Chapter 13 Probability (Additional Questions)

Correct option is (A) 0.24

We are given two events A and B with the following probabilities:

$P(A) = 0.4$

$P(B) = 0.8$

$P(B|A) = 0.6$

The probability of the intersection of events A and B, i.e., $P(A \cap B)$.

The formula for conditional probability of event B given that event A has occurred is:

To find $P(A \cap B)$, we can rearrange the above formula:

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

Now, we substitute the given values into the equation:

Therefore, the value of $P(A \cap B)$ is 0.24.

Correct option is (C) $\frac{2}{3}$

A single throw of a fair die.

The probability of getting an even number, given that the number is greater than 3.

Let S be the sample space when a fair die is thrown.

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

Total number of possible outcomes, $n(S) = 6$.

Let A be the event of getting an even number.

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

Let B be the event of getting a number greater than 3.

$B = \{4, 5, 6\}$

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

First, we find the intersection of A and B, which represents the event of getting an even number that is also greater than 3.

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

The number of outcomes in the intersection is $n(A \cap B) = 2$.

The number of outcomes in event B is $n(B) = 3$.

The formula for conditional probability is:

Substituting the values:

We are given the condition that the number obtained is greater than 3. This means our possible outcomes are no longer the entire sample space S, but are reduced to the set B.

Reduced Sample Space (outcomes greater than 3) = $\{4, 5, 6\}$.

Total number of outcomes in this reduced sample space = 3.

Now, we look for the favorable outcomes within this new sample space. The favorable outcome is getting an even number.

Favorable outcomes (even numbers from the reduced space) = $\{4, 6\}$.

Number of favorable outcomes = 2.

The required probability is the ratio of the number of favorable outcomes to the total number of outcomes in the reduced sample space.

$\text{Required Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes in reduced sample space}} = \frac{2}{3}$

Thus, the probability of getting an even number given that the number is greater than 3 is $\frac{2}{3}$.

Correct option is (C) 0.65

We are given two events A and B with the following probabilities:

$P(A) = 0.5$

$P(B) = 0.3$

Events A and B are independent.

The probability of the union of events A and B, i.e., $P(A \cup B)$.

The formula for the probability of the union of two events (also known as the addition rule of probability) is:

Since events A and B are independent, the probability of their intersection is the product of their individual probabilities:

First, let's calculate $P(A \cap B)$ using the given values:

Now, substitute the values of $P(A)$, $P(B)$, and $P(A \cap B)$ into equation (i):

Therefore, the value of $P(A \cup B)$ is 0.65.

(B) $\frac{5}{8} \times \frac{3}{7}$

Explanation:

We need to find the probability of two dependent events happening in sequence. The events are:

1. Drawing a red ball first.
2. Drawing a black ball second, without replacing the first ball.

Step 1: Find the initial total number of balls.

Number of red balls = 5

Number of black balls = 3

Total number of balls = $5 + 3 = 8$

Step 2: Calculate the probability of the first event.

The probability of drawing a red ball first is the number of red balls divided by the total number of balls.

P(first ball is red) = $\frac{\text{Number of red balls}}{\text{Total number of balls}} = \frac{5}{8}$

Step 3: Calculate the probability of the second event.

Since the first ball was drawn without replacement, there is one less ball in the bag.

Number of balls remaining = $8 - 1 = 7$

The number of black balls is still 3, as the first ball drawn was red.

The probability of drawing a black ball second, given that the first was red, is:

P(second is black | first was red) = $\frac{\text{Number of black balls}}{\text{Remaining number of balls}} = \frac{3}{7}$

Step 4: Combine the probabilities.

The probability of both events happening is the product of their individual probabilities.

P(first is red AND second is black) = P(first is red) $\times$ P(second is black | first was red)

P(first is red AND second is black) = $\frac{5}{8} \times \frac{3}{7}$

This matches option (B).

(B) $\times P(E_i)$

Explanation:

The question asks to complete the formula for the Law of Total Probability.

The law is defined under the following conditions:

• Let $S$ be the sample space.
• Let $E_1, E_2, \dots, E_n$ be a set of events that form a partition of the sample space $S$. This means two things:
  1. The events are mutually exclusive: $E_i \cap E_j = \emptyset$ for any $i \neq j$.
  2. The events are exhaustive: $E_1 \cup E_2 \cup \dots \cup E_n = S$.
• Let $A$ be any event associated with the same sample space $S$.

The probability of event $A$ can be expressed as the sum of the probabilities of its intersections with each event in the partition:

From the definition of conditional probability, we know that:

Rearranging this formula to solve for $P(A \cap E_i)$, we get:

Now, substituting this expression back into the sum, we get the Law of Total Probability:

Comparing this with the formula given in the question, $P(A) = \sum\limits_{i=1}^n P(A | E_i) \dots$, the missing part is clearly $\times P(E_i)$.

Therefore, option (B) is the correct choice.

(C) $P(A)$

Explanation:

The question asks to identify the term represented by the denominator in the formula for Bayes' Theorem.

Bayes' Theorem is given as:

The denominator of this expression is $\sum\limits_{j=1}^n P(A | E_j) P(E_j)$.

This expression is the definition of the Law of Total Probability for an event $A$, where $E_1, E_2, \dots, E_n$ form a partition of the sample space.

The Law of Total Probability states:

By comparing the denominator of Bayes' Theorem with the formula for the Law of Total Probability, we can see that they are identical.

Therefore, the denominator represents $P(A)$, the total probability of event A occurring.

The standard form of Bayes' Theorem is often written by substituting the denominator with $P(A)$:

Thus, option (C) is the correct choice.

(B) Sample space of a random experiment

Explanation:

By definition, a random variable is a function that assigns a numerical value to each possible outcome of a random experiment. Let's break this down:

• The set of all possible outcomes of a random experiment is called the sample space, often denoted by $S$.
• The random variable, say $X$, takes an outcome from the sample space and maps it to a real number.

Therefore, a random variable $X$ is a function $X: S \to \mathbb{R}$, where:

• The domain of the function is the sample space ($S$).
• The codomain of the function is the set of real numbers ($\mathbb{R}$).

Example:

Consider the random experiment of tossing two coins. The sample space is:

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

Let's define a random variable $X$ as the "number of heads".

• The domain is the sample space $S = \{HH, HT, TH, TT\}$.
• The random variable $X$ maps each outcome in $S$ to a real number:
  • $X(HH) = 2$
  • $X(HT) = 1$
  • $X(TH) = 1$
  • $X(TT) = 0$
• The range (the set of actual values taken by the function) is $\{0, 1, 2\}$, which is a subset of the real numbers.

From this definition, it is clear that the domain of a random variable is the sample space of the random experiment.

Let's look at the other options:

• (A) Set of real numbers: This is the codomain (or target set) of the random variable, not its domain.
• (C) Set of natural numbers: This could be the range of a discrete random variable, but it's not the domain.
• (D) Set of integers: This could also be the range of a discrete random variable, but it is not the domain.

