefore, the trials of drawing balls successively without replacement do not satisfy the conditions for Bernoulli trials.

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Conclusion:

The trials are Bernoulli trials only when the ball drawn is replaced after each draw.

Solution:

The experiment is tossing a fair coin 10 times. Each toss is an independent trial with two possible outcomes: Head (H) or Tail (T). Since the coin is fair, the probability of getting a head is constant for each toss, and similarly for a tail. This scenario fits the conditions for a sequence of Bernoulli trials, and the number of heads obtained in 10 tosses follows a binomial distribution.

Let $n$ be the number of trials, $p$ be the probability of getting a head (success) in a single trial, and $q$ be the probability of getting a tail (failure) in a single trial.

Given:

Number of trials, $n = 10$.

Probability of getting a head, $p = \frac{1}{2}$.

Probability of getting a tail, $q = 1 - p = 1 - \frac{1}{2} = \frac{1}{2}$.

Let $X$ be the random variable representing the number of heads in 10 tosses. The probability of getting exactly $k$ heads in $n$ trials is given by the binomial probability mass function:

$P(X=k) = \binom{n}{k} p^k q^{n-k}$

For this problem, the probability of getting exactly $k$ heads in 10 tosses is:

$P(X=k) = \binom{10}{k} (\frac{1}{2})^k (\frac{1}{2})^{10-k} = \binom{10}{k} (\frac{1}{2})^{10}$.

We know that $(\frac{1}{2})^{10} = \frac{1^{10}}{2^{10}} = \frac{1}{1024}$.

So, $P(X=k) = \binom{10}{k} \frac{1}{1024}$.

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(i) Probability of exactly six heads

We need to find $P(X=6)$. Using the formula with $k=6$:

$P(X=6) = \binom{10}{6} (\frac{1}{2})^{10}$

First, calculate the binomial coefficient $\binom{10}{6}$:

$\binom{10}{6} = \binom{10}{10-6} = \binom{10}{4} = \frac{10!}{4!(10-4)!} = \frac{10!}{4!6!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = \frac{5040}{24} = 210$.

Now, calculate the probability:

$P(X=6) = 210 \times (\frac{1}{2})^{10} = 210 \times \frac{1}{1024} = \frac{210}{1024}$.

Simplifying the fraction:

$P(X=6) = \frac{\cancel{210}^{105}}{\cancel{1024}_{512}} = \frac{105}{512}$.

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(ii) Probability of at least six heads

We need to find $P(X \geq 6)$. This includes the cases where the number of heads is 6, 7, 8, 9, or 10.

$P(X \geq 6) = P(X=6) + P(X=7) + P(X=8) + P(X=9) + P(X=10)$.

We already have $P(X=6) = \frac{210}{1024}$. Now calculate the probabilities for $k=7, 8, 9, 10$:

$P(X=7) = \binom{10}{7} (\frac{1}{2})^{10} = \binom{10}{3} (\frac{1}{2})^{10} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} \times \frac{1}{1024} = 120 \times \frac{1}{1024} = \frac{120}{1024}$.

$P(X=8) = \binom{10}{8} (\frac{1}{2})^{10} = \binom{10}{2} (\frac{1}{2})^{10} = \frac{10 \times 9}{2 \times 1} \times \frac{1}{1024} = 45 \times \frac{1}{1024} = \frac{45}{1024}$.

$P(X=9) = \binom{10}{9} (\frac{1}{2})^{10} = \binom{10}{1} (\frac{1}{2})^{10} = 10 \times \frac{1}{1024} = \frac{10}{1024}$.

$P(X=10) = \binom{10}{10} (\frac{1}{2})^{10} = 1 \times \frac{1}{1024} = \frac{1}{1024}$.

Summing these probabilities:

$P(X \geq 6) = \frac{210}{1024} + \frac{120}{1024} + \frac{45}{1024} + \frac{10}{1024} + \frac{1}{1024} = \frac{210 + 120 + 45 + 10 + 1}{1024} = \frac{386}{1024}$.

Simplifying the fraction:

$P(X \geq 6) = \frac{\cancel{386}^{193}}{\cancel{1024}_{512}} = \frac{193}{512}$.

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(iii) Probability of at most six heads

We need to find $P(X \leq 6)$. This includes the cases where the number of heads is 0, 1, 2, 3, 4, 5, or 6.

$P(X \leq 6) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6)$.

An easier way to calculate this is to use the complement rule: $P(X \leq 6) = 1 - P(X > 6)$.

$P(X > 6) = P(X=7) + P(X=8) + P(X=9) + P(X=10)$.

From part (ii), we found that $P(X=7) + P(X=8) + P(X=9) + P(X=10) = \frac{120}{1024} + \frac{45}{1024} + \frac{10}{1024} + \frac{1}{1024} = \frac{176}{1024}$.

So, $P(X \leq 6) = 1 - \frac{176}{1024} = \frac{1024}{1024} - \frac{176}{1024} = \frac{1024 - 176}{1024} = \frac{848}{1024}$.

Simplifying the fraction:

$P(X \leq 6) = \frac{\cancel{848}^{424}}{\cancel{1024}_{512}} = \frac{\cancel{424}^{212}}{\cancel{512}_{256}} = \frac{\cancel{212}^{106}}{\cancel{256}_{128}} = \frac{\cancel{106}^{53}}{\cancel{128}_{64}} = \frac{53}{64}$.

Alternate calculation for (iii):

We can also calculate each probability from $k=0$ to $k=5$ and sum them up with $P(X=6)$.

$P(X=0) = \binom{10}{0} (\frac{1}{2})^{10} = 1 \times \frac{1}{1024} = \frac{1}{1024}$.

$P(X=1) = \binom{10}{1} (\frac{1}{2})^{10} = 10 \times \frac{1}{1024} = \frac{10}{1024}$.

$P(X=2) = \binom{10}{2} (\frac{1}{2})^{10} = 45 \times \frac{1}{1024} = \frac{45}{1024}$.

$P(X=3) = \binom{10}{3} (\frac{1}{2})^{10} = 120 \times \frac{1}{1024} = \frac{120}{1024}$.

$P(X=4) = \binom{10}{4} (\frac{1}{2})^{10} = 210 \times \frac{1}{1024} = \frac{210}{1024}$.

$P(X=5) = \binom{10}{5} (\frac{1}{2})^{10} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} \times \frac{1}{1024} = 252 \times \frac{1}{1024} = \frac{252}{1024}$.

Summing $P(X=0)$ through $P(X=6)$:

$P(X \leq 6) = \frac{1}{1024} + \frac{10}{1024} + \frac{45}{1024} + \frac{120}{1024} + \frac{210}{1024} + \frac{252}{1024} + \frac{210}{1024} = \frac{1 + 10 + 45 + 120 + 210 + 252 + 210}{1024} = \frac{848}{1024} = \frac{53}{64}$.

Both methods yield the same result.

Solution:

The experiment consists of drawing an egg from a lot, noting if it is defective or not, and replacing it. This process is repeated 10 times. Since the drawing is done with replacement, the outcome of each draw is independent of the previous draws, and the probability of drawing a defective egg remains constant for each trial. This is a sequence of Bernoulli trials, and the number of defective eggs drawn in 10 trials follows a binomial distribution.

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Let $n$ be the number of trials (eggs drawn), $p$ be the probability of drawing a defective egg (success) in a single trial, and $q$ be the probability of drawing a non-defective egg (failure) in a single trial.

Given:

Number of trials, $n = 10$.

Percentage of defective eggs = 10%.

Probability of drawing a defective egg, $p = 10\% = \frac{10}{100} = 0.1$.

Probability of drawing a non-defective egg, $q = 1 - p = 1 - 0.1 = 0.9$.

Let $X$ be the random variable representing the number of defective eggs drawn in 10 trials. $X$ follows a binomial distribution with parameters $n=10$ and $p=0.1$. The probability mass function is given by:

$P(X=k) = \binom{n}{k} p^k q^{n-k} = \binom{10}{k} (0.1)^k (0.9)^{10-k}$

We want to find the probability that there is at least one defective egg, which means $P(X \geq 1)$.

Using the complement rule, this probability is $1 - P(X < 1)$, which simplifies to $1 - P(X = 0)$.

We need to calculate $P(X=0)$, the probability of drawing exactly 0 defective eggs in 10 trials.

$P(X=0) = \binom{10}{0} (0.1)^0 (0.9)^{10-0}$

We know that $\binom{10}{0} = 1$ and $(0.1)^0 = 1$.

So, $P(X=0) = 1 \times 1 \times (0.9)^{10} = (0.9)^{10}$.

Now, we can find the probability of at least one defective egg:

$P(X \geq 1) = 1 - P(X = 0) = 1 - (0.9)^{10}$.

Calculating $(0.9)^{10}$ (approx):

$(0.9)^{10} \approx 0.3486784401$

So, $P(X \geq 1) \approx 1 - 0.3486784401 \approx 0.6513215599$.

The probability that there is at least one defective egg is $1 - (0.9)^{10}$.

Solution:

The experiment is throwing a die 6 times. Let's define 'success' as getting an odd number.

In a single throw of a die, the possible outcomes are {1, 2, 3, 4, 5, 6}. The odd numbers are {1, 3, 5}.

The probability of getting an odd number (success) in a single throw is:

$p = P(\text{Success}) = P(\text{getting an odd number}) = \frac{\text{Number of odd outcomes}}{\text{Total number of outcomes}} = \frac{3}{6} = \frac{1}{2}$.

The probability of not getting an odd number (failure) in a single throw is:

$q = P(\text{Failure}) = 1 - p = 1 - \frac{1}{2} = \frac{1}{2}$.

The die is thrown 6 times. Each throw is an independent trial. The probability of success ($p$) is constant for each trial. Thus, this is a sequence of Bernoulli trials, and the number of successes in 6 trials follows a binomial distribution.

Let $n$ be the number of trials, so $n = 6$.

Let $X$ be the random variable representing the number of successes (getting an odd number) in 6 throws. The probability of getting exactly $k$ successes in $n$ trials is given by the binomial probability mass function:

$P(X=k) = \binom{n}{k} p^k q^{n-k}$

For this problem, $n=6$, $p=\frac{1}{2}$, $q=\frac{1}{2}$.

$P(X=k) = \binom{6}{k} (\frac{1}{2})^k (\frac{1}{2})^{6-k} = \binom{6}{k} (\frac{1}{2})^{k + (6-k)} = \binom{6}{k} (\frac{1}{2})^6 = \binom{6}{k} \frac{1}{64}$.

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(i) Probability of 5 successes

We need to find $P(X=5)$. Using the formula with $k=5$:

$P(X=5) = \binom{6}{5} (\frac{1}{2})^6$

Calculate the binomial coefficient $\binom{6}{5}$:

$\binom{6}{5} = \binom{6}{6-5} = \binom{6}{1} = 6$.

Now, calculate the probability:

$P(X=5) = 6 \times \frac{1}{64} = \frac{6}{64}$.

Simplifying the fraction:

$P(X=5) = \frac{\cancel{6}^{3}}{\cancel{64}_{32}} = \frac{3}{32}$.

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(ii) Probability of at least 5 successes

We need to find $P(X \geq 5)$. This includes the cases where the number of successes is 5 or 6.

$P(X \geq 5) = P(X=5) + P(X=6)$.

We already calculated $P(X=5) = \frac{6}{64}$. Now we need to calculate $P(X=6)$:

$P(X=6) = \binom{6}{6} (\frac{1}{2})^6$

Calculate the binomial coefficient $\binom{6}{6}$:

$\binom{6}{6} = 1$.

Now, calculate the probability:

$P(X=6) = 1 \times \frac{1}{64} = \frac{1}{64}$.

Summing the probabilities:

$P(X \geq 5) = \frac{6}{64} + \frac{1}{64} = \frac{6 + 1}{64} = \frac{7}{64}$.

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(iii) Probability of at most 5 successes

We need to find $P(X \leq 5)$. This includes the cases where the number of successes is 0, 1, 2, 3, 4, or 5.

$P(X \leq 5) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5)$.

It is simpler to use the complement rule: $P(X \leq 5) = 1 - P(X > 5)$.

The event $X > 5$ means getting more than 5 successes, which is only $X=6$ since the maximum number of trials is 6.

$P(X > 5) = P(X=6)$.

From part (ii), we found that $P(X=6) = \frac{1}{64}$.

So, $P(X \leq 5) = 1 - P(X=6) = 1 - \frac{1}{64}$.

$P(X \leq 5) = \frac{64}{64} - \frac{1}{64} = \frac{64 - 1}{64} = \frac{63}{64}$.

Alternate calculation for (iii):

We could calculate $P(X=k)$ for $k=0, 1, 2, 3, 4$ and sum them up along with $P(X=5)$.

$P(X=0) = \binom{6}{0} (\frac{1}{2})^6 = 1 \times \frac{1}{64} = \frac{1}{64}$.

$P(X=1) = \binom{6}{1} (\frac{1}{2})^6 = 6 \times \frac{1}{64} = \frac{6}{64}$.

$P(X=2) = \binom{6}{2} (\frac{1}{2})^6 = \frac{6 \times 5}{2 \times 1} \times \frac{1}{64} = 15 \times \frac{1}{64} = \frac{15}{64}$.

$P(X=3) = \binom{6}{3} (\frac{1}{2})^6 = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} \times \frac{1}{64} = 20 \times \frac{1}{64} = \frac{20}{64}$.

$P(X=4) = \binom{6}{4} (\frac{1}{2})^6 = \binom{6}{2} (\frac{1}{2})^6 = 15 \times \frac{1}{64} = \frac{15}{64}$.

$P(X=5) = \frac{6}{64}$ (from part i).

$P(X \leq 5) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5)$

$P(X \leq 5) = \frac{1}{64} + \frac{6}{64} + \frac{15}{64} + \frac{20}{64} + \frac{15}{64} + \frac{6}{64} = \frac{1 + 6 + 15 + 20 + 15 + 6}{64} = \frac{63}{64}$.

Both methods yield the same result.

Solution:

The experiment is throwing a pair of dice 4 times. We define 'success' as getting a doublet.

In a single throw of a pair of dice, the total number of possible outcomes is $6 \times 6 = 36$.

A doublet occurs when both dice show the same number. The possible doublets are (1,1), (2,2), (3,3), (4,4), (5,5), (6,6). There are 6 such outcomes.

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Let $p$ be the probability of getting a doublet (success) in a single throw of a pair of dice.

$p = P(\text{getting a doublet}) = \frac{\text{Number of doublets}}{\text{Total number of outcomes}} = \frac{6}{36} = \frac{1}{6}$.

Let $q$ be the probability of not getting a doublet (failure) in a single throw.

$q = 1 - p = 1 - \frac{1}{6} = \frac{5}{6}$.

The pair of dice is thrown 4 times. Each throw is an independent trial, and the probability of success $p$ is constant for each throw. Thus, this is a sequence of Bernoulli trials, and the number of successes in 4 trials follows a binomial distribution.

Let $n$ be the number of trials, so $n = 4$.

Let $X$ be the random variable representing the number of successes (getting a doublet) in 4 throws. $X$ follows a binomial distribution with parameters $n=4$ and $p=\frac{1}{6}$. The probability of getting exactly $k$ successes in $n$ trials is given by the binomial probability mass function:

$P(X=k) = \binom{n}{k} p^k q^{n-k}$

For this problem, $n=4$, $p=\frac{1}{6}$, $q=\frac{5}{6}$. We need to find the probability of two successes, i.e., $P(X=2)$.

$P(X=2) = \binom{4}{2} (\frac{1}{6})^2 (\frac{5}{6})^{4-2} = \binom{4}{2} (\frac{1}{6})^2 (\frac{5}{6})^2$.

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Calculate the components:

$\binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} = \frac{4 \times 3 \times \cancel{2!}}{\cancel{2!} \times (2 \times 1)} = \frac{4 \times 3}{2} = \frac{12}{2} = 6$.

$(\frac{1}{6})^2 = \frac{1^2}{6^2} = \frac{1}{36}$.

$(\frac{5}{6})^2 = \frac{5^2}{6^2} = \frac{25}{36}$.

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Substitute these values into the probability formula:

$P(X=2) = 6 \times \frac{1}{36} \times \frac{25}{36}$

$P(X=2) = \frac{6 \times 1 \times 25}{36 \times 36} = \frac{150}{1296}$.

Simplify the fraction:

$P(X=2) = \frac{\cancel{150}^{25}}{\cancel{1296}_{216}}$.

The simplified probability is $\frac{25}{216}$.

Solution:

The experiment is selecting a sample of 10 items from a large bulk. We define 'success' as selecting a defective item.

Given that 5% of the items are defective, the probability of selecting a defective item in a single draw is:

$p = P(\text{Defective item}) = 5\% = \frac{5}{100} = 0.05$.

The probability of selecting a non-defective item in a single draw is:

$q = P(\text{Non-defective item}) = 1 - p = 1 - 0.05 = 0.95$.

Since the bulk of items is large, drawing a small sample (10 items) successively with replacement (or effectively with replacement from a very large population) means that the trials are independent, and the probability of success $p$ is constant for each trial. This is a sequence of Bernoulli trials, and the number of defective items in the sample follows a binomial distribution.

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Let $n$ be the number of trials (items in the sample), so $n = 10$.

Let $X$ be the random variable representing the number of defective items in the sample of 10. $X$ follows a binomial distribution with parameters $n=10$ and $p=0.05$. The probability of getting exactly $k$ defective items in $n$ trials is given by the binomial probability mass function:

$P(X=k) = \binom{n}{k} p^k q^{n-k}$

For this problem, $n=10$, $p=0.05$, $q=0.95$. We need to find the probability that the sample will include not more than one defective item. This means the number of defective items is 0 or 1. We need to find $P(X \leq 1)$.

$P(X \leq 1) = P(X=0) + P(X=1)$.

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Calculate $P(X=0)$ (probability of exactly 0 defective items):

$P(X=0) = \binom{10}{0} (0.05)^0 (0.95)^{10-0} = \binom{10}{0} (0.05)^0 (0.95)^{10}$.

We know that $\binom{10}{0} = 1$ and $(0.05)^0 = 1$.

$P(X=0) = 1 \times 1 \times (0.95)^{10} = (0.95)^{10}$.

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Calculate $P(X=1)$ (probability of exactly 1 defective item):

$P(X=1) = \binom{10}{1} (0.05)^1 (0.95)^{10-1} = \binom{10}{1} (0.05)^1 (0.95)^9$.

We know that $\binom{10}{1} = 10$.

$P(X=1) = 10 \times (0.05) \times (0.95)^9 = 0.5 \times (0.95)^9$.

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Now, sum the probabilities for $X=0$ and $X=1$:

$P(X \leq 1) = P(X=0) + P(X=1) = (0.95)^{10} + 0.5 \times (0.95)^9$.

We can factor out $(0.95)^9$:

$P(X \leq 1) = (0.95)^9 [(0.95)^1 + 0.5] = (0.95)^9 [0.95 + 0.5] = 1.45 \times (0.95)^9$.

The probability that a sample of 10 items will include not more than one defective item is $(0.95)^{10} + 10(0.05)(0.95)^9$, which simplifies to $1.45 \times (0.95)^9$.

Solution:

The experiment is drawing a card from a well-shuffled deck of 52 cards and replacing it, repeated 5 times. We define 'success' as drawing a spade.

In a standard deck of 52 cards, there are 13 spades.

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Let $p$ be the probability of drawing a spade (success) in a single draw.

$p = P(\text{drawing a spade}) = \frac{\text{Number of spades}}{\text{Total number of cards}} = \frac{13}{52} = \frac{1}{4}$.

Let $q$ be the probability of not drawing a spade (failure) in a single draw.

$q = 1 - p = 1 - \frac{1}{4} = \frac{3}{4}$.

Since the cards are drawn with replacement, each draw is an independent trial, and the probability of success $p$ is constant for each draw. This is a sequence of Bernoulli trials, and the number of spades drawn in 5 trials follows a binomial distribution.

Let $n$ be the number of trials (cards drawn), so $n = 5$.

Let $X$ be the random variable representing the number of spades in 5 draws. $X$ follows a binomial distribution with parameters $n=5$ and $p=\frac{1}{4}$. The probability of getting exactly $k$ successes (spades) in $n$ trials is given by the binomial probability mass function:

$P(X=k) = \binom{n}{k} p^k q^{n-k}$

For this problem, $n=5$, $p=\frac{1}{4}$, $q=\frac{3}{4}$.

$P(X=k) = \binom{5}{k} (\frac{1}{4})^k (\frac{3}{4})^{5-k}$.

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(i) Probability that all the five cards are spades

We need to find $P(X=5)$, where $k=5$ successes (spades).

$P(X=5) = \binom{5}{5} (\frac{1}{4})^5 (\frac{3}{4})^{5-5} = \binom{5}{5} (\frac{1}{4})^5 (\frac{3}{4})^0$.

Calculate the components:

$\binom{5}{5} = 1$.

$(\frac{1}{4})^5 = \frac{1^5}{4^5} = \frac{1}{1024}$.

$(\frac{3}{4})^0 = 1$.

Substitute these values:

$P(X=5) = 1 \times \frac{1}{1024} \times 1 = \frac{1}{1024}$.

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(ii) Probability that only 3 cards are spades

We need to find $P(X=3)$, where $k=3$ successes (spades).

$P(X=3) = \binom{5}{3} (\frac{1}{4})^3 (\frac{3}{4})^{5-3} = \binom{5}{3} (\frac{1}{4})^3 (\frac{3}{4})^2$.

Calculate the components:

$\binom{5}{3} = \binom{5}{5-3} = \binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10$.

$(\frac{1}{4})^3 = \frac{1^3}{4^3} = \frac{1}{64}$.

$(\frac{3}{4})^2 = \frac{3^2}{4^2} = \frac{9}{16}$.

Substitute these values:

$P(X=3) = 10 \times \frac{1}{64} \times \frac{9}{16} = \frac{10 \times 1 \times 9}{64 \times 16} = \frac{90}{1024}$.

Simplifying the fraction:

$P(X=3) = \frac{\cancel{90}^{45}}{\cancel{1024}_{512}} = \frac{45}{512}$.

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(iii) Probability that none is a spade

We need to find $P(X=0)$, where $k=0$ successes (spades).

$P(X=0) = \binom{5}{0} (\frac{1}{4})^0 (\frac{3}{4})^{5-0} = \binom{5}{0} (\frac{1}{4})^0 (\frac{3}{4})^5$.

Calculate the components:

$\binom{5}{0} = 1$.

$(\frac{1}{4})^0 = 1$.

$(\frac{3}{4})^5 = \frac{3^5}{4^5} = \frac{243}{1024}$.

Substitute these values:

$P(X=0) = 1 \times 1 \times \frac{243}{1024} = \frac{243}{1024}$.

Solution:

The experiment is selecting 5 bulbs and checking if they fuse after 150 days of use. We define 'success' as a bulb fusing after 150 days.

Given:

Probability that a bulb fuses after 150 days, $p = 0.05$.

Probability that a bulb does not fuse after 150 days, $q = 1 - p = 1 - 0.05 = 0.95$.

The number of bulbs selected is 5. Each bulb's fusing status is independent of the others, and the probability of a bulb fusing is constant for each bulb. Thus, this is a sequence of Bernoulli trials, and the number of fused bulbs in a sample of 5 follows a binomial distribution.

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Let $n$ be the number of trials (bulbs), so $n = 5$.

Let $X$ be the random variable representing the number of bulbs that fuse after 150 days out of 5 bulbs. $X$ follows a binomial distribution with parameters $n=5$ and $p=0.05$. The probability of getting exactly $k$ successes (fused bulbs) in $n$ trials is given by the binomial probability mass function:

$P(X=k) = \binom{n}{k} p^k q^{n-k}$

For this problem, $n=5$, $p=0.05$, $q=0.95$.

$P(X=k) = \binom{5}{k} (0.05)^k (0.95)^{5-k}$.

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(i) Probability that none will fuse

We need to find the probability of exactly 0 successes (0 fused bulbs), i.e., $P(X=0)$.

$P(X=0) = \binom{5}{0} (0.05)^0 (0.95)^{5-0} = \binom{5}{0} (0.05)^0 (0.95)^5$.

We know that $\binom{5}{0} = 1$ and $(0.05)^0 = 1$.

$P(X=0) = 1 \times 1 \times (0.95)^5 = (0.95)^5$.

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(ii) Probability that not more than one will fuse

We need to find the probability that the number of fused bulbs is less than or equal to 1, i.e., $P(X \leq 1)$. This includes the cases of exactly 0 or exactly 1 fused bulb.

$P(X \leq 1) = P(X=0) + P(X=1)$.

We already found $P(X=0) = (0.95)^5$. Now we calculate $P(X=1)$:

$P(X=1) = \binom{5}{1} (0.05)^1 (0.95)^{5-1} = \binom{5}{1} (0.05)^1 (0.95)^4$.

We know that $\binom{5}{1} = 5$.

$P(X=1) = 5 \times 0.05 \times (0.95)^4 = 0.25 \times (0.95)^4$.

So, $P(X \leq 1) = (0.95)^5 + 0.25 \times (0.95)^4$.

This can also be written as:

$P(X \leq 1) = (0.95)^4 (0.95 + 0.25) = 1.20 \times (0.95)^4$.

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(iii) Probability that more than one will fuse

We need to find the probability that the number of fused bulbs is greater than 1, i.e., $P(X > 1)$.

Using the complement rule, $P(X > 1) = 1 - P(X \leq 1)$.

From part (ii), we have $P(X \leq 1) = (0.95)^5 + 0.25 \times (0.95)^4$.

So, $P(X > 1) = 1 - [(0.95)^5 + 0.25 \times (0.95)^4]$.

Alternatively, $P(X > 1) = 1 - [1.20 \times (0.95)^4]$.

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(iv) Probability that at least one will fuse

We need to find the probability that the number of fused bulbs is greater than or equal to 1, i.e., $P(X \geq 1)$.

Using the complement rule, $P(X \geq 1) = 1 - P(X < 1)$.

The event $X < 1$ means exactly 0 fused bulbs, i.e., $X=0$.

$P(X \geq 1) = 1 - P(X=0)$.

From part (i), we found $P(X=0) = (0.95)^5$.

So, $P(X \geq 1) = 1 - (0.95)^5$.

Solution:

The experiment is drawing a ball from a bag containing 10 balls marked with digits 0 to 9, and replacing it, repeated 4 times. We define 'success' as drawing a ball NOT marked with the digit 0.

The 10 balls are marked with digits {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}.

The balls not marked with the digit 0 are those marked with {1, 2, 3, 4, 5, 6, 7, 8, 9}. There are 9 such balls.

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Let $p$ be the probability of drawing a ball NOT marked with the digit 0 (success) in a single draw.

$p = P(\text{drawing a digit other than 0}) = \frac{\text{Number of balls not marked 0}}{\text{Total number of balls}} = \frac{9}{10} = 0.9$.

Let $q$ be the probability of drawing a ball marked with the digit 0 (failure) in a single draw.

$q = 1 - p = 1 - 0.9 = 0.1$.

The balls are drawn successively with replacement, so each draw is an independent trial, and the probability of success $p$ is constant for each draw. This is a sequence of Bernoulli trials, and the number of successes in 4 trials follows a binomial distribution.

Let $n$ be the number of trials (balls drawn), so $n = 4$.

Let $X$ be the random variable representing the number of successes (balls not marked with 0) in 4 draws. $X$ follows a binomial distribution with parameters $n=4$ and $p=0.9$. The probability of getting exactly $k$ successes in $n$ trials is given by the binomial probability mass function:

$P(X=k) = \binom{n}{k} p^k q^{n-k}$

For this problem, $n=4$, $p=0.9$, $q=0.1$. We need to find the probability that none is marked with the digit 0. This means all 4 balls drawn are NOT marked with 0, which corresponds to 4 successes. So, we need to find $P(X=4)$.

$P(X=4) = \binom{4}{4} (0.9)^4 (0.1)^{4-4} = \binom{4}{4} (0.9)^4 (0.1)^0$.

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Calculate the components:

$\binom{4}{4} = 1$.

$(0.9)^4 = 0.9 \times 0.9 \times 0.9 \times 0.9 = 0.81 \times 0.81 = 0.6561$.

$(0.1)^0 = 1$ (any non-zero number raised to the power of 0 is 1).

Substitute these values:

$P(X=4) = 1 \times 0.6561 \times 1 = 0.6561$.

The probability that none of the four balls is marked with the digit 0 is 0.6561.

Solution:

The experiment consists of answering 20 true-false questions by tossing a fair coin for each. For each question, there are two possible outcomes for the student's answer: 'True' or 'False'. Since it is a true-false question, there is only one correct answer out of two options.

If the student answers 'True' when the correct answer is 'True', or 'False' when the correct answer is 'False', he gets the question correct. Given that he uses a fair coin (Heads for True, Tails for False), the probability of his chosen answer matching the correct answer for any single question is $\frac{1}{2}$, assuming the correct answer is equally likely to be True or False, or simply that his random strategy gives him a 50% chance of getting the answer right for any question.

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Let's define 'success' as the student answering a question correctly.

The probability of answering a question correctly (success) in a single trial (one question) is $p = \frac{1}{2}$ (since it's a fair coin and a true-false question).

The probability of answering a question incorrectly (failure) is $q = 1 - p = 1 - \frac{1}{2} = \frac{1}{2}$.

The number of questions is 20. Each question is answered independently. The probability of success $p$ is constant for each question. Thus, this is a sequence of Bernoulli trials, and the number of correct answers in 20 questions follows a binomial distribution.

Let $n$ be the number of trials (questions), so $n = 20$.

Let $X$ be the random variable representing the number of questions answered correctly out of 20. $X$ follows a binomial distribution with parameters $n=20$ and $p=\frac{1}{2}$. The probability of answering exactly $k$ questions correctly in $n$ trials is given by the binomial probability mass function:

$P(X=k) = \binom{n}{k} p^k q^{n-k}$

For this problem, $n=20$, $p=\frac{1}{2}$, $q=\frac{1}{2}$.

$P(X=k) = \binom{20}{k} (\frac{1}{2})^k (\frac{1}{2})^{20-k} = \binom{20}{k} (\frac{1}{2})^{20} = \frac{\binom{20}{k}}{2^{20}}$.

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We need to find the probability that the student answers at least 12 questions correctly. This means the number of correct answers is 12 or more. We need to find $P(X \geq 12)$.

$P(X \geq 12) = P(X=12) + P(X=13) + P(X=14) + \dots + P(X=20)$.

Using the formula for $P(X=k)$:

$P(X \geq 12) = \frac{\binom{20}{12}}{2^{20}} + \frac{\binom{20}{13}}{2^{20}} + \frac{\binom{20}{14}}{2^{20}} + \dots + \frac{\binom{20}{20}}{2^{20}}$.

We can factor out the common term $\frac{1}{2^{20}}$:

$P(X \geq 12) = \frac{1}{2^{20}} \left[ \binom{20}{12} + \binom{20}{13} + \binom{20}{14} + \binom{20}{15} + \binom{20}{16} + \binom{20}{17} + \binom{20}{18} + \binom{20}{19} + \binom{20}{20} \right]$.

Alternatively, using summation notation:

$P(X \geq 12) = \sum_{k=12}^{20} P(X=k) = \sum_{k=12}^{20} \binom{20}{k} (\frac{1}{2})^{20} = (\frac{1}{2})^{20} \sum_{k=12}^{20} \binom{20}{k}$.

Calculating the sum of binomial coefficients:

$\binom{20}{12} = 125970$

$\binom{20}{13} = 77520$

$\binom{20}{14} = 38760$

$\binom{20}{15} = 15504$

$\binom{20}{16} = 4845$

$\binom{20}{17} = 1140$

$\binom{20}{18} = 190$

$\binom{20}{19} = 20$

$\binom{20}{20} = 1$

Sum $= 125970 + 77520 + 38760 + 15504 + 4845 + 1140 + 190 + 20 + 1 = 263950$. No, let me recalculate. $125970+77520+38760+15504+4845+1140+190+20+1 = 263950$. Wait, $1+20=21$, $21+190=211$, $211+1140=1351$, $1351+4845=6196$, $6196+15504=21700$, $21700+38760=60460$, $60460+77520=137980$, $137980+125970=263950$. The sum is 263950.

$2^{20} = 1048576$.

The probability is $\frac{263950}{1048576}$.

Simplifying the fraction by dividing by 2:

$\frac{263950}{1048576} = \frac{\cancel{263950}^{131975}}{\cancel{1048576}_{524288}} = \frac{131975}{524288}$.