Thus, option (B) is the correct choice.

(C) $\sum\limits_{i=1}^n p_i = 1$ and $p_i \ge 0$ for all $i$

Explanation:

The set of probabilities associated with the possible values of a discrete random variable is called its probability distribution or probability mass function (PMF). For a function to be a valid probability distribution for a discrete random variable $X$ with values $x_1, x_2, \dots, x_n$ and corresponding probabilities $P(X=x_i) = p_i$, it must satisfy two fundamental conditions:

1. Non-negativity: The probability of any outcome cannot be negative. Therefore, each probability $p_i$ must be greater than or equal to zero.

   $p_i \ge 0$ for all $i=1, 2, \dots, n$

2. Normalization: The sum of the probabilities of all possible outcomes must be equal to 1. Since $x_1, x_2, \dots, x_n$ represent all the possible values the random variable can take, their probabilities must sum to 1.

   $\sum\limits_{i=1}^n p_i = p_1 + p_2 + \dots + p_n = 1$

Now let's evaluate the given options based on these two rules:

• (A) $\sum\limits_{i=1}^n p_i < 1$: This is incorrect. If the sum is less than 1, it implies that not all possible outcomes have been accounted for, which contradicts the problem statement.
• (B) $p_i < 0$ for some $i$: This is incorrect. Probability can never be negative.
• (C) $\sum\limits_{i=1}^n p_i = 1$ and $p_i \ge 0$ for all $i$: This statement correctly includes both conditions required for a valid probability distribution.
• (D) $\sum\limits_{i=1}^n p_i > 1$: This is incorrect. The sum of probabilities for all possible, mutually exclusive outcomes cannot exceed 1.

Therefore, the correct statemen

(C) $\sum\limits_{i=1}^n x_i p_i$

Explanation:

The mean or expected value of a discrete random variable $X$, denoted as $E(X)$ or $\mu$, is a measure of the central tendency of its probability distribution. It is calculated as the weighted average of the possible values that the random variable can take, where the weights are their corresponding probabilities.

Let $X$ be a discrete random variable with possible values $x_1, x_2, \dots, x_n$ and corresponding probabilities $P(X=x_i) = p_i$. The formula for the mean (expected value) is:

This means we multiply each possible value of the random variable by its probability and then sum all these products.

Let's analyze the other options to see why they are incorrect:

• (A) $\sum\limits_{i=1}^n x_i^2 p_i$: This is the formula for the expected value of $X^2$, i.e., $E(X^2)$. It is used to calculate the variance, but it is not the mean of $X$.
• (B) $(\sum\limits_{i=1}^n x_i p_i)^2$: This represents the square of the mean, $(E(X))^2$.
• (D) $\sqrt{\sum\limits_{i=1}^n x_i^2 p_i - (\sum\limits_{i=1}^n x_i p_i)^2}$: This is the formula for the standard deviation of the random variable $X$. The expression inside the square root, $\sum x_i^2 p_i - (\sum x_i p_i)^2$, is the formula for the variance of $X$, i.e., $Var(X) = E(X^2) - [E(X)]^2$.

Therefore, the correct formula for the mean (expected value) is given in option (C).

(B) $0 \times 0.2 + 1 \times 0.5 + 2 \times 0.3 = 0.5 + 0.6 = 1.1$

Explanation:

The mean or expected value, $E(X)$, of a discrete random variable is calculated by summing the product of each possible value of the variable and its corresponding probability.

The formula for the mean is:

From the given probability distribution table:

• The possible values of $X$ ($x_i$) are 0, 1, and 2.
• The corresponding probabilities ($p_i$) are $P(X=0)=0.2$, $P(X=1)=0.5$, and $P(X=2)=0.3$.

We can now apply the formula:

$E(X) = (0 \times P(X=0)) + (1 \times P(X=1)) + (2 \times P(X=2))$

$E(X) = (0 \times 0.2) + (1 \times 0.5) + (2 \times 0.3)$

$E(X) = 0 + 0.5 + 0.6$

$E(X) = 1.1$

This calculation matches the one presented in option (B).

• Option (A) is the sum of probabilities, which must equal 1 for a valid distribution, but it is not the mean.
• Option (C) is the simple average of the values, which is incorrect because it doesn't account for their different probabilities.
• Option (D) is an irrelevant calculation.

Therefore, the correct calculation for the mean is shown in option (B).

(A) The number of trials is fixed.

(B) There are exactly two outcomes for each trial (success/failure).

(D) The probability of success is constant for each trial.

Explanation:

A statistical experiment is classified as a Binomial experiment if it satisfies the following four conditions. An easy way to remember them is the acronym BINS:

• Binary: Each trial can result in only one of two possible outcomes. These are typically called "success" and "failure".
• Independent: The outcome of any trial is independent of the outcomes of all other trials.
• Number: The experiment consists of a fixed number of trials, denoted by $n$.
• Success: The probability of a "success," denoted by $p$, remains the same for every trial. The probability of a "failure" is therefore also constant and is equal to $q = 1-p$.

Let's evaluate the given options based on these characteristics:

• (A) The number of trials is fixed. - This corresponds to the 'N' in BINS. This is a core requirement for a binomial experiment. This statement is correct.
• (B) There are exactly two outcomes for each trial (success/failure). - This corresponds to the 'B' in BINS. This is the "binary" nature of a binomial trial. This statement is correct.
• (C) The trials are dependent. - This contradicts the 'I' in BINS. The trials in a binomial experiment must be independent. This statement is incorrect.
• (D) The probability of success is constant for each trial. - This corresponds to the 'S' in BINS. This is another fundamental condition. This statement is correct.

Therefore, options (A), (B), and (D) are all characteristics of a Binomial experiment.

(A) Both A and R are true and R is the correct explanation of A.

Explanation:

To evaluate this question, we must first check the validity of both the Assertion (A) and the Reason (R), and then determine if the Reason correctly explains the Assertion.

Analysis of Assertion (A):

The assertion states that the probability of getting exactly 3 heads in 5 tosses of a fair coin follows a Binomial distribution with parameters $n=5$ and $p=0.5$.

A Binomial distribution models the number of successes in a fixed number of independent trials. Let's check if the experiment (tossing a coin 5 times) fits the criteria:

• Fixed number of trials (n): Yes, the coin is tossed a fixed number of times, so $n=5$.
• Two possible outcomes: Yes, each toss results in either a Head (let's call it a "success") or a Tail ("failure").
• Independent trials: Yes, the outcome of one coin toss does not influence the outcome of the others.
• Constant probability of success (p): Yes, the coin is fair, so the probability of getting a head is $p=0.5$ for every toss.

Since all four conditions are met, the experiment is a binomial experiment. The parameters are indeed $n=5$ and $p=0.5$. Therefore, Assertion (A) is true.