The probability is $\frac{131975}{524288}$.

Solution:

The random variable $X$ follows a binomial distribution with parameters $n=6$ and $p=\frac{1}{2}$. We write this as $X \sim B(6, \frac{1}{2})$.

The number of trials is $n = 6$.

The probability of success in a single trial is $p = \frac{1}{2}$.

The probability of failure in a single trial is $q = 1 - p = 1 - \frac{1}{2} = \frac{1}{2}$.

The possible values for the random variable $X$ (the number of successes in 6 trials) are $k = 0, 1, 2, 3, 4, 5, 6$.

The probability mass function for a binomial distribution is given by:

$P(X=k) = \binom{n}{k} p^k q^{n-k}$

Substituting the given values $n=6$, $p=\frac{1}{2}$, and $q=\frac{1}{2}$:

$P(X=k) = \binom{6}{k} (\frac{1}{2})^k (\frac{1}{2})^{6-k} = \binom{6}{k} (\frac{1}{2})^{k + (6-k)} = \binom{6}{k} (\frac{1}{2})^6$.

Since $(\frac{1}{2})^6 = \frac{1^6}{2^6} = \frac{1}{64}$, the probability is $P(X=k) = \frac{\binom{6}{k}}{64}$.

To find the most likely outcome, we need to calculate $P(X=k)$ for each possible value of $k$ (from 0 to 6) and find which value gives the maximum probability. This is equivalent to finding the value of $k$ for which the binomial coefficient $\binom{6}{k}$ is maximum.

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Let's calculate $P(X=k)$ for each $k$:

$P(X=0) = \binom{6}{0} (\frac{1}{2})^6 = 1 \times \frac{1}{64} = \frac{1}{64}$.

$P(X=1) = \binom{6}{1} (\frac{1}{2})^6 = 6 \times \frac{1}{64} = \frac{6}{64}$.

$P(X=2) = \binom{6}{2} (\frac{1}{2})^6 = \frac{6 \times 5}{2 \times 1} \times \frac{1}{64} = 15 \times \frac{1}{64} = \frac{15}{64}$.

$P(X=3) = \binom{6}{3} (\frac{1}{2})^6 = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} \times \frac{1}{64} = 20 \times \frac{1}{64} = \frac{20}{64}$.

$P(X=4) = \binom{6}{4} (\frac{1}{2})^6 = \binom{6}{6-4} (\frac{1}{2})^6 = \binom{6}{2} (\frac{1}{2})^6 = 15 \times \frac{1}{64} = \frac{15}{64}$.

$P(X=5) = \binom{6}{5} (\frac{1}{2})^6 = \binom{6}{6-5} (\frac{1}{2})^6 = \binom{6}{1} (\frac{1}{2})^6 = 6 \times \frac{1}{64} = \frac{6}{64}$.

$P(X=6) = \binom{6}{6} (\frac{1}{2})^6 = 1 \times \frac{1}{64} = \frac{1}{64}$.

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Comparing the probabilities for $k=0, 1, 2, 3, 4, 5, 6$:

$P(X=0) = \frac{1}{64}$

$P(X=1) = \frac{6}{64}$

$P(X=2) = \frac{15}{64}$

$P(X=3) = \frac{20}{64}$

$P(X=4) = \frac{15}{64}$

$P(X=5) = \frac{6}{64}$

$P(X=6) = \frac{1}{64}$

The maximum probability is $\frac{20}{64}$, which occurs when $X=3$.

Thus, $P(X=3)$ is the maximum among all $P(X=k)$ for $k=0, 1, \dots, 6$.

Therefore, $X=3$ is the most likely outcome.

Solution:

The examination has 5 multiple-choice questions, and for each question, there are three possible answers. A candidate is guessing the answer for each question. We define 'success' as getting a question correct by guessing.

For a single question, there are 3 possible answers, only one of which is correct. Since the candidate is guessing, the probability of choosing the correct answer is $\frac{1}{3}$.

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Let $p$ be the probability of getting a question correct by guessing (success) in a single trial (one question).

$p = P(\text{correct answer}) = \frac{1}{3}$.

Let $q$ be the probability of getting a question incorrect by guessing (failure) in a single trial.

$q = 1 - p = 1 - \frac{1}{3} = \frac{2}{3}$.

The candidate answers 5 questions. Each question is answered independently by guessing, and the probability of success $p$ is constant for each question. Thus, this is a sequence of Bernoulli trials, and the number of correct answers in 5 questions follows a binomial distribution.

Let $n$ be the number of trials (questions), so $n = 5$.

Let $X$ be the random variable representing the number of correct answers out of 5. $X$ follows a binomial distribution with parameters $n=5$ and $p=\frac{1}{3}$. The probability of getting exactly $k$ successes (correct answers) in $n$ trials is given by the binomial probability mass function:

$P(X=k) = \binom{n}{k} p^k q^{n-k}$

For this problem, $n=5$, $p=\frac{1}{3}$, $q=\frac{2}{3}$.

$P(X=k) = \binom{5}{k} (\frac{1}{3})^k (\frac{2}{3})^{5-k}$.

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We need to find the probability that the candidate would get four or more correct answers. This means the number of correct answers is 4 or 5. We need to find $P(X \geq 4)$.

$P(X \geq 4) = P(X=4) + P(X=5)$.

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Calculate $P(X=4)$ (probability of exactly 4 correct answers):

$P(X=4) = \binom{5}{4} (\frac{1}{3})^4 (\frac{2}{3})^{5-4} = \binom{5}{4} (\frac{1}{3})^4 (\frac{2}{3})^1$.

Calculate the components:

$\binom{5}{4} = \binom{5}{5-4} = \binom{5}{1} = 5$.

$(\frac{1}{3})^4 = \frac{1^4}{3^4} = \frac{1}{81}$.

$(\frac{2}{3})^1 = \frac{2}{3}$.

Substitute these values:

$P(X=4) = 5 \times \frac{1}{81} \times \frac{2}{3} = \frac{5 \times 1 \times 2}{81 \times 3} = \frac{10}{243}$.

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Calculate $P(X=5)$ (probability of exactly 5 correct answers):

$P(X=5) = \binom{5}{5} (\frac{1}{3})^5 (\frac{2}{3})^{5-5} = \binom{5}{5} (\frac{1}{3})^5 (\frac{2}{3})^0$.

Calculate the components:

$\binom{5}{5} = 1$.

$(\frac{1}{3})^5 = \frac{1^5}{3^5} = \frac{1}{243}$.

$(\frac{2}{3})^0 = 1$.

Substitute these values:

$P(X=5) = 1 \times \frac{1}{243} \times 1 = \frac{1}{243}$.

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Now, sum the probabilities for $X=4$ and $X=5$:

$P(X \geq 4) = P(X=4) + P(X=5) = \frac{10}{243} + \frac{1}{243} = \frac{10 + 1}{243} = \frac{11}{243}$.

The probability that the candidate would get four or more correct answers just by guessing is $\frac{11}{243}$.

Solution:

The experiment involves buying a lottery ticket in 50 lotteries. For each lottery, there are two outcomes: winning a prize or not winning a prize. The chance of winning a prize in each lottery is given and is constant. Since the lotteries are independent, this scenario fits the conditions for a sequence of Bernoulli trials, and the number of prizes won follows a binomial distribution.

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Let $n$ be the number of lotteries (trials), so $n = 50$.

Let $p$ be the probability of winning a prize in a single lottery (success), so $p = \frac{1}{100} = 0.01$.

Let $q$ be the probability of not winning a prize in a single lottery (failure), so $q = 1 - p = 1 - 0.01 = 0.99$.

Let $X$ be the random variable representing the number of prizes won in 50 lotteries. $X$ follows a binomial distribution with parameters $n=50$ and $p=0.01$. The probability of winning exactly $k$ prizes in $n$ trials is given by the binomial probability mass function:

$P(X=k) = \binom{n}{k} p^k q^{n-k}$

For this problem, $n=50$, $p=0.01$, $q=0.99$.

$P(X=k) = \binom{50}{k} (0.01)^k (0.99)^{50-k}$.

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(a) Probability that he will win a prize at least once

We need to find the probability that the number of prizes won is greater than or equal to 1, i.e., $P(X \geq 1)$.

Using the complement rule, this probability is $1 - P(X < 1)$, which simplifies to $1 - P(X = 0)$.

We need to calculate $P(X=0)$, the probability of winning exactly 0 prizes in 50 lotteries.

$P(X=0) = \binom{50}{0} (0.01)^0 (0.99)^{50-0} = \binom{50}{0} (0.01)^0 (0.99)^{50}$.

We know that $\binom{50}{0} = 1$ and $(0.01)^0 = 1$.

$P(X=0) = 1 \times 1 \times (0.99)^{50} = (0.99)^{50}$.

So, the probability of winning at least once is:

$P(X \geq 1) = 1 - P(X = 0) = 1 - (0.99)^{50}$.

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(b) Probability that he will win a prize exactly once

We need to find the probability of winning exactly 1 prize, i.e., $P(X = 1)$.

Using the binomial probability mass function with $k=1$:

$P(X=1) = \binom{50}{1} (0.01)^1 (0.99)^{50-1} = \binom{50}{1} (0.01)^1 (0.99)^{49}$.

We know that $\binom{50}{1} = 50$.

So, the probability of winning exactly once is:

$P(X=1) = 50 \times 0.01 \times (0.99)^{49} = 0.50 \times (0.99)^{49}$.

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(c) Probability that he will win a prize at least twice

We need to find the probability that the number of prizes won is greater than or equal to 2, i.e., $P(X \geq 2)$.

Using the complement rule, this probability is $1 - P(X < 2)$. The event $X < 2$ includes the cases $X=0$ and $X=1$.

$P(X \geq 2) = 1 - [P(X=0) + P(X=1)]$.

From part (a), we have $P(X=0) = (0.99)^{50}$.

From part (b), we have $P(X=1) = 50 \times (0.01) \times (0.99)^{49} = 0.5 \times (0.99)^{49}$.

So, the probability of winning at least twice is:

$P(X \geq 2) = 1 - [(0.99)^{50} + 0.5 \times (0.99)^{49}]$.

This can also be written as:

$P(X \geq 2) = 1 - (0.99)^{49} [0.99 + 0.5] = 1 - 1.49 \times (0.99)^{49}$.

Solution:

The experiment is throwing a die 7 times. We define 'success' as getting the digit 5 in a single throw.

In a single throw of a standard die, the possible outcomes are {1, 2, 3, 4, 5, 6}. Getting a 5 is one specific outcome.

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Let $p$ be the probability of getting a 5 (success) in a single throw of a die.

$p = P(\text{getting a 5}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{1}{6}$.

Let $q$ be the probability of not getting a 5 (failure) in a single throw.

$q = 1 - p = 1 - \frac{1}{6} = \frac{5}{6}$.

The die is thrown 7 times. Each throw is an independent trial, and the probability of success $p$ is constant for each throw. Thus, this is a sequence of Bernoulli trials, and the number of successes in 7 trials follows a binomial distribution.

Let $n$ be the number of trials (throws), so $n = 7$.

Let $X$ be the random variable representing the number of successes (getting a 5) in 7 throws. $X$ follows a binomial distribution with parameters $n=7$ and $p=\frac{1}{6}$. The probability of getting exactly $k$ successes in $n$ trials is given by the binomial probability mass function:

$P(X=k) = \binom{n}{k} p^k q^{n-k}$

For this problem, $n=7$, $p=\frac{1}{6}$, $q=\frac{5}{6}$. We need to find the probability of getting '5' exactly twice, which means we need to find $P(X=2)$, where $k=2$ successes.

$P(X=2) = \binom{7}{2} (\frac{1}{6})^2 (\frac{5}{6})^{7-2} = \binom{7}{2} (\frac{1}{6})^2 (\frac{5}{6})^5$.

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Calculate the components:

$\binom{7}{2} = \frac{7!}{2!(7-2)!} = \frac{7!}{2!5!} = \frac{7 \times 6 \times \cancel{5!}}{(\cancel{2 \times 1}) \times \cancel{5!}} = \frac{42}{2} = 21$.

$(\frac{1}{6})^2 = \frac{1^2}{6^2} = \frac{1}{36}$.

$(\frac{5}{6})^5 = \frac{5^5}{6^5} = \frac{5 \times 5 \times 5 \times 5 \times 5}{6 \times 6 \times 6 \times 6 \times 6} = \frac{3125}{7776}$.

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Substitute these values into the probability formula:

$P(X=2) = 21 \times \frac{1}{36} \times \frac{3125}{7776}$.

$P(X=2) = \frac{21 \times 1 \times 3125}{36 \times 7776}$.

Simplify the fraction by canceling out common factors:

$P(X=2) = \frac{\cancel{21}^{7}}{\cancel{36}_{12}} \times \frac{3125}{7776} = \frac{7}{12} \times \frac{3125}{7776}$.

$P(X=2) = \frac{7 \times 3125}{12 \times 7776} = \frac{21875}{93312}$.

The probability of getting 5 exactly twice in 7 throws of a die is $\frac{21875}{93312}$.

Solution:

The experiment is throwing a single die 6 times. We define 'success' as getting a six in a single throw.

In a single throw of a standard die, the possible outcomes are {1, 2, 3, 4, 5, 6}. Getting a 6 is one specific outcome.

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Let $p$ be the probability of getting a six (success) in a single throw of a die.

$p = P(\text{getting a 6}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{1}{6}$.

Let $q$ be the probability of not getting a six (failure) in a single throw.

$q = 1 - p = 1 - \frac{1}{6} = \frac{5}{6}$.

The die is thrown 6 times. Each throw is an independent trial, and the probability of success $p$ is constant for each throw. Thus, this is a sequence of Bernoulli trials, and the number of successes in 6 trials follows a binomial distribution.

Let $n$ be the number of trials (throws), so $n = 6$.

Let $X$ be the random variable representing the number of successes (getting a six) in 6 throws. $X$ follows a binomial distribution with parameters $n=6$ and $p=\frac{1}{6}$. The probability of getting exactly $k$ successes in $n$ trials is given by the binomial probability mass function:

$P(X=k) = \binom{n}{k} p^k q^{n-k}$

For this problem, $n=6$, $p=\frac{1}{6}$, $q=\frac{5}{6}$.

$P(X=k) = \binom{6}{k} (\frac{1}{6})^k (\frac{5}{6})^{6-k} = \frac{\binom{6}{k} 5^{6-k}}{6^6}$.

We need to find the probability of throwing at most 2 sixes. This means the number of sixes is 0, 1, or 2. We need to find $P(X \leq 2)$.

$P(X \leq 2) = P(X=0) + P(X=1) + P(X=2)$.

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Calculate $P(X=0)$ (probability of exactly 0 sixes):

$P(X=0) = \binom{6}{0} (\frac{1}{6})^0 (\frac{5}{6})^{6-0} = \binom{6}{0} (\frac{1}{6})^0 (\frac{5}{6})^6$.

$\binom{6}{0} = 1$. $(\frac{1}{6})^0 = 1$. $(\frac{5}{6})^6 = \frac{5^6}{6^6} = \frac{15625}{46656}$.

$P(X=0) = 1 \times 1 \times \frac{15625}{46656} = \frac{15625}{46656}$.

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Calculate $P(X=1)$ (probability of exactly 1 six):

$P(X=1) = \binom{6}{1} (\frac{1}{6})^1 (\frac{5}{6})^{6-1} = \binom{6}{1} (\frac{1}{6})^1 (\frac{5}{6})^5$.

$\binom{6}{1} = 6$. $(\frac{1}{6})^1 = \frac{1}{6}$. $(\frac{5}{6})^5 = \frac{5^5}{6^5} = \frac{3125}{7776}$.

$P(X=1) = 6 \times \frac{1}{6} \times \frac{3125}{7776} = 1 \times \frac{3125}{7776} = \frac{3125}{7776}$.

To add this to fractions with denominator $6^6=46656$, we multiply numerator and denominator by 6:

$P(X=1) = \frac{3125 \times 6}{7776 \times 6} = \frac{18750}{46656}$.

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Calculate $P(X=2)$ (probability of exactly 2 sixes):

$P(X=2) = \binom{6}{2} (\frac{1}{6})^2 (\frac{5}{6})^{6-2} = \binom{6}{2} (\frac{1}{6})^2 (\frac{5}{6})^4$.