Analysis of Reason (R):

The reason states that the experiment consists of a fixed number of independent trials with two possible outcomes, and the probability of success is constant. As shown above, these are precisely the defining characteristics of a binomial experiment.

Therefore, Reason (R) is true.

Connecting A and R:

The Assertion (A) makes a claim that the experiment follows a Binomial distribution. The Reason (R) lists the exact conditions that must be met for an experiment to be considered binomial. Since the experiment in A satisfies all the conditions listed in R, R provides the correct and logical justification for why A is true.

Thus, both A and R are true, and R is the correct explanation of A.

(B) $np$

Explanation:

The notation $X \sim B(n, p)$ signifies that the random variable $X$ follows a Binomial distribution.

The parameters of this distribution are:

• $n$: the fixed number of trials in the experiment.
• $p$: the constant probability of success in each trial.

For a binomial distribution, the key measures of central tendency and dispersion have well-defined formulas. The mean (or expected value), denoted by $E(X)$ or $\mu$, represents the average number of successes one would expect if the experiment were repeated many times.

The formula for the mean of a binomial distribution is:

Let's analyze the given options:

• (A) $\sqrt{np(1-p)}$: This is the formula for the standard deviation of a binomial distribution.
• (B) $np$: This is the correct formula for the mean of a binomial distribution.
• (C) $n(1-p)$: This is the expected number of failures, often written as $nq$ where $q=1-p$. It is not the mean number of successes.
• (D) $np^2$: This formula does not represent a standard measure for a binomial distribution.

Therefore, the mean of a binomial distribution with parameters $n$ and $p$ is correctly given by $np$.

(B) $10 \times 0.2 \times 0.8 = 1.6$

Explanation:

The notation $X \sim B(10, 0.2)$ indicates that the random variable $X$ follows a Binomial distribution with the following parameters:

• Number of trials, $n = 10$
• Probability of success, $p = 0.2$

The question asks for the variance of $X$, denoted as $Var(X)$ or $\sigma^2$.

The formula for the variance of a binomial distribution is:

First, we calculate the probability of failure, which is $(1-p)$:

$1-p = 1 - 0.2 = 0.8$

Now, we substitute the values of $n$, $p$, and $(1-p)$ into the variance formula:

$Var(X) = 10 \times 0.2 \times 0.8$

$Var(X) = 2 \times 0.8$

$Var(X) = 1.6$

This calculation directly corresponds to option (B).

• (A) $10 \times 0.2 = 2$: This is the formula for the mean ($np$), not the variance.
• (C) $\sqrt{1.6}$: This is the standard deviation ($\sqrt{np(1-p)}$), which is the square root of the variance.
• (D) $10 \times 0.8 = 8$: This represents the expected number of failures ($n(1-p)$).

Therefore, the correct variance is 1.6.

(B) Independent events

Explanation:

The question asks to identify the relationship between two events $A$ and $B$ given the condition $P(A \cap B) = P(A)P(B)$. Let's analyze the definitions of the terms in the options.

• (A) Mutually exclusive events: Two events are mutually exclusive if they cannot occur at the same time. This means their intersection is empty, so the probability of their intersection is zero, i.e., $P(A \cap B) = 0$. However, the problem states that $P(A)>0$ and $P(B)>0$, which implies $P(A)P(B) > 0$. Since $P(A \cap B) = P(A)P(B)$, it means $P(A \cap B) > 0$. This contradicts the condition for mutual exclusivity. So, this option is incorrect.

• (B) Independent events: Two events are defined as independent if the occurrence of one event does not affect the probability of the other event occurring. The mathematical definition for the independence of two events $A$ and $B$ is precisely:

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

  This exactly matches the condition given in the question. Therefore, this option is correct.

• (C) Dependent events: Two events are dependent if they are not independent. This means the occurrence of one event does affect the probability of the other. Mathematically, this would be expressed as $P(A \cap B) \neq P(A)P(B)$. This is the opposite of the given condition. So, this option is incorrect.

• (D) Exhaustive events: Two events are exhaustive if their union covers the entire sample space, meaning at least one of them must occur. Mathematically, this is $P(A \cup B) = 1$. The condition for independence, $P(A \cap B) = P(A)P(B)$, does not imply that the events are exhaustive. For example, two independent tosses of a coin are independent events, but they are not exhaustive. So, this option is incorrect.

Based on the standard definition of probability, the given equation is the rule for independent events.

(B) $\frac{4}{10} \times \frac{4}{10}$

Explanation:

This problem involves calculating the probability of two consecutive events where the sampling is done with replacement.

Step 1: Define the initial conditions.

Number of white balls = 4

Number of black balls = 6

Total number of balls = $4 + 6 = 10$

Step 2: Calculate the probability of the first event.

The probability of drawing a white ball on the first draw is:

P(first ball is white) = $\frac{\text{Number of white balls}}{\text{Total number of balls}} = \frac{4}{10}$

Step 3: Analyze the effect of replacement on the second event.

The problem states that the first ball is returned to the bag after its colour is noted. This means that for the second draw, the total number of balls and the number of white balls are the same as they were for the first draw.

Total number of balls for the second draw = 10

Number of white balls for the second draw = 4

Because the conditions are unchanged, the two draws are independent events.

Step 4: Calculate the probability of the second event.

The probability of drawing a white ball on the second draw is:

P(second ball is white) = $\frac{\text{Number of white balls}}{\text{Total number of balls}} = \frac{4}{10}$

Step 5: Calculate the combined probability.

The probability of two independent events both occurring is the product of their individual probabilities.

P(both are white) = P(first is white) $\times$ P(second is white)

P(both are white) = $\frac{4}{10} \times \frac{4}{10}$

This matches the expression in option (B).

• Option (A) $\frac{4}{10} \times \frac{3}{9}$ would be correct if the drawing was done without replacement.
• Option (C) $\frac{4}{10} + \frac{4}{10}$ is incorrect as probabilities are multiplied for independent "AND" events, not added.

The correct matching is as follows:

(i) → (c)

(ii) → (b)

(iii) → (d)

(iv) → (a)

Explanation:

(i) Conditional Probability $P(A|B)$ → (c) $\frac{P(A \cap B)}{P(B)}$

This is the fundamental definition of the conditional probability of event A occurring, given that event B has already occurred. It is calculated as the ratio of the probability of both events A and B occurring ($P(A \cap B)$) to the probability of the given event B occurring ($P(B)$), assuming $P(B)>0$.

(ii) Variance of $X$ → (b) $E(X^2) - [E(X)]^2$

The variance of a random variable X, denoted as $Var(X)$ or $\sigma^2$, measures the spread or dispersion of its probability distribution. A common computational formula for variance is the "mean of the squares minus the square of the mean". Here, $E(X^2)$ is the expected value of $X^2$ (the second moment about the origin), and $E(X)$ is the mean of X (the first moment).