$\binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15$. $(\frac{1}{6})^2 = \frac{1}{36}$. $(\frac{5}{6})^4 = \frac{5^4}{6^4} = \frac{625}{1296}$.

$P(X=2) = 15 \times \frac{1}{36} \times \frac{625}{1296} = \frac{15}{36} \times \frac{625}{1296} = \frac{5}{12} \times \frac{625}{1296} = \frac{5 \times 625}{12 \times 1296} = \frac{3125}{15552}$.

To add this to fractions with denominator $6^6=46656$, we multiply numerator and denominator by 3:

$P(X=2) = \frac{3125 \times 3}{15552 \times 3} = \frac{9375}{46656}$.

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Now, sum the probabilities for $X=0, X=1,$ and $X=2$:

$P(X \leq 2) = P(X=0) + P(X=1) + P(X=2)$

$P(X \leq 2) = \frac{15625}{46656} + \frac{18750}{46656} + \frac{9375}{46656}$

$P(X \leq 2) = \frac{15625 + 18750 + 9375}{46656}$

$P(X \leq 2) = \frac{43750}{46656}$.

Simplifying the fraction:

$P(X \leq 2) = \frac{\cancel{43750}^{21875}}{\cancel{46656}_{23328}} = \frac{21875}{23328}$.

The probability of throwing at most 2 sixes in 6 throws of a single die is $\frac{21875}{23328}$.

Solution:

The experiment consists of selecting a random sample of 12 articles and checking if each article is defective or not. We define 'success' as an article being defective.

Given that 10% of the articles manufactured are defective, the probability of selecting a defective article in a single trial is:

$p = P(\text{Defective article}) = 10\% = \frac{10}{100} = 0.1$.

The probability of selecting a non-defective article in a single trial is:

$q = P(\text{Non-defective article}) = 1 - p = 1 - 0.1 = 0.9$.

A random sample of 12 articles is taken. Assuming the sample is small compared to the total production or the sampling is done with replacement (implied by "random sample" from a large bulk), each selection is an independent trial, and the probability of success $p$ is constant for each trial. This is a sequence of Bernoulli trials, and the number of defective articles in the sample follows a binomial distribution.

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Let $n$ be the number of trials (articles in the sample), so $n = 12$.

Let $X$ be the random variable representing the number of defective articles in the sample of 12. $X$ follows a binomial distribution with parameters $n=12$ and $p=0.1$. The probability of getting exactly $k$ successes (defective articles) in $n$ trials is given by the binomial probability mass function:

$P(X=k) = \binom{n}{k} p^k q^{n-k}$

For this problem, $n=12$, $p=0.1$, $q=0.9$. We need to find the probability that exactly 9 articles are defective, i.e., $P(X=9)$, where $k=9$ successes.

$P(X=9) = \binom{12}{9} (0.1)^9 (0.9)^{12-9} = \binom{12}{9} (0.1)^9 (0.9)^3$.

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Calculate the components:

$\binom{12}{9} = \binom{12}{12-9} = \binom{12}{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = \frac{1320}{6} = 220$.

$(0.1)^9 = (10^{-1})^9 = 10^{-9}$.

$(0.9)^3 = 0.9 \times 0.9 \times 0.9 = 0.81 \times 0.9 = 0.729$.

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Substitute these values into the probability formula:

$P(X=9) = 220 \times (0.1)^9 \times (0.9)^3$.

$P(X=9) = 220 \times 10^{-9} \times 0.729$.

$P(X=9) = 220 \times 0.000000001 \times 0.729$.

$P(X=9) = 160.38 \times 10^{-9}$.

$P(X=9) = 0.00000016038$.

The probability that in a random sample of 12 such articles, 9 are defective is $220 \times (0.1)^9 \times (0.9)^3$ or $0.00000016038$.

Solution:

The experiment is selecting a sample of 5 bulbs from a box containing 100 bulbs, where 10 are defective. We want to find the probability that none of the selected bulbs are defective.

We assume that the sample is drawn with replacement, or the total number of bulbs is large enough that the probability of drawing a defective bulb remains practically constant for each draw. In this case, the draws are independent trials, and we can model this using a binomial distribution (Bernoulli trials).

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Let $n$ be the number of bulbs in the sample (number of trials), so $n = 5$.

Let 'success' be drawing a defective bulb.

The total number of bulbs is 100.

The number of defective bulbs is 10.

The probability of drawing a defective bulb (success) in a single draw is $p = \frac{\text{Number of defective bulbs}}{\text{Total number of bulbs}} = \frac{10}{100} = \frac{1}{10}$.

The probability of drawing a non-defective bulb (failure) in a single draw is $q = 1 - p = 1 - \frac{1}{10} = \frac{9}{10}$.

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Let $X$ be the random variable representing the number of defective bulbs in the sample of 5. $X$ follows a binomial distribution with parameters $n=5$ and $p=\frac{1}{10}$. The probability of getting exactly $k$ defective bulbs in $n$ trials is given by the binomial probability mass function:

$P(X=k) = \binom{n}{k} p^k q^{n-k}$

For this problem, we want to find the probability that none of the bulbs are defective, which means exactly 0 defective bulbs. So, we need to find $P(X=0)$, where $k=0$ successes.

$P(X=0) = \binom{5}{0} (\frac{1}{10})^0 (\frac{9}{10})^{5-0}$.

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Calculate the components:

$\binom{5}{0} = 1$.

$(\frac{1}{10})^0 = 1$.

$(\frac{9}{10})^{5} = \frac{9^5}{10^5}$.

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Substitute these values into the probability formula:

$P(X=0) = 1 \times 1 \times (\frac{9}{10})^5 = (\frac{9}{10})^5$.

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Comparing this result with the given options:

(A) $10^{-1}$

(B) $\left( \frac{1}{2} \right)^5$

(C) $\left( \frac{9}{10} \right)^5$

(D) $\frac{9}{10}$

The calculated probability is $(\frac{9}{10})^5$, which matches option (C).

The correct answer is (C).

Solution:

We are given the probability that a student is not a swimmer. We need to find the probability that out of five students, exactly four are swimmers.

This problem involves a fixed number of independent trials (selecting 5 students), each with two possible outcomes (being a swimmer or not being a swimmer), and a constant probability of success. This is a Bernoulli trial scenario, and the number of swimmers in a sample of 5 follows a binomial distribution.

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Let's define 'success' as a student being a swimmer.

Given the probability that a student is not a swimmer is $\frac{1}{5}$.

The probability that a student is a swimmer (success) is $p = 1 - P(\text{not a swimmer}) = 1 - \frac{1}{5} = \frac{4}{5}$.

The probability that a student is not a swimmer (failure) is $q = \frac{1}{5}$.

The number of students in the sample (number of trials) is $n = 5$.

We want to find the probability that exactly four students are swimmers. Let $X$ be the random variable representing the number of swimmers in the sample of 5. We need to find $P(X=4)$, where $k=4$ successes.

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Using the binomial probability mass function:

$P(X=k) = \binom{n}{k} p^k q^{n-k}$

Substitute the values $n=5$, $k=4$, $p=\frac{4}{5}$, and $q=\frac{1}{5}$:

$P(X=4) = \binom{5}{4} (\frac{4}{5})^4 (\frac{1}{5})^{5-4}$

$P(X=4) = \binom{5}{4} (\frac{4}{5})^4 (\frac{1}{5})^1$.

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Calculate the binomial coefficient $\binom{5}{4}$:

$\binom{5}{4} = \binom{5}{5-4} = \binom{5}{1} = 5$.

So, $P(X=4) = 5 \times (\frac{4}{5})^4 \times \frac{1}{5}$.

This expression matches option (A).

Note that $\binom{5}{1} = 5$, so the expression in option (C), $^5C_1 \frac{1}{5} \left( \frac{4}{5} \right)^4$, is also mathematically equal to $5 \times \frac{1}{5} \times (\frac{4}{5})^4 = 5 \times (\frac{4}{5})^4 \times \frac{1}{5}$, which is the same as option (A). However, option (A) is written in the standard form $\binom{n}{k} p^k q^{n-k}$ where $k=4$ and $p=4/5$ (probability of the characteristic being counted, which is 'swimmer'). Option (C) is equivalent but uses $\binom{5}{1}$ and the probability $q=1/5$ raised to the power of 1 (number of non-swimmers). Given the question asks for the probability of four swimmers, option (A) is the direct application of the formula for counting swimmers as successes.

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Comparing the result with the given options, option (A) matches our derived probability expression.

The correct answer is (A).

Solution:

This is a problem involving conditional probability, specifically using Bayes' Theorem.

Let's define the events:

Let $B_i$ be the event that Box $i$ is selected, for $i = 1, 2, 3, 4$.

Let $A$ be the event that a black ball is drawn.

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Since a box is selected at random from the four boxes, the probability of selecting any specific box is equal:

$P(B_1) = P(B_2) = P(B_3) = P(B_4) = \frac{1}{4}$.

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Now, let's find the probability of drawing a black ball given that a particular box is selected. We can read the number of black balls and the total number of balls in each box from the table:

Total balls in Box I = $3 + 4 + 5 + 6 = 18$. Number of black balls = 3. Probability of drawing black from Box I: $P(A|B_1) = \frac{3}{18} = \frac{1}{6}$.

Total balls in Box II = $2 + 2 + 2 + 2 = 8$. Number of black balls = 2. Probability of drawing black from Box II: $P(A|B_2) = \frac{2}{8} = \frac{1}{4}$.

Total balls in Box III = $1 + 2 + 3 + 1 = 7$. Number of black balls = 1. Probability of drawing black from Box III: $P(A|B_3) = \frac{1}{7}$.

Total balls in Box IV = $4 + 3 + 1 + 5 = 13$. Number of black balls = 4. Probability of drawing black from Box IV: $P(A|B_4) = \frac{4}{13}$.

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We want to find the probability that the ball drawn is from Box III, given that it is black. This is $P(B_3|A)$. Using Bayes' Theorem:

$P(B_3|A) = \frac{P(A|B_3)P(B_3)}{P(A)}$.

We need to find the total probability of drawing a black ball, $P(A)$, using the Law of Total Probability:

$P(A) = P(A|B_1)P(B_1) + P(A|B_2)P(B_2) + P(A|B_3)P(B_3) + P(A|B_4)P(B_4)$.

Substitute the known values:

$P(A) = (\frac{1}{6})(\frac{1}{4}) + (\frac{1}{4})(\frac{1}{4}) + (\frac{1}{7})(\frac{1}{4}) + (\frac{4}{13})(\frac{1}{4})$.

$P(A) = \frac{1}{24} + \frac{1}{16} + \frac{1}{28} + \frac{4}{52}$.

Simplify the last term: $\frac{4}{52} = \frac{1}{13}$.

$P(A) = \frac{1}{24} + \frac{1}{16} + \frac{1}{28} + \frac{1}{13}$.

To sum these fractions, find the Least Common Multiple (LCM) of the denominators 24, 16, 28, and 13.

$24 = 2^3 \times 3$

$16 = 2^4$

$28 = 2^2 \times 7$

$13$ is a prime number.

LCM$(24, 16, 28, 13) = 2^4 \times 3 \times 7 \times 13 = 16 \times 3 \times 7 \times 13 = 48 \times 91 = 4368$.

Convert the fractions to the common denominator:

$\frac{1}{24} = \frac{1 \times 182}{24 \times 182} = \frac{182}{4368}$.

$\frac{1}{16} = \frac{1 \times 273}{16 \times 273} = \frac{273}{4368}$.

$\frac{1}{28} = \frac{1 \times 156}{28 \times 156} = \frac{156}{4368}$.

$\frac{1}{13} = \frac{1 \times 336}{13 \times 336} = \frac{336}{4368}$.

Summing the fractions:

$P(A) = \frac{182}{4368} + \frac{273}{4368} + \frac{156}{4368} + \frac{336}{4368} = \frac{182 + 273 + 156 + 336}{4368} = \frac{947}{4368}$.

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Now, apply Bayes' Theorem to find $P(B_3|A)$:

$P(B_3|A) = \frac{P(A|B_3)P(B_3)}{P(A)}$.

$P(B_3|A) = \frac{(\frac{1}{7})(\frac{1}{4})}{\frac{947}{4368}}$.

$P(B_3|A) = \frac{\frac{1}{28}}{\frac{947}{4368}}$.

Invert the denominator fraction and multiply:

$P(B_3|A) = \frac{1}{28} \times \frac{4368}{947}$.

Simplify the fraction $\frac{4368}{28}$:

$\frac{4368}{28} = 156$.

So, $P(B_3|A) = 1 \times \frac{156}{947} = \frac{156}{947}$.

The probability that the ball drawn is from Box III, given that the ball is black, is $\frac{156}{947}$.

Solution:

A binomial distribution is denoted by $B(n, p)$, where $n$ is the number of trials and $p$ is the probability of success in a single trial.

The given binomial distribution is $B \left( 4, \frac{1}{3} \right)$.

Comparing this with the standard notation $B(n, p)$, we can identify the parameters:

Number of trials, $n = 4$.

Probability of success, $p = \frac{1}{3}$.

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The mean ($\mu$) of a binomial distribution $B(n, p)$ is given by the formula:

$\mu = np$

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Substitute the values of $n$ and $p$ into the formula:

$\mu = 4 \times \frac{1}{3}$

$\mu = \frac{4}{3}$.

The mean of the given binomial distribution is $\frac{4}{3}$.

Solution:

Let $n$ be the number of times the shooter fires. Each fire is an independent trial with two possible outcomes: hitting the target (success) or not hitting the target (failure).

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Let $p$ be the probability of hitting the target in a single fire.

Given, $p = \frac{3}{4}$.

Let $q$ be the probability of not hitting the target in a single fire.

$q = 1 - p = 1 - \frac{3}{4} = \frac{1}{4}$.

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The number of hits in $n$ fires follows a binomial distribution, $X \sim B(n, p)$. The probability of hitting the target exactly $k$ times in $n$ fires is $P(X=k) = \binom{n}{k} p^k q^{n-k}$.

We are interested in the probability of hitting the target at least once, which is $P(X \geq 1)$.

Using the complement rule, $P(X \geq 1) = 1 - P(X < 1)$. The event $X < 1$ means getting exactly 0 hits, i.e., $X=0$.

$P(X \geq 1) = 1 - P(X=0)$.

The probability of getting 0 hits in $n$ fires is:

$P(X=0) = \binom{n}{0} p^0 q^{n-0} = 1 \times 1 \times q^n = q^n$.

So, $P(X \geq 1) = 1 - q^n$.

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We are given that the probability of hitting the target at least once is more than 0.99.

$P(X \geq 1) > 0.99$.

Substitute the expression for $P(X \geq 1)$ and the value of $q$:

$1 - (\frac{1}{4})^n > 0.99$.

Rearrange the inequality:

$1 - 0.99 > (\frac{1}{4})^n$.

$0.01 > (\frac{1}{4})^n$.

Write 0.01 as a fraction:

$\frac{1}{100} > (\frac{1}{4})^n$.

This inequality can be rewritten as:

$(\frac{1}{4})^n < \frac{1}{100}$.

Taking the reciprocal of both sides and reversing the inequality sign:

$4^n > 100$.

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We need to find the minimum integer value of $n$ that satisfies the inequality $4^n > 100$. Let's check values of $n$:

If $n=1$, $4^1 = 4$. $4 \ngtr 100$.

If $n=2$, $4^2 = 16$. $16 \ngtr 100$.

If $n=3$, $4^3 = 64$. $64 \ngtr 100$.

If $n=4$, $4^4 = 256$. $256 > 100$.

The smallest integer value of $n$ for which the inequality holds is 4.

Therefore, the shooter must fire at least 4 times.

The minimum number of times he/she must fire is 4.

Solution:

The game involves two players, A and B, throwing a die alternatively. The first player to get a '6' wins the game.

Let 'success' be the event of getting a '6' in a single throw of a fair die.

The possible outcomes when throwing a standard die are {1, 2, 3, 4, 5, 6}. The outcome '6' is one of these.

The probability of getting a '6' (success) in a single throw is:

$p = P(\text{getting a '6'}) = \frac{1}{6}$.

The probability of not getting a '6' (failure) in a single throw is:

$q = P(\text{not getting a '6'}) = 1 - p = 1 - \frac{1}{6} = \frac{5}{6}$.

The throws are independent trials.