(iii) Binomial Probability $P(X=k)$ → (d) $\binom{n}{k} p^k (1-p)^{n-k}$

This is the probability mass function (PMF) for a binomial distribution. It calculates the probability of achieving exactly $k$ successes in $n$ independent Bernoulli trials. In the formula, $p$ is the probability of success in a single trial, and $\binom{n}{k}$ is the binomial coefficient, which counts the number of ways to choose $k$ successful trials from a total of $n$ trials.

(iv) Law of Total Probability → (a) $\sum\limits_{i=1}^n P(A | E_i) P(E_i)$

The Law of Total Probability provides a way to find the probability of an event $A$ by considering a partition of the sample space into events $E_1, E_2, \dots, E_n$. It states that the probability of $A$ is the sum of the probabilities of $A$ occurring given each event $E_i$, weighted by the probability of each $E_i$. The formula gives $P(A)$ and is notably used as the denominator in Bayes' Theorem.

(A) $\frac{0.95 \times 0.005}{0.95 \times 0.005 + 0.01 \times 0.995}$

Explanation:

This problem requires the application of Bayes' Theorem to find a conditional probability. We need to find the probability that a person has the disease given that they tested positive, which is denoted as $P(D|T)$.

Step 1: Identify the given probabilities.

Let's define the events:

• $D$: The event that a person has the disease.
• $D'$: The event that a person does not have the disease (is healthy).
• $T$: The event that the test result is positive.

From the problem statement, we have:

• The probability of having the disease (prior probability): $P(D) = 0.005$.
• The probability of not having the disease is the complement: $P(D') = 1 - P(D) = 1 - 0.005 = 0.995$.
• The probability of a positive test given the person has the disease (test accuracy): $P(T|D) = 0.95$.
• The probability of a positive test given the person is healthy (false positive rate): $P(T|D') = 0.01$.

Step 2: State Bayes' Theorem.

Bayes' Theorem for finding $P(D|T)$ is:

Step 3: Calculate the denominator, $P(T)$, using the Law of Total Probability.

The total probability of testing positive, $P(T)$, is the sum of probabilities of testing positive whether you have the disease or not. $P(T) = P(T \text{ and } D) + P(T \text{ and } D')$ Using the conditional probability rule, this becomes:

Step 4: Substitute the Law of Total Probability into Bayes' Theorem.

This gives the full formula for this problem:

Step 5: Substitute the given values into the formula.

$P(D|T) = \frac{(0.95) \times (0.005)}{(0.95) \times (0.005) + (0.01) \times (0.995)}$

This expression exactly matches the one given in option (A). The numerator represents the probability of a true positive result, and the denominator represents the total probability of a positive result (true positives plus false positives).

(C) {0, 1, 2}

Explanation:

A random variable is a function that assigns a real number to each outcome in the sample space of a random experiment.

Step 1: Identify the sample space.

The random experiment is "two tosses of a fair coin". The set of all possible outcomes (the sample space, S) is:

S = {HH, HT, TH, TT}

where H represents Heads and T represents Tails.

Step 2: Define the random variable X.

The random variable $X$ is defined as "the number of heads" in each outcome.

Step 3: Determine the value of X for each outcome.

We apply the definition of $X$ to each element of the sample space:

• For the outcome HH, the number of heads is 2. So, $X = 2$.
• For the outcome HT, the number of heads is 1. So, $X = 1$.
• For the outcome TH, the number of heads is 1. So, $X = 1$.
• For the outcome TT, the number of heads is 0. So, $X = 0$.

Step 4: List the possible values of X.

The set of all unique numerical values that $X$ can take is therefore {0, 1, 2}. This set is the range of the random variable.

• Option (A) {H, T} represents the outcomes for a single toss.
• Option (B) {HH, HT, TH, TT} is the sample space, not the numerical values of the random variable.
• Option (D) {0.25, 0.5, 0.25} represents the probabilities associated with the values {0, 1, 2}, not the values themselves.

Thus, the correct set of possible values for the random variable $X$ is {0, 1, 2}.

(B) $E(X^2) - (E(X))^2$

Explanation:

The question provides the definition of variance and asks for an equivalent formula. The variance of a random variable $X$, denoted $Var(X)$, is defined as the expected value of the squared deviation of $X$ from its mean $E(X)$.

Definition: $Var(X) = E((X - E(X))^2)$

To find the equivalent formula, we can expand the expression inside the expectation. For simplicity, let's denote the mean $E(X)$ by $\mu$.

So, $Var(X) = E((X - \mu)^2)$.

Step 1: Expand the squared term.

$(X - \mu)^2 = X^2 - 2\mu X + \mu^2$

Step 2: Apply the expectation operator to the expanded expression.

$Var(X) = E(X^2 - 2\mu X + \mu^2)$

Step 3: Use the linearity property of expectation.

The linearity property states that $E(aX + bY + c) = aE(X) + bE(Y) + c$ for random variables $X, Y$ and constants $a, b, c$. In our case, $\mu$ and $\mu^2$ are constants.

$E(X^2 - 2\mu X + \mu^2) = E(X^2) - E(2\mu X) + E(\mu^2)$

$= E(X^2) - 2\mu E(X) + \mu^2$

Step 4: Substitute $\mu = E(X)$ back into the equation.

$Var(X) = E(X^2) - 2E(X) \cdot E(X) + (E(X))^2$

Step 5: Simplify the expression.

$Var(X) = E(X^2) - 2(E(X))^2 + (E(X))^2$

$Var(X) = E(X^2) - (E(X))^2$

This is a very common computational formula for variance, often described as "the mean of the squares minus the square of the mean".

Comparing this result with the given options:

• (A) $E(X^2)$ is the second moment of $X$ about the origin, not the variance.
• (B) $E(X^2) - (E(X))^2$ is the formula we derived.
• (C) $(E(X))^2 - E(X^2)$ is the negative of the variance.
• (D) $E(X) - E(X^2)$ is an incorrect formula.

Therefore, option (B) is the correct answer.

(A) $P(B|A) = P(B)$

(B) $A$ and $B$ are independent events.

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

Explanation:

The given condition is $P(A|B) = P(A)$. This is the intuitive definition of independent events. It means that the knowledge that event B has occurred does not change the probability of event A occurring. Let's analyze each option based on this condition.

(B) $A$ and $B$ are independent events.

This is true by definition. The condition $P(A|B) = P(A)$ is one way of stating that event A is independent of event B. Therefore, A and B are independent events. This statement is correct.

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

This is the formal mathematical definition of two independent events. We can derive it from the given condition and the formula for conditional probability.

The definition of conditional probability is:

We are given that $P(A|B) = P(A)$. Substituting this into the formula:

Multiplying both sides by $P(B)$, we get:

This shows that the given condition implies the formal definition of independence. This statement is correct.

(A) $P(B|A) = P(B)$

This statement says that event B is independent of event A. Independence is a symmetric property. If A is independent of B, then B must be independent of A. We can prove this using the result from (C).