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Probability of A winning:

Player A starts the game. A can win on their first turn, or their second turn, or their third turn, and so on.

A wins on the 1st throw (A's 1st turn) if A gets a '6'. The probability is $p$.

A wins on the 3rd throw (A's 2nd turn) if A fails on the 1st throw, B fails on the 2nd throw, and A gets a '6' on the 3rd throw. The probability is $q \times q \times p = q^2 p$.

A wins on the 5th throw (A's 3rd turn) if A fails (1st), B fails (2nd), A fails (3rd), B fails (4th), and A gets a '6' (5th). The probability is $q \times q \times q \times q \times p = q^4 p$.

In general, A wins on the $(2k-1)$th throw (A's $k$th turn) if the first $2(k-1)$ throws result in failures (A fails $k-1$ times, B fails $k-1$ times) and A succeeds on the $(2k-1)$th throw. The probability is $q^{2(k-1)} p$.

The total probability of A winning is the sum of the probabilities of A winning on each of their turns:

$P(\text{A wins}) = p + q^2 p + q^4 p + q^6 p + \dots$

This is an infinite geometric series with the first term $a = p$ and the common ratio $r = q^2$.

Since $q = \frac{5}{6}$, the common ratio is $r = q^2 = (\frac{5}{6})^2 = \frac{25}{36}$. As $|r| = |\frac{25}{36}| < 1$, the sum of this infinite geometric series is given by the formula $S = \frac{a}{1-r}$.

$P(\text{A wins}) = \frac{p}{1 - q^2}$.

Substitute the values of $p$ and $q$:

$P(\text{A wins}) = \frac{\frac{1}{6}}{1 - (\frac{5}{6})^2} = \frac{\frac{1}{6}}{1 - \frac{25}{36}} = \frac{\frac{1}{6}}{\frac{36 - 25}{36}} = \frac{\frac{1}{6}}{\frac{11}{36}}$.

$P(\text{A wins}) = \frac{1}{6} \times \frac{36}{11} = \frac{\cancel{6} \times 6}{\cancel{6} \times 11} = \frac{6}{11}$.

The probability of A winning is $\frac{6}{11}$.

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Probability of B winning:

Player B can win on their first turn (which is the 2nd throw overall), or their second turn (which is the 4th throw overall), and so on.

B wins on the 2nd throw (B's 1st turn) if A fails on the 1st throw and B gets a '6' on the 2nd throw. The probability is $q \times p = qp$.

B wins on the 4th throw (B's 2nd turn) if A fails (1st), B fails (2nd), A fails (3rd), and B gets a '6' (4th). The probability is $q \times q \times q \times p = q^3 p$.

In general, B wins on the $(2k)$th throw (B's $k$th turn) if the first $(2k-1)$ throws result in failures and B succeeds on the $(2k)$th throw. The probability is $q^{2k-1} p$.

The total probability of B winning is the sum of the probabilities of B winning on each of their turns:

$P(\text{B wins}) = qp + q^3 p + q^5 p + \dots$

This is an infinite geometric series with the first term $a = qp$ and the common ratio $r = q^2$.

$P(\text{B wins}) = \frac{qp}{1 - q^2}$.

Substitute the values of $p$ and $q$:

$P(\text{B wins}) = \frac{(\frac{5}{6})(\frac{1}{6})}{1 - (\frac{5}{6})^2} = \frac{\frac{5}{36}}{1 - \frac{25}{36}} = \frac{\frac{5}{36}}{\frac{11}{36}} = \frac{5}{11}$.

The probability of B winning is $\frac{5}{11}$.

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Alternate method for B's probability:

Since the game ends when either A or B gets a '6', one of them must win. Therefore, the sum of their probabilities of winning must be equal to 1.

$P(\text{A wins}) + P(\text{B wins}) = 1$.

We can find the probability of B winning by subtracting the probability of A winning from 1:

$P(\text{B wins}) = 1 - P(\text{A wins})$.

$P(\text{B wins}) = 1 - \frac{6}{11} = \frac{11}{11} - \frac{6}{11} = \frac{11 - 6}{11} = \frac{5}{11}$.

This confirms the result obtained by summing the series for B.

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Final Answer:

The probability that A wins is $\frac{6}{11}$.

The probability that B wins is $\frac{5}{11}$.

Solution:

This problem can be solved using Bayes' Theorem.

Let's define the events:

Let $C$ be the event that the machine is correctly set up.

Let $I$ be the event that the machine is incorrectly set up.

Let $A$ be the event that the machine produces 2 acceptable items.

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We are given the following probabilities:

Prior probability of a correct setup: $P(C) = 80\% = 0.80$.

Prior probability of an incorrect setup: $P(I) = 1 - P(C) = 1 - 0.80 = 0.20$.

Probability of producing an acceptable item given a correct setup: $p_C = 90\% = 0.90$.

Probability of producing an acceptable item given an incorrect setup: $p_I = 40\% = 0.40$.

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The statement "the machine produces 2 acceptable items" implies that a sample of 2 items was produced and both were found to be acceptable. This can be modeled as a binomial experiment with $n=2$ trials, where a 'success' is producing an acceptable item.

The probability of producing $k$ acceptable items in $n$ trials, given a certain setup state, follows a binomial distribution. For $n=2$ and $k=2$, the probability is $\binom{2}{2} (p')^2 (1-p')^{2-2} = (p')^2$, where $p'$ is the probability of producing an acceptable item for that setup state.

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Probability of producing 2 acceptable items given the machine is correctly set up ($C$):

$P(A|C) = (\text{Probability of acceptable item given } C)^2 = (0.90)^2 = 0.81$.

Probability of producing 2 acceptable items given the machine is incorrectly set up ($I$):

$P(A|I) = (\text{Probability of acceptable item given } I)^2 = (0.40)^2 = 0.16$.

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We want to find the probability that the machine is correctly set up given that 2 acceptable items were produced, i.e., $P(C|A)$.

Using Bayes' Theorem:

$P(C|A) = \frac{P(A|C) P(C)}{P(A)}$

First, we need to calculate the total probability of event $A$ (producing 2 acceptable items) using the Law of Total Probability:

$P(A) = P(A|C) P(C) + P(A|I) P(I)$.

Substitute the calculated and given values:

$P(A) = (0.81)(0.80) + (0.16)(0.20)$.

$P(A) = 0.648 + 0.032$.

$P(A) = 0.680$.

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Now, substitute the values into the Bayes' Theorem formula:

$P(C|A) = \frac{(0.81)(0.80)}{0.680}$.

$P(C|A) = \frac{0.648}{0.680}$.

To simplify the fraction, we can multiply the numerator and denominator by 1000 to remove decimals:

$P(C|A) = \frac{648}{680}$.

Divide both numerator and denominator by their greatest common divisor. Both are divisible by 8:

$\frac{\cancel{648}^{81}}{\cancel{680}_{85}}$.

$P(C|A) = \frac{81}{85}$.

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The probability that the machine is correctly set up, given that it produced 2 acceptable items, is $\frac{81}{85}$.

Solution:

We are asked to find the conditional probability $P(B|A)$ given that $P(A) \neq 0$. The formula for conditional probability is:

$P(B|A) = \frac{P(A \cap B)}{P(A)}$.

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(i) A is a subset of B ($A \subset B$)

If A is a subset of B, it means that every outcome in event A is also an outcome in event B. In this case, the intersection of A and B, $A \cap B$, consists of all outcomes that are in both A and B. Since all outcomes in A are also in B, the intersection $A \cap B$ is simply the set of outcomes in A.

Therefore, $A \cap B = A$.

Consequently, the probability of the intersection is equal to the probability of A:

$P(A \cap B) = P(A)$.

Now, substitute this into the formula for $P(B|A)$:

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{P(A)}{P(A)}$.

Since we are given that $P(A) \neq 0$, we can simplify the expression:

$P(B|A) = 1$.

This means if A is a subset of B, and event A has occurred, then event B must also have occurred (since every outcome in A is also in B), so the probability of B given A is 1.

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(ii) A ∩ B = φ

If the intersection of A and B is the empty set ($\phi$), it means that events A and B are mutually exclusive. They cannot occur at the same time.

The probability of the empty set is 0:

$P(A \cap B) = P(\phi) = 0$.

Now, substitute this into the formula for $P(B|A)$:

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0}{P(A)}$.

Since we are given that $P(A) \neq 0$, the denominator is not zero.

$P(B|A) = 0$.

This means if events A and B are mutually exclusive, and event A has occurred, then event B cannot have occurred, so the probability of B given A is 0.

Solution:

Let's denote 'Male' by M and 'Female' by F. We consider the gender of the children in the order of their birth (elder child first, then younger child). The sample space of the gender of two children is:

$S = \{MM, MF, FM, FF\}$.

Assuming that the probability of having a male child is equal to the probability of having a female child, and the gender of the children are independent, the probability of each outcome in the sample space is equally likely, i.e., $\frac{1}{4}$.

$P(MM) = \frac{1}{4}$

$P(MF) = \frac{1}{4}$

$P(FM) = \frac{1}{4}$

$P(FF) = \frac{1}{4}$

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(i) Find the probability that both children are males, if it is known that at least one of the children is male.

Let $A$ be the event that both children are males: $A = \{MM\}$.

Let $B$ be the event that at least one of the children is male. The outcomes where at least one child is male are $\{MM, MF, FM\}$. So, $B = \{MM, MF, FM\}$.

We need to find the conditional probability $P(A|B) = \frac{P(A \cap B)}{P(B)}$.

First, find the intersection of events A and B:

$A \cap B = \{MM\} \cap \{MM, MF, FM\} = \{MM\}$.

The probability of the intersection is $P(A \cap B) = P(\{MM\}) = \frac{1}{4}$.

Next, find the probability of event B:

$P(B) = P(\{MM, MF, FM\}) = P(MM) + P(MF) + P(FM) = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \frac{3}{4}$.

Now, apply the conditional probability formula:

$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{1}{4}}{\frac{3}{4}}$.

$P(A|B) = \frac{1}{4} \times \frac{4}{3} = \frac{1}{3}$.

The probability that both children are males, given that at least one is male, is $\frac{1}{3}$.

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(ii) Find the probability that both children are females, if it is known that the elder child is a female.

Let $A'$ be the event that both children are females: $A' = \{FF\}$.

Let $C$ be the event that the elder child is a female. The outcomes where the elder child is female are $\{FM, FF\}$. So, $C = \{FM, FF\}$.

We need to find the conditional probability $P(A'|C) = \frac{P(A' \cap C)}{P(C)}$.

First, find the intersection of events A' and C:

$A' \cap C = \{FF\} \cap \{FM, FF\} = \{FF\}$.

The probability of the intersection is $P(A' \cap C) = P(\{FF\}) = \frac{1}{4}$.

Next, find the probability of event C:

$P(C) = P(\{FM, FF\}) = P(FM) + P(FF) = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$.

Now, apply the conditional probability formula:

$P(A'|C) = \frac{P(A' \cap C)}{P(C)} = \frac{\frac{1}{4}}{\frac{1}{2}}$.

$P(A'|C) = \frac{1}{4} \times \frac{2}{1} = \frac{2}{4} = \frac{1}{2}$.

The probability that both children are females, given that the elder child is female, is $\frac{1}{2}$.

Solution:

This is a problem that can be solved using Bayes' Theorem.

Let's define the events:

Let $M$ be the event that the selected person is male.

Let $W$ be the event that the selected person is female.

Let $G$ be the event that the selected person has grey hair.

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We are given the following probabilities:

Percentage of men with grey hair: $P(G|M) = 5\% = 0.05$.

Percentage of women with grey hair: $P(G|W) = 0.25\% = 0.0025$.

It is assumed that there are equal numbers of males and females in the population from which the person is selected. Therefore, the prior probabilities of selecting a male or a female are:

$P(M) = \frac{1}{2} = 0.5$.

$P(W) = \frac{1}{2} = 0.5$.

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We want to find the probability that the selected person is male, given that the person has grey hair. This is the conditional probability $P(M|G)$.

Using Bayes' Theorem, the formula is:

$P(M|G) = \frac{P(G|M) P(M)}{P(G)}$.

First, we need to calculate the total probability of event $G$ (selecting a person with grey hair) using the Law of Total Probability:

$P(G) = P(G|M) P(M) + P(G|W) P(W)$.

Substitute the given values into the formula for $P(G)$:

$P(G) = (0.05)(0.5) + (0.0025)(0.5)$.

$P(G) = 0.025 + 0.00125$.

$P(G) = 0.02625$.

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Now, substitute the values $P(G|M)$, $P(M)$, and $P(G)$ into the Bayes' Theorem formula for $P(M|G)$:

$P(M|G) = \frac{(0.05)(0.5)}{0.02625}$.

$P(M|G) = \frac{0.025}{0.02625}$.

To simplify the fraction, multiply the numerator and the denominator by 10000 to remove the decimals:

$P(M|G) = \frac{0.025 \times 10000}{0.02625 \times 10000} = \frac{250}{262.5 \times 10} = \frac{2500}{2625}$.

Now, simplify the fraction $\frac{2500}{2625}$ by dividing both numerator and denominator by their greatest common divisor. Both are divisible by 25:

$2500 \div 25 = 100$.

$2625 \div 25 = 105$.

$P(M|G) = \frac{\cancel{2500}^{100}}{\cancel{2625}_{105}}$.

The fraction is now $\frac{100}{105}$. Both are divisible by 5:

$100 \div 5 = 20$.

$105 \div 5 = 21$.

$P(M|G) = \frac{\cancel{100}^{20}}{\cancel{105}_{21}} = \frac{20}{21}$.

The probability that the grey-haired person selected at random is male is $\frac{20}{21}$.

Solution:

The experiment is drawing a ball from an urn, noting its mark, and replacing it, repeated 6 times. This is sampling with replacement from a finite population, which constitutes a sequence of Bernoulli trials.

Total number of balls in the urn = 25.

Number of balls with mark 'X' = 10.

Number of balls with mark 'Y' = 15.

The number of draws (trials) is $n = 6$.

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Let's define the probabilities for a single draw:

Probability of drawing a ball with mark 'X', $p_X = \frac{\text{Number of 'X' balls}}{\text{Total balls}} = \frac{10}{25} = \frac{2}{5}$.

Probability of drawing a ball with mark 'Y', $p_Y = \frac{\text{Number of 'Y' balls}}{\text{Total balls}} = \frac{15}{25} = \frac{3}{5}$.

Note that $p_X + p_Y = \frac{2}{5} + \frac{3}{5} = 1$.

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We will use the binomial probability mass function $P(k) = \binom{n}{k} p^k q^{n-k}$, where $n=6$. The parameter $p$ will depend on whether we are counting 'X' marks or 'Y' marks as 'success'.

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(i) Probability that all will bear 'X' mark.

Here, 'success' is getting an 'X' mark. $p = p_X = \frac{2}{5}$ and $q = p_Y = \frac{3}{5}$. We want exactly 6 'X' marks, so $k=6$.

$P(X=6) = \binom{6}{6} (p_X)^6 (p_Y)^{6-6} = \binom{6}{6} (\frac{2}{5})^6 (\frac{3}{5})^0$.

$\binom{6}{6} = 1$. $(\frac{3}{5})^0 = 1$.

$P(X=6) = 1 \times (\frac{2}{5})^6 \times 1 = (\frac{2}{5})^6 = \frac{2^6}{5^6} = \frac{64}{15625}$.

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(ii) Probability that not more than 2 will bear 'Y' mark.

Here, 'success' is getting a 'Y' mark. $p = p_Y = \frac{3}{5}$ and $q = p_X = \frac{2}{5}$. Let $Y$ be the number of 'Y' marks. We want $P(Y \leq 2)$, which means $P(Y=0) + P(Y=1) + P(Y=2)$.

$P(Y=k) = \binom{6}{k} (p_Y)^k (p_X)^{6-k} = \binom{6}{k} (\frac{3}{5})^k (\frac{2}{5})^{6-k}$.

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$P(Y=0) = \binom{6}{0} (\frac{3}{5})^0 (\frac{2}{5})^{6-0} = 1 \times 1 \times (\frac{2}{5})^6 = (\frac{2}{5})^6 = \frac{64}{15625}$.

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$P(Y=1) = \binom{6}{1} (\frac{3}{5})^1 (\frac{2}{5})^{6-1} = 6 \times (\frac{3}{5})^1 (\frac{2}{5})^5 = 6 \times \frac{3}{5} \times \frac{32}{3125} = \frac{18}{5} \times \frac{32}{3125} = \frac{576}{15625}$.