The definition of $P(B|A)$ is:

Since $P(B \cap A) = P(A \cap B)$ and we know from (C) that $P(A \cap B) = P(A)P(B)$, we can substitute:

Canceling $P(A)$ from the numerator and denominator, we get:

This statement is correct.

(D) $A$ and $B$ are mutually exclusive.

Two events are mutually exclusive if they cannot happen at the same time, which means $P(A \cap B) = 0$. From (C), we know that for independent events, $P(A \cap B) = P(A)P(B)$. For these to be compatible, we would need $P(A)P(B) = 0$, which implies either $P(A)=0$ or $P(B)=0$. However, independence is a meaningful concept primarily when the events have a non-zero probability. If $P(A)>0$ and $P(B)>0$, then $P(A \cap B) > 0$, so the events cannot be mutually exclusive. Thus, independence does not imply mutual exclusivity. This statement is incorrect.

(A) Both A and R are true and R is the correct explanation of A.

Explanation:

To answer this question, we need to evaluate the truthfulness of both the Assertion (A) and the Reason (R) and then determine if R correctly explains A.

Analysis of the logical argument:

Let's consider two events, A and B.

1. If A and B are mutually exclusive, it means they cannot happen at the same time. The mathematical consequence is that the probability of their intersection is zero.

   $P(A \cap B) = 0$

   ... (i)

2. If A and B are independent, the probability of their intersection is the product of their individual probabilities.

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

   ... (ii)

Now, let's assume both conditions are true simultaneously. By equating (i) and (ii), we get:

For the product of two probabilities to be zero, at least one of the probabilities must be zero. This means:

Either $P(A) = 0$ or $P(B) = 0$.

An event with a probability of 0 is called an impossible event.

Conclusion on Assertion (A):

Assertion (A) states that if A and B are mutually exclusive, they cannot be independent unless one is an impossible event. Our analysis shows this is precisely the case. Therefore, Assertion (A) is true.

Conclusion on Reason (R):

Reason (R) lays out the exact logical and mathematical steps we followed above: it correctly states the formulas for mutual exclusivity and independence, equates them, and draws the correct conclusion that either $P(A)=0$ or $P(B)=0$. Therefore, Reason (R) is true.

Connecting A and R:

Since Reason (R) provides the complete and correct mathematical proof for the statement made in Assertion (A), R is the correct explanation of A.

(C) Any value within a given interval

Explanation:

The key distinction between a discrete and a continuous random variable lies in the set of values they can assume.

• A discrete random variable is one which can take on only a finite or countably infinite number of distinct values. For example, the number of defects in a product can be 0, 1, 2, ... (countably infinite), or the outcome of a dice roll can be 1, 2, 3, 4, 5, 6 (finite).

• A continuous random variable is one which can take on an uncountably infinite number of possible values. It can assume any value within a specified range or interval. For example, the height of a student could be 165 cm, 165.1 cm, 165.11 cm, or any value between, say, 150 cm and 180 cm.

Let's analyze the given options:

• (A) A finite number of values: This describes a discrete random variable.
• (B) A countably infinite number of values: This also describes a discrete random variable.
• (C) Any value within a given interval: This is the correct definition for a continuous random variable.
• (D) Only integer values: This is a specific case of a discrete random variable.

Therefore, a continuous random variable is characterized by its ability to take any value within a given interval.

(D) Both (B) and (C) are correct.

Explanation:

The notation $X \sim B(n, p)$ means that the random variable $X$ follows a Binomial distribution, where $X$ typically represents the number of successes in $n$ trials, and $p$ is the probability of success in a single trial.

The question asks for the probability of getting $k$ failures. We can approach this in two equivalent ways.

Method 1: By considering the number of successes

If there are $n$ total trials and we have $k$ failures, then the number of successes must be $n-k$. We can use the standard binomial probability formula, $P(X=x) = \binom{n}{x} p^x (1-p)^{n-x}$, where $x$ is the number of successes.

In our case, the number of successes is $x = n-k$. Substituting this into the formula:

$P(\text{k failures}) = P(X = n-k) = \binom{n}{n-k} p^{n-k} (1-p)^{n-(n-k)}$

$P(\text{k failures}) = \binom{n}{n-k} p^{n-k} (1-p)^k$

This expression matches option (C).

Method 2: By treating "failure" as the event of interest

We can define a new random variable, say $Y$, that represents the number of failures. The parameters for this new variable would be:

• Number of trials = $n$
• Probability of a "failure" = $q = 1-p$

We want to find the probability of $k$ failures, which is $P(Y=k)$. Using the standard binomial formula with these new parameters:

$P(Y=k) = \binom{n}{k} (\text{prob of failure})^k (\text{prob of success})^{n-k}$

$P(Y=k) = \binom{n}{k} (1-p)^k p^{n-k}$

This expression matches option (B).

Equivalence of (B) and (C)

The expressions in (B) and (C) are mathematically identical. The probability terms are the same: $p^{n-k}(1-p)^k$. We only need to show that the binomial coefficients are equal.

By the symmetry property of binomial coefficients:

Since $\binom{n}{k} = \binom{n}{n-k}$, the expressions in (B) and (C) are equivalent.

Therefore, both (B) and (C) are correct ways to express the probability of getting $k$ failures, making (D) the best answer.

(A) $\frac{0.6 \times 0.3}{0.6 \times 0.3 + 0.2 \times 0.7}$

Explanation:

This problem is a classic application of Bayes' Theorem. We are asked to find the probability that a customer is loyal, given that their order was for more than $\textsf{₹} 5000$.

Step 1: Define the events and list the given probabilities.

As defined in the case study:

• $L$: The event that the customer is loyal.
• $N$: The event that the customer is new.
• $B$: The event that the order is for more than $\textsf{₹} 5000$.

The given probabilities are:

• $P(L) = 0.3$ (The probability that a customer is loyal)
• $P(N) = 0.7$ (The probability that a customer is new)
• $P(B|L) = 0.6$ (The probability of a big order, given the customer is loyal)
• $P(B|N) = 0.2$ (The probability of a big order, given the customer is new)

Step 2: Identify the probability we need to find.

We need to find the probability that the order came from a loyal customer, given that the order was for more than $\textsf{₹} 5000$. In probability notation, this is $P(L|B)$.

Step 3: Apply Bayes' Theorem.

Bayes' Theorem states:

The denominator, $P(B)$, is the total probability of an order being big. We can find this using the Law of Total Probability, since any big order must come from either a loyal customer or a new customer.

Step 4: Substitute the expression for $P(B)$ into Bayes' Theorem.

This gives us the complete formula for our problem:

Step 5: Substitute the numerical values into the formula.