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$P(Y=2) = \binom{6}{2} (\frac{3}{5})^2 (\frac{2}{5})^{6-2} = 15 \times (\frac{3}{5})^2 (\frac{2}{5})^4 = 15 \times \frac{9}{25} \times \frac{16}{625}$.

$P(Y=2) = \frac{\cancel{15}^{3}}{\cancel{25}_{5}} \times \frac{9}{1} \times \frac{16}{625} = \frac{3 \times 9 \times 16}{5 \times 625} = \frac{432}{3125}$.

To add this to fractions with denominator 15625, multiply numerator and denominator by 5: $\frac{432 \times 5}{3125 \times 5} = \frac{2160}{15625}$.

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$P(Y \leq 2) = P(Y=0) + P(Y=1) + P(Y=2) = \frac{64}{15625} + \frac{576}{15625} + \frac{2160}{15625}$.

$P(Y \leq 2) = \frac{64 + 576 + 2160}{15625} = \frac{2800}{15625}$.

Simplify the fraction by dividing by 25:

$\frac{\cancel{2800}^{112}}{\cancel{15625}_{625}} = \frac{112}{625}$.

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(iii) Probability that at least one ball will bear 'Y' mark.

Using 'success' as getting a 'Y' mark ($p_Y = \frac{3}{5}$), we want $P(Y \geq 1)$.

Using the complement rule, $P(Y \geq 1) = 1 - P(Y < 1) = 1 - P(Y=0)$.

From part (ii), $P(Y=0) = (\frac{2}{5})^6 = \frac{64}{15625}$.

$P(Y \geq 1) = 1 - \frac{64}{15625} = \frac{15625 - 64}{15625} = \frac{15561}{15625}$.

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(iv) The number of balls with 'X' mark and 'Y' mark will be equal.

In 6 draws, having an equal number of 'X' and 'Y' marks means having 3 'X' marks and 3 'Y' marks. We can count either 'X' or 'Y' as success. Let's use 'X' as success, so $p = p_X = \frac{2}{5}$ and $q = p_Y = \frac{3}{5}$. We want $P(X=3)$, where $k=3$ successes.

$P(X=3) = \binom{6}{3} (p_X)^3 (p_Y)^{6-3} = \binom{6}{3} (\frac{2}{5})^3 (\frac{3}{5})^3$.

Calculate the components:

$\binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$.

$(\frac{2}{5})^3 = \frac{2^3}{5^3} = \frac{8}{125}$.

$(\frac{3}{5})^3 = \frac{3^3}{5^3} = \frac{27}{125}$.

Substitute these values:

$P(X=3) = 20 \times \frac{8}{125} \times \frac{27}{125} = \frac{20 \times 8 \times 27}{125 \times 125} = \frac{160 \times 27}{15625} = \frac{4320}{15625}$.

Simplify the fraction by dividing by 5:

$\frac{\cancel{4320}^{864}}{\cancel{15625}_{3125}} = \frac{864}{3125}$.

Solution:

The race involves crossing 10 hurdles. For each hurdle, there are two possible outcomes for the player: either he clears it or he knocks it down. We are given the probability of clearing a hurdle.

Let's define 'success' as the player knocking down a hurdle.

The probability of clearing a hurdle is given as $\frac{5}{6}$.

The probability of knocking down a hurdle (success) in a single attempt is $p = 1 - P(\text{clearing a hurdle}) = 1 - \frac{5}{6} = \frac{1}{6}$.

The probability of not knocking down a hurdle (failure, which means clearing it) is $q = \frac{5}{6}$.

The player attempts to cross 10 hurdles. Each attempt is an independent trial, and the probability of success $p$ is constant for each hurdle. Thus, this is a sequence of Bernoulli trials, and the number of hurdles knocked down in 10 attempts follows a binomial distribution.

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Let $n$ be the number of trials (hurdles), so $n = 10$.

Let $X$ be the random variable representing the number of hurdles knocked down out of 10. $X$ follows a binomial distribution with parameters $n=10$ and $p=\frac{1}{6}$. The probability of knocking down exactly $k$ hurdles in $n$ trials is given by the binomial probability mass function:

$P(X=k) = \binom{n}{k} p^k q^{n-k}$

For this problem, $n=10$, $p=\frac{1}{6}$, $q=\frac{5}{6}$.

$P(X=k) = \binom{10}{k} (\frac{1}{6})^k (\frac{5}{6})^{10-k}$.

We need to find the probability that he will knock down fewer than 2 hurdles. This means the number of knocked down hurdles is 0 or 1. We need to find $P(X < 2)$.

$P(X < 2) = P(X=0) + P(X=1)$.

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Calculate $P(X=0)$ (probability of exactly 0 hurdles knocked down):

$P(X=0) = \binom{10}{0} (\frac{1}{6})^0 (\frac{5}{6})^{10-0} = \binom{10}{0} (\frac{1}{6})^0 (\frac{5}{6})^{10}$.

We know that $\binom{10}{0} = 1$ and $(\frac{1}{6})^0 = 1$.

$P(X=0) = 1 \times 1 \times (\frac{5}{6})^{10} = (\frac{5}{6})^{10} = \frac{5^{10}}{6^{10}}$.

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Calculate $P(X=1)$ (probability of exactly 1 hurdle knocked down):

$P(X=1) = \binom{10}{1} (\frac{1}{6})^1 (\frac{5}{6})^{10-1} = \binom{10}{1} (\frac{1}{6})^1 (\frac{5}{6})^9$.

We know that $\binom{10}{1} = 10$.

$P(X=1) = 10 \times \frac{1}{6} \times (\frac{5}{6})^9 = \frac{10}{6} \times \frac{5^9}{6^9}$.

Simplify $\frac{10}{6} = \frac{5}{3}$.

$P(X=1) = \frac{5}{3} \times \frac{5^9}{6^9}$. To express this with the denominator $6^{10}$, we can write:

$P(X=1) = 10 \times \frac{1}{6} \times \frac{5^9}{6^9} = \frac{10 \times 5^9}{6 \times 6^9} = \frac{10 \times 5^9}{6^{10}}$.

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Now, sum the probabilities for $X=0$ and $X=1$:

$P(X < 2) = P(X=0) + P(X=1) = \frac{5^{10}}{6^{10}} + \frac{10 \times 5^9}{6^{10}}$.

Combine the terms with the common denominator:

$P(X < 2) = \frac{5^{10} + 10 \times 5^9}{6^{10}}$.

Factor out $5^9$ from the numerator:

$5^{10} + 10 \times 5^9 = (5 \times 5^9) + (10 \times 5^9) = (5 + 10) \times 5^9 = 15 \times 5^9$.

So, $P(X < 2) = \frac{15 \times 5^9}{6^{10}}$.

This expression can be simplified further by expressing 15 and 6 in terms of their prime factors:

$15 = 3 \times 5$.

$6^{10} = (2 \times 3)^{10} = 2^{10} \times 3^{10}$.

$P(X < 2) = \frac{(3 \times 5) \times 5^9}{2^{10} \times 3^{10}} = \frac{3^1 \times 5^{10}}{2^{10} \times 3^{10}}$.

Cancel out a factor of $3^1$ from the numerator and denominator:

$P(X < 2) = \frac{5^{10}}{2^{10} \times 3^{10-1}} = \frac{5^{10}}{2^{10} \times 3^9}$.

The probability that the player will knock down fewer than 2 hurdles is $\frac{5^{10}}{2^{10} \times 3^9}$ or $\frac{15 \times 5^9}{6^{10}}$.

Solution:

The experiment is throwing a die repeatedly until a certain number of successes (getting a six) is achieved. We are specifically interested in the event where the third six occurs on the sixth throw.

Let 'success' be the event of getting a '6' in a single throw of a fair die.

The probability of getting a '6' (success) in a single throw is $p = \frac{1}{6}$.

The probability of not getting a '6' (failure) in a single throw is $q = 1 - p = 1 - \frac{1}{6} = \frac{5}{6}$.

The throws are independent trials.

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We want the probability that the third six is obtained on the sixth throw. This means two conditions must be met:

1. In the first 5 throws, there must be exactly two sixes.

2. The 6th throw must be a six.

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Let's calculate the probability of the first condition: getting exactly 2 sixes in the first 5 throws. This is a standard binomial probability problem with $n'=5$ trials and $k'=2$ successes.

The probability of getting exactly $k'$ successes in $n'$ independent trials is given by the binomial probability formula: $P(k') = \binom{n'}{k'} p^{k'} q^{n'-k'}$.

For this part, $n'=5$, $k'=2$, $p=\frac{1}{6}$, and $q=\frac{5}{6}$.

$P(\text{2 sixes in first 5 throws}) = \binom{5}{2} (\frac{1}{6})^2 (\frac{5}{6})^{5-2} = \binom{5}{2} (\frac{1}{6})^2 (\frac{5}{6})^3$.

Calculate the binomial coefficient $\binom{5}{2}$:

$\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5!}{2!3!} = \frac{5 \times 4}{2 \times 1} = 10$.

Substitute the values:

$P(\text{2 sixes in first 5 throws}) = 10 \times (\frac{1}{6})^2 \times (\frac{5}{6})^3 = 10 \times \frac{1}{36} \times \frac{125}{216}$.

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Now, consider the second condition: the 6th throw must be a six.

The probability of getting a six on the 6th throw is $p = \frac{1}{6}$. This throw is independent of the first 5 throws.

---

The probability of both conditions occurring (getting exactly 2 sixes in the first 5 throws AND getting a six on the 6th throw) is the product of their individual probabilities due to independence:

$P(\text{3rd six on 6th throw}) = P(\text{2 sixes in first 5}) \times P(\text{six on 6th})$.

$P(\text{3rd six on 6th throw}) = \left( 10 \times \frac{1}{36} \times \frac{125}{216} \right) \times \frac{1}{6}$.

$P(\text{3rd six on 6th throw}) = 10 \times \frac{1 \times 125 \times 1}{36 \times 216 \times 6}$.

Calculate the denominator: $36 \times 216 \times 6 = 7776 \times 6 = 46656$.

Calculate the numerator: $10 \times 1 \times 125 \times 1 = 1250$.

$P(\text{3rd six on 6th throw}) = \frac{1250}{46656}$.

Simplify the fraction by dividing both numerator and denominator by 2:

$P(\text{3rd six on 6th throw}) = \frac{\cancel{1250}^{625}}{\cancel{46656}_{23328}} = \frac{625}{23328}$.

The probability of obtaining the third six in the sixth throw of the die is $\frac{625}{23328}$.

Solution:

A leap year has 366 days.

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We know that there are 7 days in a week.

To find the number of full weeks and remaining days in a leap year, we divide the total number of days by 7:

$366 \div 7$

We can express 366 as:

$366 = (52 \times 7) + 2$

This means a leap year consists of 52 full weeks and an additional 2 days.

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The 52 full weeks guarantee 52 occurrences of each day of the week (52 Mondays, 52 Tuesdays, ..., 52 Sundays).

To have 53 Tuesdays, one of the two additional days must be a Tuesday.

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The two additional days must be consecutive days of the week. The possible pairs of consecutive days at the end of the year are:

(Monday, Tuesday)

(Tuesday, Wednesday)

(Wednesday, Thursday)

(Thursday, Friday)

(Friday, Saturday)

(Saturday, Sunday)

(Sunday, Monday)

There are 7 possible pairs for the two additional days, and each pair is equally likely to occur in a randomly selected leap year.

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For the leap year to contain 53 Tuesdays, the pair of additional days must contain a Tuesday. The pairs that contain a Tuesday are:

(Monday, Tuesday)

(Tuesday, Wednesday)

There are 2 favorable outcomes for the pair of additional days.

---

The probability of having 53 Tuesdays in a randomly selected leap year is the ratio of the number of favorable outcomes to the total number of possible outcomes for the pair of additional days.

$P(\text{53 Tuesdays}) = \frac{\text{Number of favorable pairs}}{\text{Total number of possible pairs}} = \frac{2}{7}$

Solution:

Let 'success' be the event that the experiment succeeds in a single trial.

Let 'failure' be the event that the experiment fails in a single trial.

Let $p$ be the probability of success in a single trial.

Let $q$ be the probability of failure in a single trial.

We know that the sum of probabilities of success and failure in a trial is 1:

$p + q = 1$.

We are given that the experiment succeeds twice as often as it fails. This means the probability of success is twice the probability of failure:

$p = 2q$.

---

Substitute the relationship $p = 2q$ into the equation $p + q = 1$:

$2q + q = 1$

$3q = 1$

$q = \frac{1}{3}$.

Now, find the probability of success $p$:

$p = 1 - q = 1 - \frac{1}{3} = \frac{2}{3}$.

So, the probability of success in a single trial is $p = \frac{2}{3}$, and the probability of failure is $q = \frac{1}{3}$.

---

The experiment is performed for 6 trials. Each trial is independent. This scenario fits the conditions for a sequence of Bernoulli trials, and the number of successes in 6 trials follows a binomial distribution.

Let $n$ be the number of trials, so $n = 6$.

Let $X$ be the random variable representing the number of successes in 6 trials. $X$ follows a binomial distribution with parameters $n=6$ and $p=\frac{2}{3}$. The probability of getting exactly $k$ successes in $n$ trials is given by the binomial probability mass function:

$P(X=k) = \binom{n}{k} p^k q^{n-k}$

For this problem, $n=6$, $p=\frac{2}{3}$, $q=\frac{1}{3}$.

$P(X=k) = \binom{6}{k} (\frac{2}{3})^k (\frac{1}{3})^{6-k} = \binom{6}{k} \frac{2^k}{3^k} \frac{1^{6-k}}{3^{6-k}} = \binom{6}{k} \frac{2^k}{3^{k + (6-k)}} = \frac{\binom{6}{k} 2^k}{3^6}$.

We need to find the probability of having at least 4 successes. This means the number of successes is 4, 5, or 6. We need to find $P(X \geq 4)$.

$P(X \geq 4) = P(X=4) + P(X=5) + P(X=6)$.

---

Calculate $P(X=4)$ (probability of exactly 4 successes):

$P(X=4) = \binom{6}{4} (\frac{2}{3})^4 (\frac{1}{3})^{6-4} = \binom{6}{4} (\frac{2}{3})^4 (\frac{1}{3})^2$.

$\binom{6}{4} = \binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15$.

$(\frac{2}{3})^4 = \frac{16}{81}$. $(\frac{1}{3})^2 = \frac{1}{9}$.

$P(X=4) = 15 \times \frac{16}{81} \times \frac{1}{9} = \frac{15 \times 16}{81 \times 9} = \frac{240}{729}$.

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Calculate $P(X=5)$ (probability of exactly 5 successes):

$P(X=5) = \binom{6}{5} (\frac{2}{3})^5 (\frac{1}{3})^{6-5} = \binom{6}{5} (\frac{2}{3})^5 (\frac{1}{3})^1$.

$\binom{6}{5} = \binom{6}{1} = 6$.

$(\frac{2}{3})^5 = \frac{32}{243}$. $(\frac{1}{3})^1 = \frac{1}{3}$.

$P(X=5) = 6 \times \frac{32}{243} \times \frac{1}{3} = \frac{6 \times 32}{243 \times 3} = \frac{192}{729}$.

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Calculate $P(X=6)$ (probability of exactly 6 successes):

$P(X=6) = \binom{6}{6} (\frac{2}{3})^6 (\frac{1}{3})^{6-6} = \binom{6}{6} (\frac{2}{3})^6 (\frac{1}{3})^0$.

$\binom{6}{6} = 1$. $(\frac{2}{3})^6 = \frac{64}{729}$. $(\frac{1}{3})^0 = 1$.

$P(X=6) = 1 \times \frac{64}{729} \times 1 = \frac{64}{729}$.

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Now, sum the probabilities for $X=4, X=5,$ and $X=6$:

$P(X \geq 4) = P(X=4) + P(X=5) + P(X=6)$

$P(X \geq 4) = \frac{240}{729} + \frac{192}{729} + \frac{64}{729}$.

$P(X \geq 4) = \frac{240 + 192 + 64}{729} = \frac{496}{729}$.

The probability that in the next six trials, there will be at least 4 successes is $\frac{496}{729}$.