Plugging in the given probabilities:

$P(L|B) = \frac{0.6 \times 0.3}{0.6 \times 0.3 + 0.2 \times 0.7}$

This expression exactly matches option (A). The numerator represents the probability of the joint event that a customer is loyal AND places a big order. The denominator represents the total probability of a big order, regardless of customer type.

(B) $1-p$

Explanation:

In a binomial experiment, each individual trial is called a Bernoulli trial. A fundamental characteristic of a Bernoulli trial is that it has exactly two possible outcomes.

These two outcomes are:

1. Success
2. Failure

These two outcomes are mutually exclusive (if one occurs, the other cannot) and exhaustive (one of the two must occur). Therefore, they are complementary events.

By the axioms of probability, the sum of the probabilities of a set of exhaustive and mutually exclusive events must equal 1.

So, we can write:

P(Success) + P(Failure) = 1

The standard notation used is:

• P(Success) = $p$
• P(Failure) = $q$

Substituting this notation into the equation, we get:

The question asks for the probability of failure, $q$. To solve for $q$, we simply rearrange the equation:

This matches option (B).

• Option (D) $p(1-p)$ represents the product $pq$, which is used in the formula for the variance ($npq$) but is not the probability of failure itself.

(B) $\sum\limits x_i^2 p_i$

Explanation:

The question asks for the expected value of the square of a discrete random variable $X$, which is denoted as $E(X^2)$.

The general rule for finding the expected value of a function of a discrete random variable, say $g(X)$, is given by the formula:

This means we take each possible value of the random variable, apply the function $g$ to it, multiply by the corresponding probability, and then sum up all these products.

In this specific question, the function is $g(X) = X^2$. Therefore, $g(x_i) = x_i^2$.

Substituting this into the general formula, we get:

Given that $P(X=x_i) = p_i$, the formula becomes:

This expression is often called the "mean of the squares" and is a key component in calculating the variance of $X$.

Let's analyze the other options:

• (A) $(\sum\limits x_i p_i)^2$: This represents $(E(X))^2$, which is the "square of the mean". This is different from the "mean of the squares".
• (C) $\sum\limits x_i p_i^2$: In this formula, the probabilities are squared, which is incorrect.
• (D) $\sum\limits x_i^2 p_i^2$: In this formula, both the values and the probabilities are squared, which is also incorrect.

Therefore, the correct formula for $E(X^2)$ is given in option (B).

(B) $\sqrt{np(1-p)}$

Explanation:

The notation $X \sim B(n, p)$ indicates that the random variable $X$ follows a Binomial distribution with parameters $n$ (the number of trials) and $p$ (the probability of success).

The question asks for the standard deviation of $X$, which is a measure of the amount of variation or dispersion of the random variable. The standard deviation is denoted by $\sigma$ or $SD(X)$.

The standard deviation is defined as the square root of the variance ($Var(X)$).

Step 1: Recall the formula for the variance of a binomial distribution.

The variance of a binomial distribution is given by the formula:

where $q = 1-p$ is the probability of failure.

Step 2: Calculate the standard deviation.

The standard deviation is the positive square root of the variance:

Now let's examine the options:

• (A) $np(1-p)$: This is the formula for the variance, not the standard deviation.
• (B) $\sqrt{np(1-p)}$: This is the correct formula for the standard deviation.
• (C) $np$: This is the formula for the mean (expected value).
• (D) $\sqrt{np}$: This is the square root of the mean, which is not the standard deviation.

Therefore, the correct formula for the standard deviation of a binomial distribution is $\sqrt{np(1-p)}$.

(D) All of the above are steps/values to calculate the probability.

Explanation:

This problem requires us to calculate a probability using the Binomial distribution formula. The experiment fits the binomial criteria: a fixed number of independent trials, each with two outcomes (head/tail) and a constant probability of success.

Step 1: Identify the parameters of the binomial distribution.

• Number of trials, $n = 4$.
• Probability of success (getting a head), $p = 0.6$.
• Probability of failure (getting a tail), $q = 1 - p = 1 - 0.6 = 0.4$.
• Number of successes we want to find, $k = 2$.

Step 2: Apply the Binomial Probability Formula.

The formula for getting exactly $k$ successes in $n$ trials is:

Substituting our values into this formula:

$P(X=2) = \binom{4}{2} (0.6)^2 (0.4)^{4-2}$

$P(X=2) = \binom{4}{2} (0.6)^2 (0.4)^2$

This expression exactly matches option (A).

Step 3: Calculate the value of the expression.

First, calculate the binomial coefficient:

$\binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} = \frac{4 \times 3}{2 \times 1} = 6$

Next, calculate the powers:

$(0.6)^2 = 0.36$

$(0.4)^2 = 0.16$

Now, multiply these values together:

$P(X=2) = 6 \times 0.36 \times 0.16$

This expression matches the first part of option (B).

Continuing the calculation:

$P(X=2) = 6 \times (0.36 \times 0.16) = 6 \times 0.0576 = 0.3456$

This shows that the rest of option (B) is the correct calculation and that option (C) is the correct final value.

Since options (A), (B), and (C) all represent correct stages in finding the solution, the most complete answer is (D).

(C) $p_1 = -0.2, p_2 = 0.6, p_3 = 0.6$

Explanation:

For a set of values to represent a valid probability distribution for a discrete random variable, two fundamental conditions must be met:

1. Non-negativity: The probability of any event must be greater than or equal to zero. For all values $p_i$, it must be that $p_i \ge 0$.
2. Normalization: The sum of the probabilities of all possible outcomes must be equal to 1. In this case, $p_1 + p_2 + p_3 = 1$.

We will check each option against these two rules.

(A) $p_1 = 0.1, p_2 = 0.5, p_3 = 0.4$

• Non-negativity: All probabilities are $\ge 0$. (Condition met)
• Normalization: $0.1 + 0.5 + 0.4 = 1.0$. (Condition met)
• This is a possible distribution.

(B) $p_1 = 0.3, p_2 = 0.7, p_3 = 0$

• Non-negativity: All probabilities are $\ge 0$. (Condition met)
• Normalization: $0.3 + 0.7 + 0 = 1.0$. (Condition met)
• This is a possible distribution.

(C) $p_1 = -0.2, p_2 = 0.6, p_3 = 0.6$

• Non-negativity: The value $p_1 = -0.2$ is negative. Probability cannot be negative. (Condition failed)
• Normalization: $-0.2 + 0.6 + 0.6 = 1.0$. (Even though the sum is 1, the first condition is not met).
• This is NOT a possible distribution because it violates the non-negativity rule.

(D) $p_1 = 0.4, p_2 = 0.4, p_3 = 0.2$

• Non-negativity: All probabilities are $\ge 0$. (Condition met)
• Normalization: $0.4 + 0.4 + 0.2 = 1.0$. (Condition met)
• This is a possible distribution.

Therefore, the set of values in option (C) is not possible for a probability distribution.

(B) $c E(X)$

Explanation:

This question asks about a fundamental property of the expected value operator, $E()$. The expected value is a linear operator, which means it has specific properties when dealing with constants and sums.

The property relevant to this question is that a constant factor can be pulled out of the expectation.

We can prove this using the definition of expected value for a discrete random variable.

Let $X$ be a discrete random variable with possible values $x_1, x_2, \dots, x_n$ and corresponding probabilities $p_1, p_2, \dots, p_n$.

The expected value of $X$ is defined as:

Now, let's find the expected value of $cX$, where $c$ is a constant. The new random variable is $cX$, and its possible values are $cx_1, cx_2, \dots, cx_n$. The probabilities remain the same.

By the definition of expectation:

$E(cX) = \sum\limits_i (cx_i) p_i$

Since $c$ is a constant, we can factor it out of the summation:

$E(cX) = c \sum\limits_i x_i p_i$

We recognize that the summation part, $\sum\limits_i x_i p_i$, is the definition of $E(X)$. Therefore, we can substitute $E(X)$ back into the equation:

This proves the property.

• Option (A) $c + E(X)$ is the result for $E(c+X)$.
• Option (D) $c^2 E(X)$ is incorrect. The property $Var(cX) = c^2 Var(X)$ involves the square of the constant, but this question is about expectation, not variance.

Thus, the correct answer is (B).

(C) $c^2 \text{Var}(X)$

Explanation:

The question asks for a property of the variance operator, $\text{Var}()$, when applied to a random variable $X$ multiplied by a constant $c$.

The definition of the variance of a random variable $Y$ is:

We want to find $\text{Var}(cX)$. Let's set $Y = cX$ and substitute this into the definition.

Step 1: Find the expected value of $cX$.

Using the properties of expectation, a constant factor can be pulled out:

Step 2: Substitute into the variance definition.

We replace $Y$ with $cX$ and $E(Y)$ with $cE(X)$:

$\text{Var}(cX) = E((cX - cE(X))^2)$

Step 3: Factor out the constant $c$ from the inner expression.

$\text{Var}(cX) = E((c(X - E(X)))^2)$

Step 4: Apply the square to both parts inside the parenthesis.

$\text{Var}(cX) = E(c^2(X - E(X))^2)$

Step 5: Pull the constant factor $c^2$ out of the expectation.

Since $c^2$ is a constant, we can move it outside the expectation operator:

$\text{Var}(cX) = c^2 E((X - E(X))^2)$

Step 6: Recognize the definition of $\text{Var}(X)$.

The term $E((X - E(X))^2)$ is the definition of $\text{Var}(X)$. Therefore, we can substitute it back:

This shows that when a random variable is multiplied by a constant, its variance is multiplied by the square of that constant.

• Option (A) $c + \text{Var}(X)$ is incorrect. The property related to adding a constant is $\text{Var}(X+c) = \text{Var}(X)$.
• Option (B) $c \text{Var}(X)$ is incorrect as it misses the square on the constant.

Thus, the correct answer is (C).

(B) 1

Explanation:

A probability distribution for a discrete random variable lists all the possible values the variable can take, along with their corresponding probabilities. For a set of probabilities to form a valid distribution, it must satisfy two fundamental axioms of probability:

1. Non-negativity: The probability of any outcome must be non-negative (i.e., greater than or equal to zero).

2. Normalization: The set of all possible outcomes is exhaustive, meaning one of the outcomes must occur. Therefore, the sum of the probabilities of all possible outcomes must be equal to 1.

If a discrete random variable $X$ can take values $x_1, x_2, \dots, x_n$ with probabilities $p_1, p_2, \dots, p_n$, then the normalization condition is expressed mathematically as:

Thus, the sum of probabilities in any probability distribution must be equal to 1.

(A) $10 \times 0.1 = 1$

Explanation:

This problem can be modeled using a Binomial distribution because it meets the required criteria:

• There is a fixed number of trials (10 bulbs are selected).
• Each trial has only two outcomes (a bulb is either defective or not defective).
• The trials are independent (the state of one bulb does not affect the others).
• The probability of "success" (a bulb being defective) is constant for each trial.

The question asks for the expected number of defective bulbs, which is the mean of the binomial distribution, denoted as $E(X)$.

Step 1: Identify the binomial parameters.

• Number of trials, $n = 10$.
• Probability of success (a bulb being defective), $p = 0.1$.

Step 2: Use the formula for the mean of a binomial distribution.

The mean (expected value) is given by the formula:

Step 3: Substitute the values and calculate.

$E(X) = 10 \times 0.1 = 1$

So, the expected number of defective bulbs in a sample of 10 is 1.

Let's look at the other options:

• (B) $10 \times 0.9 = 9$: This calculates $n(1-p)$, which is the expected number of non-defective bulbs.
• (C) $\sqrt{10 \times 0.1 \times 0.9} = \sqrt{0.9}$: This is the formula for the standard deviation, not the mean.
• (D) $10$: This is simply the total number of trials, $n$.

Therefore, the correct calculation for the expected number of defective bulbs is given in option (A).

(B) 0.6

Explanation:

To find the probability of event B, we use the Addition Rule of Probability, which relates the probabilities of the union and intersection of two events.

The general formula for the addition rule is:

We are given the following values:

• $P(A \cup B) = 0.8$
• $P(A \cap B) = 0.3$
• $P(A) = 0.5$

We need to find $P(B)$. We can rearrange the formula to isolate $P(B)$ on one side of the equation:

Now, we substitute the given values into this rearranged formula:

$P(B) = 0.8 - 0.5 + 0.3$

Performing the calculation:

$P(B) = 0.3 + 0.3$

$P(B) = 0.6$

Therefore, the probability of event B is 0.6, which corresponds to option (B).

(C) $1 - (0.4 \times 0.3) = 1 - 0.12 = 0.88$

(D) $0.6 + 0.7 - 0.6 \times 0.7 = 1.3 - 0.42 = 0.88$

Note: Both (C) and (D) represent correct methods to solve the problem and arrive at the same answer. We will explain both.

Explanation:

Let's define the events:

• $A$: The event that student A solves the problem. $P(A) = 0.6$.
• $B$: The event that student B solves the problem. $P(B) = 0.7$.

The problem states that they try to solve the problem independently. This is a key piece of information.

We need to find the probability that "the problem is solved". This means that at least one of them solves the problem, which is the probability of the union of the two events, $P(A \cup B)$.

There are two common methods to solve this.

Method 1: Using the Complement Rule (as shown in option C)

1. The opposite (complement) of "the problem is solved" is "the problem is NOT solved".
2. "The problem is not solved" means that student A fails AND student B fails.
3. First, find the probabilities of failure for each student:
   • Probability that A fails: $P(A') = 1 - P(A) = 1 - 0.6 = 0.4$.
   • Probability that B fails: $P(B') = 1 - P(B) = 1 - 0.7 = 0.3$.
4. Since the events are independent, the probability that both fail is the product of their individual failure probabilities:

   $P(\text{both fail}) = P(A' \cap B') = P(A') \times P(B') = 0.4 \times 0.3 = 0.12$.