Solution:

Let $n$ be the number of times the man tosses a fair coin. Each toss is an independent trial with two possible outcomes: Head (H) or Tail (T). Since the coin is fair, the probability of getting a head is constant for each toss, and similarly for a tail. This is a sequence of Bernoulli trials.

---

Let 'success' be the event of getting a head in a single toss.

The probability of getting a head (success) is $p = \frac{1}{2}$.

The probability of getting a tail (failure) is $q = 1 - p = 1 - \frac{1}{2} = \frac{1}{2}$.

Let $X$ be the random variable representing the number of heads in $n$ tosses. $X$ follows a binomial distribution $B(n, p)$ with $p=\frac{1}{2}$. The probability of getting exactly $k$ heads in $n$ tosses is $P(X=k) = \binom{n}{k} p^k q^{n-k}$.

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We want to find the minimum number of tosses $n$ such that the probability of having at least one head is more than 90% (or 0.9).

The probability of having at least one head is $P(X \geq 1)$.

Using the complement rule, $P(X \geq 1) = 1 - P(X < 1)$. The event $X < 1$ means getting exactly 0 heads, i.e., $X=0$.

$P(X=0) = \binom{n}{0} (\frac{1}{2})^0 (\frac{1}{2})^{n-0} = 1 \times 1 \times (\frac{1}{2})^n = (\frac{1}{2})^n$.

So, $P(X \geq 1) = 1 - (\frac{1}{2})^n$.

---

We are given the condition that this probability is more than 0.90:

$P(X \geq 1) > 0.90$.

Substitute the expression for $P(X \geq 1)$:

$1 - (\frac{1}{2})^n > 0.90$.

Rearrange the inequality to isolate the term with $n$:

$1 - 0.90 > (\frac{1}{2})^n$.

$0.10 > (\frac{1}{2})^n$.

Convert the decimal to a fraction:

$\frac{1}{10} > (\frac{1}{2})^n$.

This inequality can be written as:

$(\frac{1}{2})^n < \frac{1}{10}$.

Taking the reciprocal of both sides and reversing the inequality sign:

$2^n > 10$.

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We need to find the smallest integer value of $n$ that satisfies the inequality $2^n > 10$. Let's test consecutive integer values for $n$ starting from 1:

For $n=1$, $2^1 = 2$. Is $2 > 10$? No.

For $n=2$, $2^2 = 4$. Is $4 > 10$? No.

For $n=3$, $2^3 = 8$. Is $8 > 10$? No.

For $n=4$, $2^4 = 16$. Is $16 > 10$? Yes.

The smallest integer $n$ that satisfies the inequality is 4.

Therefore, the man must toss the coin at least 4 times.

The minimum number of times he must toss the fair coin is 4.

Solution:

Let the event of getting a '6' be a Success (S), and the event of not getting a '6' be a Failure (F).

When a fair die is thrown, the probability of getting a '6' is $p = P(S) = \frac{1}{6}$.

The probability of not getting a '6' is $q = P(F) = 1 - p = 1 - \frac{1}{6} = \frac{5}{6}$.

The man throws the die at most 3 times and quits as soon as he gets a six. We need to find the expected value of the amount he wins or loses.

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Let's consider the possible outcomes and the amount won/lost in each case:

1. The man gets a '6' on the first throw (S).

Sequence: S

Probability: $P(S) = p = \frac{1}{6}$.

Amount won: $+\textsf{₹}1$ (wins $\textsf{₹}1$ and quits).

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2. The man does not get a '6' on the first throw (F) and gets a '6' on the second throw (S).

Sequence: FS

Probability: $P(FS) = P(F) \times P(S) = q \times p = \frac{5}{6} \times \frac{1}{6} = \frac{5}{36}$.

Amount won: $-\textsf{₹}1$ (loses $\textsf{₹}1$ on 1st throw) $+ \textsf{₹}1$ (wins $\textsf{₹}1$ on 2nd throw) $= \textsf{₹}0$.

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3. The man does not get a '6' on the first throw (F), does not get a '6' on the second throw (F), and gets a '6' on the third throw (S).

Sequence: FFS

Probability: $P(FFS) = P(F) \times P(F) \times P(S) = q \times q \times p = q^2 p = (\frac{5}{6})^2 \times \frac{1}{6} = \frac{25}{36} \times \frac{1}{6} = \frac{25}{216}$.

Amount won: $-\textsf{₹}1$ (loses $\textsf{₹}1$ on 1st) $-\textsf{₹}1$ (loses $\textsf{₹}1$ on 2nd) $+\textsf{₹}1$ (wins $\textsf{₹}1$ on 3rd) $= -\textsf{₹}1$.

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4. The man does not get a '6' on the first (F), second (F), or third throw (F). He stops after the third throw as per his decision.

Sequence: FFF

Probability: $P(FFF) = P(F) \times P(F) \times P(F) = q^3 = (\frac{5}{6})^3 = \frac{125}{216}$.

Amount won: $-\textsf{₹}1$ (loses $\textsf{₹}1$ on 1st) $-\textsf{₹}1$ (loses $\textsf{₹}1$ on 2nd) $-\textsf{₹}1$ (loses $\textsf{₹}1$ on 3rd) $= -\textsf{₹}3$.

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Let $X$ be the random variable representing the amount the man wins/loses. The possible values of $X$ and their probabilities are:

Amount ($x_i$) | Probability ($P(X=x_i)$)

----------------|--------------------------

$\textsf{₹}1$ | $\frac{1}{6} = \frac{36}{216}$

$\textsf{₹}0$ | $\frac{5}{36} = \frac{30}{216}$

$-\textsf{₹}1$ | $\frac{25}{216}$

$-\textsf{₹}3$ | $\frac{125}{216}$

Note: The sum of probabilities is $\frac{36+30+25+125}{216} = \frac{216}{216} = 1$.

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The expected value of the amount won/lost is given by $E(X) = \sum x_i P(X=x_i)$.

$E(X) = (1) \times P(S) + (0) \times P(FS) + (-1) \times P(FFS) + (-3) \times P(FFF)$.

$E(X) = (1) \times \frac{36}{216} + (0) \times \frac{30}{216} + (-1) \times \frac{25}{216} + (-3) \times \frac{125}{216}$.

$E(X) = \frac{36}{216} + 0 - \frac{25}{216} - \frac{375}{216}$.

$E(X) = \frac{36 - 25 - 375}{216}$.

$E(X) = \frac{11 - 375}{216} = \frac{-364}{216}$.

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Simplify the fraction $\frac{-364}{216}$. Both numerator and denominator are divisible by 4:

$\frac{\cancel{-364}^{-91}}{\cancel{216}_{54}}$.

$E(X) = -\frac{91}{54}$.

The expected value is $-\frac{91}{54}$. Since the expected value is negative, it represents an expected loss.

$-\frac{91}{54} \approx -1.685$.

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The expected value of the amount he wins/loses is $-\textsf{₹}\frac{91}{54}$. This means, on average over many such games, the man is expected to lose approximately $\textsf{₹}1.69$ per game.

Solution:

This is a problem of finding posterior probabilities using Bayes' Theorem, given that a certain event (drawing a red marble) has occurred.

Let's define the events:

Let $E_A$ be the event that Box A is selected.

Let $E_B$ be the event that Box B is selected.

Let $E_C$ be the event that Box C is selected.

Let $E_D$ be the event that Box D is selected.

Let $R$ be the event that a red marble is drawn.

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Since one of the four boxes is selected at random, the prior probability of selecting any specific box is equal:

$P(E_A) = P(E_B) = P(E_C) = P(E_D) = \frac{1}{4}$.

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Next, we need to find the conditional probability of drawing a red marble given that a particular box is selected. We can obtain the number of red marbles and the total number of marbles from the table:

Total marbles in Box A = $1 + 6 + 3 = 10$. Number of red marbles = 1.

$P(R|E_A) = \frac{\text{Number of red marbles in A}}{\text{Total marbles in A}} = \frac{1}{10}$.

Total marbles in Box B = $6 + 2 + 2 = 10$. Number of red marbles = 6.

$P(R|E_B) = \frac{\text{Number of red marbles in B}}{\text{Total marbles in B}} = \frac{6}{10}$.

Total marbles in Box C = $8 + 1 + 1 = 10$. Number of red marbles = 8.

$P(R|E_C) = \frac{\text{Number of red marbles in C}}{\text{Total marbles in C}} = \frac{8}{10}$.

Total marbles in Box D = $0 + 6 + 4 = 10$. Number of red marbles = 0.

$P(R|E_D) = \frac{\text{Number of red marbles in D}}{\text{Total marbles in D}} = \frac{0}{10} = 0$.

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We are given that a red marble is drawn, and we want to find the probability that it was drawn from Box A, Box B, or Box C. This is the conditional probability of selecting a specific box given that a red marble was drawn, i.e., $P(E_A|R)$, $P(E_B|R)$, and $P(E_C|R)$.

According to Bayes' Theorem, for any event $E_i$ and event $R$ (where $P(R) \neq 0$):

$P(E_i|R) = \frac{P(R|E_i)P(E_i)}{P(R)}$.

First, we need to find the total probability of drawing a red marble, $P(R)$, using the Law of Total Probability:

$P(R) = P(R|E_A)P(E_A) + P(R|E_B)P(E_B) + P(R|E_C)P(E_C) + P(R|E_D)P(E_D)$.

Substitute the calculated prior and conditional probabilities:

$P(R) = (\frac{1}{10})(\frac{1}{4}) + (\frac{6}{10})(\frac{1}{4}) + (\frac{8}{10})(\frac{1}{4}) + (0)(\frac{1}{4})$.

$P(R) = \frac{1}{40} + \frac{6}{40} + \frac{8}{40} + 0$.

$P(R) = \frac{1 + 6 + 8}{40} = \frac{15}{40}$.

Simplify the fraction: $\frac{\cancel{15}^{3}}{\cancel{40}_{8}} = \frac{3}{8}$.

So, the total probability of drawing a red marble is $P(R) = \frac{3}{8}$.

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Now, we can find the probability that the red marble was drawn from each specified box using Bayes' Theorem:

Probability that the marble was drawn from box A given it is red ($P(E_A|R)$):

$P(E_A|R) = \frac{P(R|E_A)P(E_A)}{P(R)} = \frac{(\frac{1}{10})(\frac{1}{4})}{\frac{3}{8}}$.

$P(E_A|R) = \frac{\frac{1}{40}}{\frac{3}{8}}$.

$P(E_A|R) = \frac{1}{40} \times \frac{8}{3} = \frac{8}{120}$.

Simplify the fraction: $\frac{\cancel{8}^{1}}{\cancel{120}_{15}} = \frac{1}{15}$.

The probability that the red marble was drawn from box A is $\frac{1}{15}$.

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Probability that the marble was drawn from box B given it is red ($P(E_B|R)$):

$P(E_B|R) = \frac{P(R|E_B)P(E_B)}{P(R)} = \frac{(\frac{6}{10})(\frac{1}{4})}{\frac{3}{8}}$.

$P(E_B|R) = \frac{\frac{6}{40}}{\frac{3}{8}}$.

$P(E_B|R) = \frac{6}{40} \times \frac{8}{3} = \frac{48}{120}$.

Simplify the fraction: $\frac{\cancel{48}^{2}}{\cancel{120}_{5}} = \frac{2}{5}$.

The probability that the red marble was drawn from box B is $\frac{2}{5}$.

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Probability that the marble was drawn from box C given it is red ($P(E_C|R)$):

$P(E_C|R) = \frac{P(R|E_C)P(E_C)}{P(R)} = \frac{(\frac{8}{10})(\frac{1}{4})}{\frac{3}{8}}$.

$P(E_C|R) = \frac{\frac{8}{40}}{\frac{3}{8}}$.

$P(E_C|R) = \frac{8}{40} \times \frac{8}{3} = \frac{64}{120}$.

Simplify the fraction: $\frac{\cancel{64}^{8}}{\cancel{120}_{15}} = \frac{8}{15}$.

The probability that the red marble was drawn from box C is $\frac{8}{15}$.

Solution:

This is a problem that can be solved using Bayes' Theorem.

Let's define the events:

Let $H$ be the event that a patient suffers a heart attack.

Let $M$ be the event that the patient chose the meditation and yoga course.

Let $D$ be the event that the patient chose the prescription drug.

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We are given the following probabilities:

Base probability of a heart attack (without any intervention): $P(\text{Heart Attack}) = 40\% = 0.40$.

The patient can choose either option with equal probability. These are the prior probabilities of choosing each treatment:

$P(M) = 0.5$.

$P(D) = 0.5$.

The meditation and yoga course reduces the risk of heart attack by 30%. This means the probability of suffering a heart attack given the meditation and yoga course ($P(H|M)$) is the original probability reduced by 30% of its value.

$P(H|M) = P(\text{Heart Attack}) \times (1 - \text{Risk Reduction})$.

$P(H|M) = 0.40 \times (1 - 0.30) = 0.40 \times 0.70 = 0.28$.

The prescription drug reduces the chances by 25%. This means the probability of suffering a heart attack given the drug ($P(H|D)$) is the original probability reduced by 25% of its value.

$P(H|D) = P(\text{Heart Attack}) \times (1 - \text{Risk Reduction})$.

$P(H|D) = 0.40 \times (1 - 0.25) = 0.40 \times 0.75 = 0.30$.

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We are given that the patient selected at random suffers a heart attack (event $H$ has occurred). We want to find the probability that this patient followed a course of meditation and yoga, given that they suffered a heart attack. This is the posterior probability $P(M|H)$.

Using Bayes' Theorem:

$P(M|H) = \frac{P(H|M) P(M)}{P(H)}$.

First, we need to calculate the total probability of event $H$ (suffering a heart attack). This can happen if the patient chose meditation/yoga AND had a heart attack ($M \cap H$), or if they chose the drug AND had a heart attack ($D \cap H$). Using the Law of Total Probability:

$P(H) = P(H \cap M) + P(H \cap D)$.

Since $P(H \cap M) = P(H|M) P(M)$ and $P(H \cap D) = P(H|D) P(D)$, we have:

$P(H) = P(H|M) P(M) + P(H|D) P(D)$.

Substitute the calculated and given values:

$P(H) = (0.28)(0.5) + (0.30)(0.5)$.

$P(H) = 0.14 + 0.15$.

$P(H) = 0.29$.

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Now, substitute the values $P(H|M)$, $P(M)$, and $P(H)$ into the Bayes' Theorem formula for $P(M|H)$:

$P(M|H) = \frac{(0.28)(0.5)}{0.29}$.

$P(M|H) = \frac{0.14}{0.29}$.

To express this probability as a fraction, we can remove the decimals by multiplying the numerator and denominator by 100:

$P(M|H) = \frac{0.14 \times 100}{0.29 \times 100} = \frac{14}{29}$.

The fraction $\frac{14}{29}$ cannot be simplified further as 29 is a prime number and 14 is not a multiple of 29.

---

The probability that the patient followed a course of meditation and yoga, given that they suffered a heart attack, is $\frac{14}{29}$.

Solution:

Let the second order determinant be represented by a matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$, where each element $a, b, c, d$ can independently take a value of either 0 or 1.

The probability that an individual element is 0 is $P(0) = \frac{1}{2}$.

The probability that an individual element is 1 is $P(1) = \frac{1}{2}$.

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Since there are 4 elements and each can be chosen in 2 ways (0 or 1) independently, the total number of possible second order determinants is $2^4 = 16$.

Each specific determinant (combination of 0s and 1s) is equally likely, with a probability of $(\frac{1}{2})^4 = \frac{1}{16}$.

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The value of the determinant is given by $ad - bc$.

We want to find the probability that the value of the determinant is positive, i.e., $ad - bc > 0$, which means $ad > bc$.

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Since the elements $a, b, c, d$ can only be 0 or 1, the products $ad$ and $bc$ can only take values from $\{0 \times 0, 0 \times 1, 1 \times 0, 1 \times 1\} = \{0, 1\}$.

For the inequality $ad > bc$ to hold with $ad, bc \in \{0, 1\}$, the only possibility is when $ad = 1$ and $bc = 0$.

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Let's analyze the conditions $ad=1$ and $bc=0$ in terms of the elements $a, b, c, d$:

1. $ad = 1$: This is only possible if both $a=1$ AND $d=1$.

2. $bc = 0$: This is possible if $b=0$ (and $c$ is 0 or 1) OR if $c=0$ (and $b$ is 0 or 1). This means at least one of $b$ or $c$ must be 0. The combinations for $(b, c)$ that result in $bc=0$ are $(0, 0), (0, 1), (1, 0)$.

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So, we need the combinations where $a=1$, $d=1$, AND ($b=0$ or $c=0$).

Let's list the possible $(a, b, c, d)$ combinations that satisfy these conditions:

Case 1: $a=1, d=1$, and $bc=0$. The pairs for $(b, c)$ where $bc=0$ are $(0, 0), (0, 1), (1, 0)$.

- $a=1, b=0, c=0, d=1$. Determinant: $\begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} = 1 \times 1 - 0 \times 0 = 1 > 0$.

- $a=1, b=0, c=1, d=1$. Determinant: $\begin{vmatrix} 1 & 0 \\ 1 & 1 \end{vmatrix} = 1 \times 1 - 0 \times 1 = 1 > 0$.

- $a=1, b=1, c=0, d=1$. Determinant: $\begin{vmatrix} 1 & 1 \\ 0 & 1 \end{vmatrix} = 1 \times 1 - 1 \times 0 = 1 > 0$.

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These are the only 3 combinations of elements (out of 16 total possible combinations) that result in a positive determinant value.

Since each specific combination of the 4 elements has a probability of $\frac{1}{16}$, the probability of getting a positive determinant is the sum of the probabilities of these 3 favorable outcomes.

Probability = $P(a=1,b=0,c=0,d=1) + P(a=1,b=0,c=1,d=1) + P(a=1,b=1,c=0,d=1)$.

Probability = $\frac{1}{16} + \frac{1}{16} + \frac{1}{16} = \frac{3}{16}$.

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The probability that the value of the determinant is positive is $\frac{3}{16}$.

Solution:

Let's define the events:

Let A be the event that subsystem A fails.

Let B be the event that subsystem B fails.

We are given the following probabilities:

$P(A) = 0.2$

$P(\text{B fails alone}) = P(B \cap A^c) = 0.15$

$P(\text{A and B fail}) = P(A \cap B) = 0.15$

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(i) P(A fails|B has failed)

We need to find the conditional probability $P(A|B)$. The formula for conditional probability is:

$P(A|B) = \frac{P(A \cap B)}{P(B)}$.

We are given $P(A \cap B) = 0.15$. We need to find $P(B)$, the probability that subsystem B fails.

The event B can be partitioned into two mutually exclusive events: (B fails and A fails) and (B fails and A does not fail). In set notation, $B = (B \cap A) \cup (B \cap A^c)$.

Using the axiom of total probability for disjoint events:

$P(B) = P(B \cap A) + P(B \cap A^c)$.

We are given $P(A \cap B) = 0.15$ and $P(B \cap A^c) = 0.15$.

Substitute these values to find $P(B)$:

$P(B) = 0.15 + 0.15 = 0.30$.

Now, we can calculate $P(A|B)$:

$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.15}{0.30}$.

$P(A|B) = \frac{15}{30} = \frac{1}{2} = 0.5$.

The probability that A fails given B has failed is 0.5.

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(ii) P(A fails alone)

The event "A fails alone" means that subsystem A fails AND subsystem B does not fail. In set notation, this is the event $A \cap B^c$.

The event A (A fails) can be partitioned into two mutually exclusive events: (A fails and B fails) and (A fails and B does not fail). In set notation, $A = (A \cap B) \cup (A \cap B^c)$.

Using the axiom of total probability for disjoint events:

$P(A) = P(A \cap B) + P(A \cap B^c)$.

We are given $P(A) = 0.2$ and $P(A \cap B) = 0.15$.

Substitute these values into the equation:

$0.2 = 0.15 + P(A \cap B^c)$.

Solve for $P(A \cap B^c)$:

$P(A \cap B^c) = 0.2 - 0.15 = 0.05$.

The probability that A fails alone is 0.05.

Solution:

This is a problem involving conditional probability and Bayes' Theorem.

Let's define the events:

Let $T_R$ be the event that a red ball is transferred from Bag I to Bag II.

Let $T_B$ be the event that a black ball is transferred from Bag I to Bag II.

Let $R_{II}$ be the event that a red ball is drawn from Bag II after the transfer.

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First, let's determine the initial composition of the bags:

Bag I contains 3 red balls and 4 black balls. Total balls in Bag I = $3 + 4 = 7$.

Bag II initially contains 4 red balls and 5 black balls. Initial total balls in Bag II = $4 + 5 = 9$.

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We need to find the probabilities of the events $T_R$ and $T_B$. Since one ball is transferred from Bag I, the probability depends on the composition of Bag I.

$P(T_R) = P(\text{drawing a red ball from Bag I}) = \frac{\text{Number of red balls in Bag I}}{\text{Total balls in Bag I}} = \frac{3}{7}$.

$P(T_B) = P(\text{drawing a black ball from Bag I}) = \frac{\text{Number of black balls in Bag I}}{\text{Total balls in Bag I}} = \frac{4}{7}$.

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Next, we need to find the conditional probability of drawing a red ball from Bag II, given the type of ball that was transferred.

Case 1: A red ball was transferred from Bag I to Bag II ($T_R$).

Bag II now contains (4+1) = 5 red balls and 5 black balls. Total balls in Bag II = $5 + 5 = 10$.

$P(R_{II}|T_R) = P(\text{drawing a red ball from Bag II | } T_R) = \frac{\text{Number of red balls in Bag II}}{\text{Total balls in Bag II}} = \frac{5}{10} = \frac{1}{2}$.

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Case 2: A black ball was transferred from Bag I to Bag II ($T_B$).

Bag II now contains 4 red balls and (5+1) = 6 black balls. Total balls in Bag II = $4 + 6 = 10$.

$P(R_{II}|T_B) = P(\text{drawing a red ball from Bag II | } T_B) = \frac{\text{Number of red balls in Bag II}}{\text{Total balls in Bag II}} = \frac{4}{10} = \frac{2}{5}$.

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We are given that the ball drawn from Bag II is red ($R_{II}$ has occurred), and we want to find the probability that the transferred ball was black ($T_B$). This is the posterior probability $P(T_B|R_{II})$.

Using Bayes' Theorem:

$P(T_B|R_{II}) = \frac{P(R_{II}|T_B) P(T_B)}{P(R_{II})}$.

First, we need to calculate the total probability of drawing a red ball from Bag II, $P(R_{II})$, using the Law of Total Probability. This can happen in two ways: (a red ball was transferred AND a red ball was drawn from Bag II) OR (a black ball was transferred AND a red ball was drawn from Bag II).

$P(R_{II}) = P(R_{II} \cap T_R) + P(R_{II} \cap T_B)$.

Using the multiplication rule for conditional probability, $P(R_{II} \cap T_R) = P(R_{II}|T_R)P(T_R)$ and $P(R_{II} \cap T_B) = P(R_{II}|T_B)P(T_B)$.

$P(R_{II}) = P(R_{II}|T_R) P(T_R) + P(R_{II}|T_B) P(T_B)$.

Substitute the probabilities we calculated:

$P(R_{II}) = (\frac{1}{2})(\frac{3}{7}) + (\frac{2}{5})(\frac{4}{7})$.

$P(R_{II}) = \frac{3}{14} + \frac{8}{35}$.

Find a common denominator (LCM of 14 and 35 is 70):

$P(R_{II}) = \frac{3 \times 5}{14 \times 5} + \frac{8 \times 2}{35 \times 2} = \frac{15}{70} + \frac{16}{70} = \frac{15 + 16}{70} = \frac{31}{70}$.

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Now, substitute the values into the Bayes' Theorem formula for $P(T_B|R_{II})$:

$P(T_B|R_{II}) = \frac{P(R_{II}|T_B) P(T_B)}{P(R_{II})}$.

$P(T_B|R_{II}) = \frac{(\frac{2}{5}) (\frac{4}{7})}{\frac{31}{70}}$.

Calculate the numerator: $(\frac{2}{5}) (\frac{4}{7}) = \frac{8}{35}$.

$P(T_B|R_{II}) = \frac{\frac{8}{35}}{\frac{31}{70}}$.

Multiply by the reciprocal of the denominator:

$P(T_B|R_{II}) = \frac{8}{35} \times \frac{70}{31}$.

Simplify by canceling out common factors (35 and 70):

$P(T_B|R_{II}) = \frac{8}{\cancel{35}_{1}} \times \frac{\cancel{70}^{2}}{31} = \frac{8 \times 2}{1 \times 31} = \frac{16}{31}$.

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The probability that the transferred ball was black, given that the ball drawn from Bag II was red, is $\frac{16}{31}$.

Solution:

We are given that A and B are two events such that $P(A) \neq 0$ and $P(B | A) = 1$.

The formula for the conditional probability of event B occurring given that event A has occurred is:

$P(B|A) = \frac{P(A \cap B)}{P(A)}$.

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We are given that $P(B|A) = 1$. Substitute this into the formula:

$1 = \frac{P(A \cap B)}{P(A)}$.

Since we are given that $P(A) \neq 0$, we can multiply both sides of the equation by $P(A)$:

$1 \times P(A) = P(A \cap B)$.

$P(A) = P(A \cap B)$.

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This result tells us that the probability of the intersection of events A and B is equal to the probability of event A. The event $A \cap B$ represents the outcomes that are common to both A and B. The event A represents all outcomes in A.

If $P(A \cap B) = P(A)$, it means that all outcomes in A must also be in the intersection $A \cap B$. This can only happen if all outcomes in A are also outcomes in B.

In set theory terms, if every element of set A is also an element of set B, then A is a subset of B.

Thus, the condition $P(A \cap B) = P(A)$ implies that $A \subset B$.

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Let's consider the given options:

(A) $A \subset B$: If A is a subset of B, then $A \cap B = A$. Therefore, $P(A \cap B) = P(A)$. This matches our derived condition.

(B) $B \subset A$: If B is a subset of A, then $A \cap B = B$. Therefore, $P(A \cap B) = P(B)$. This would imply $P(A) = P(B)$, which is not necessarily true given $P(A \cap B) = P(A)$.

(C) $B = \phi$: If B is the empty set, then $A \cap B = \phi$. $P(A \cap B) = P(\phi) = 0$. This would imply $P(A) = 0$, which contradicts the given condition $P(A) \neq 0$.

(D) $A = \phi$: If A is the empty set, then $P(A) = 0$. This contradicts the given condition $P(A) \neq 0$.

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Based on the analysis, the correct statement is that A is a subset of B.

The correct answer is (A).

Solution:

We are given the inequality $P(A|B) > P(A)$.

By the definition of conditional probability, $P(A|B) = \frac{P(A \cap B)}{P(B)}$, provided $P(B) \neq 0$. We assume $P(B) > 0$ for $P(A|B)$ to be well-defined and the inequality to be meaningful in this context.

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Substitute the formula for $P(A|B)$ into the given inequality:

$\frac{P(A \cap B)}{P(B)} > P(A)$.

Multiply both sides by $P(B)$ (which is positive):

$P(A \cap B) > P(A) \times P(B)$.

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Now consider the conditional probability $P(B|A)$. By definition, $P(B|A) = \frac{P(A \cap B)}{P(A)}$, provided $P(A) \neq 0$. Given the options involve $P(B|A)$, we assume $P(A) > 0$.

We know that $P(A \cap B) > P(A) \times P(B)$. Let's substitute this into the numerator of the expression for $P(B|A)$:

$P(B|A) = \frac{P(A \cap B)}{P(A)} > \frac{P(A) \times P(B)}{P(A)}$.

Since $P(A) > 0$, we can cancel $P(A)$ from the numerator and denominator on the right side of the inequality:

$P(B|A) > P(B)$.

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This result shows that if $P(A|B) > P(A)$, then $P(B|A) > P(B)$ (assuming $P(A) > 0$ and $P(B) > 0$). This means that if the occurrence of B makes A more likely, then the occurrence of A also makes B more likely. Events A and B are positively correlated.

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Let's compare our result with the given options:

(A) $P(B|A) < P(B)$ - This contradicts our finding.

(B) $P(A \cap B) < P(A) . P(B)$ - This contradicts the intermediate inequality we derived, $P(A \cap B) > P(A) \times P(B)$.

(C) $P(B|A) > P(B)$ - This matches our finding.

(D) $P(B|A) = P(B)$ - This is the condition for independence, but the given inequality $P(A|B) > P(A)$ implies dependence.

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Thus, the correct statement is $P(B|A) > P(B)$.

The correct answer is (C).

Solution:

We are given that A and B are two events such that $P(A) \neq 0$ and $P(A) + P(B) – P(A \text{ and } B) = P(A)$.

The given equation can be simplified:

$P(A) + P(B) - P(A \cap B) = P(A)$

Subtract $P(A)$ from both sides:

$P(B) - P(A \cap B) = 0$

This gives us the condition:

The event $A \cap B$ is a subset of event B ($A \cap B \subseteq B$). The equality of their probabilities, $P(A \cap B) = P(B)$, implies that the probability of outcomes in B that are not in $A \cap B$ must be zero. The outcomes in B that are not in $A \cap B$ constitute the set $B \setminus (A \cap B)$, which is equivalent to $B \cap A^c$.

So, the condition $P(B) = P(A \cap B)$ implies $P(B \cap A^c) = 0$.

This means that the event "B occurs and A does not occur" has a probability of 0. In other words, if event B occurs, event A must occur (with probability 1).

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Now let's consider the conditional probabilities in the options. The conditional probability $P(X|Y)$ is defined as $\frac{P(X \cap Y)}{P(Y)}$ when $P(Y) \neq 0$.

Consider option (B): $P(A|B)$. This is the probability of A occurring given that B has occurred.

$P(A|B) = \frac{P(A \cap B)}{P(B)}$.

From condition (i), we have $P(A \cap B) = P(B)$.

Substitute this into the formula for $P(A|B)$:

$P(A|B) = \frac{P(B)}{P(B)}$.

If $P(B) \neq 0$, then $P(A|B) = 1$.

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What if $P(B) = 0$? If $P(B)=0$, then from condition (i), $P(A \cap B) = 0$. The original equation $P(A) + P(B) - P(A \cap B) = P(A)$ becomes $P(A) + 0 - 0 = P(A)$, which is true, and the condition $P(A) \neq 0$ is given. So $P(B)=0$ is a possible scenario.

If $P(B) = 0$, the expression $P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0}{0}$ is undefined according to the standard definition of conditional probability.

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However, the condition $P(B) = P(A \cap B)$ implies that the set of outcomes in B is effectively contained within A (in terms of probability measure). That is, the event B can only happen where A also happens ($P(B \cap A^c)=0$). If B occurs (and $P(B)>0$), then it must be an outcome within $A \cap B$, which means A has also occurred. This is precisely what $P(A|B)=1$ means when $P(B)>0$. In the context of such multiple-choice questions, the option $P(A|B)=1$ is typically considered correct as a consequence of the probabilistic relationship $P(B) = P(A \cap B)$, assuming the conditional probability is well-defined ($P(B) > 0$).

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Let's examine other options for completeness:

(A) $P(B|A) = 1$. $P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{P(B)}{P(A)}$. This is equal to 1 only if $P(B) = P(A)$ (and $P(A) \neq 0$), which is not guaranteed by the given conditions.

(C) $P(B|A) = 0$. $P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{P(B)}{P(A)}$. This is equal to 0 if $P(B) = 0$ (and $P(A) \neq 0$). While $P(B)=0$ is possible, $P(B|A)=0$ is not guaranteed if $P(B)>0$. For example, if $A=\{1,2\}, B=\{1\}$ with $P(\{1\})=P(\{2\})=1/2$, $P(A)=1$, $P(B)=1/2$. $P(A \cap B)=1/2$. $P(A)+P(B)-P(A \cap B) = 1+1/2-1/2=1=P(A)$. $P(B|A) = P(A \cap B)/P(A) = (1/2)/1 = 1/2 \neq 0$.

(D) $P(A|B) = 0$. As shown for option (B), if $P(B)>0$, $P(A|B)=1$, which is not 0. If $P(B)=0$, $P(A|B)$ is undefined.

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The condition $P(B) = P(A \cap B)$ is equivalent to stating that whenever event B occurs, event A must also occur (up to sets of measure zero). This is captured by $P(A|B) = 1$ when $P(B) > 0$. Option (B) is the most fitting description of the relationship implied by the given equation, assuming the context where the conditional probability is defined.

The correct answer is (B).