5. The probability that the problem is solved is 1 minus the probability that both fail:

   $P(\text{problem is solved}) = 1 - P(\text{both fail}) = 1 - 0.12 = 0.88$.

This matches the calculation in option (C).

Method 2: Using the Addition Rule for Probability (as shown in option D)

1. The probability that the problem is solved is $P(A \cup B)$.
2. The general addition rule is $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
3. Since the events A and B are independent, the probability of their intersection is the product of their individual probabilities: $P(A \cap B) = P(A) \times P(B)$.
4. Substituting this into the addition rule:

   $P(A \cup B) = P(A) + P(B) - (P(A) \times P(B))$.

5. Now, plug in the given values:

   $P(A \cup B) = 0.6 + 0.7 - (0.6 \times 0.7)$

   $P(A \cup B) = 1.3 - 0.42$

   $P(A \cup B) = 0.88$.

This matches the calculation in option (D).

Since both options (C) and (D) show correct calculations that lead to the correct final probability of 0.88, both are valid representations of the solution.

(B) Mutually exclusive

Explanation:

We are given two conditions about events A and B:

Let's start by expanding the first condition using the definition of conditional probability:

The definition of conditional probability states:

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

Substituting these into equation (i), we get:

Since the intersection is commutative, $P(A \cap B) = P(B \cap A)$. Let's cross-multiply the equation (assuming $P(A)>0$ and $P(B)>0$):

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

Now, let's move all terms to one side:

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

Factor out the common term $P(A \cap B)$:

For the product of two terms to be zero, at least one of the terms must be zero. This gives us two possibilities:

1. $P(A \cap B) = 0$
2. $P(A) - P(B) = 0 \implies P(A) = P(B)$

Now we use the second given condition, $P(A) \neq P(B)$. This condition directly contradicts the second possibility.

Therefore, we are left with only one possibility that must be true:

$P(A \cap B) = 0$

An event where the probability of the intersection is zero is the definition of mutually exclusive events. They cannot occur at the same time.

Let's check the other options:

• (A) Independent: If A and B were independent, then $P(A|B) = P(A)$ and $P(B|A) = P(B)$. The given condition $P(A|B) = P(B|A)$ would imply $P(A) = P(B)$, which is contradicted by the problem statement.
• (C) Equally likely: This means $P(A) = P(B)$, which is also directly contradicted by the problem statement.

Thus, the only valid conclusion is that A and B are mutually exclusive.

(C) Binomial distribution

Explanation:

The problem describes a scenario that perfectly fits the characteristics of a Binomial experiment. Let's verify the four conditions for a binomial distribution (often remembered by the acronym BINS):

1. Binary: Each trial has only two possible outcomes. In this case, each child can be either a boy (let's call this a "success") or a girl ("failure").

2. Independent: The outcome of each trial is independent of the others. The problem states that the gender of each child is independent.

3. Number: There is a fixed number of trials. Here, the number of trials is the number of children, which is fixed at $n=3$.

4. Success: The probability of success is constant for each trial. The probability of having a boy is given as $p=0.5$ for each child.

Since all four conditions are met, the random variable $X$, which represents the number of successes (boys) in the fixed number of trials, follows a Binomial distribution with parameters $n=3$ and $p=0.5$.

The probabilities provided in the table confirm this. They are calculated using the binomial probability formula $P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$:

• $P(X=0) = \binom{3}{0}(0.5)^0(0.5)^3 = 1 \times 1 \times 0.125 = 1/8$
• $P(X=1) = \binom{3}{1}(0.5)^1(0.5)^2 = 3 \times 0.5 \times 0.25 = 0.375 = 3/8$
• $P(X=2) = \binom{3}{2}(0.5)^2(0.5)^1 = 3 \times 0.25 \times 0.5 = 0.375 = 3/8$
• $P(X=3) = \binom{3}{3}(0.5)^3(0.5)^0 = 1 \times 0.125 \times 1 = 1/8$

Let's consider the other options:

• (A) Uniform distribution: This would require all outcomes to have the same probability. Here, $P(X=0) \neq P(X=1)$, so it's not uniform.
• (B) Poisson distribution: This models the number of events in a fixed interval, not the number of successes in a fixed number of trials.
• (D) Normal distribution: This is a continuous distribution, whereas the number of boys is a discrete random variable.

Therefore, the correct classification is a Binomial distribution.

(D) All of the above are correct ways to find the mean, and the value is 1.5.

Explanation:

The question asks for the mean (or expected value) of the random variable $X$, which represents the number of boys. There are several ways to calculate this, and the options present three of them. Let's analyze each one.

Method 1: Using the definition of Expected Value (as in option B)

The fundamental way to calculate the mean of any discrete random variable is to sum the products of each possible value and its corresponding probability. The formula is $E(X) = \sum x_i P(X=x_i)$.

Using the values from the probability distribution table:

$E(X) = (0 \times \frac{1}{8}) + (1 \times \frac{3}{8}) + (2 \times \frac{3}{8}) + (3 \times \frac{1}{8})$

$E(X) = \frac{0}{8} + \frac{3}{8} + \frac{6}{8} + \frac{3}{8}$

$E(X) = \frac{0 + 3 + 6 + 3}{8} = \frac{12}{8} = 1.5$

This is a correct method, and the calculation in option (B) is correct.

Method 2: Using the Binomial Distribution Mean Formula (as in option C)

As established in the previous question, this scenario follows a Binomial distribution with parameters $n=3$ (number of trials/children) and $p=0.5$ (probability of success/boy). The formula for the mean of a binomial distribution is simply $E(X) = np$.

$E(X) = 3 \times 0.5 = 1.5$

This is also a correct and efficient method, and the calculation in option (C) is correct.

Method 3: Using the simple average of values (as in option A)

The calculation is $(0+1+2+3)/4 = 1.5$. While the numerical result is correct, this method is generally not valid for finding the mean of a random variable. It only works in this specific case because the probability distribution is symmetric (since $p=0.5$). For any other value of $p$, this simple average would give an incorrect answer. However, in the context of this question, the calculation itself leads to the correct value.

Conclusion

Since the calculations presented in options (A), (B), and (C) all result in the correct mean of 1.5, and options (B) and (C) represent fundamentally correct methods, option (D) is the best and most comprehensive answer, as it acknowledges that all the listed calculations lead to the correct final value.

(C) $\{2, 3, \dots, 12\}$

Explanation:

A random variable assigns a numerical value to each outcome of a random experiment. In this case, the random variable $X$ is the sum of the numbers that appear when two fair dice are rolled.

Step 1: Determine the range of possible outcomes for a single die.

A single fair die can show any integer value from 1 to 6.

Step 2: Determine the minimum possible sum.

The minimum sum occurs when both dice show the smallest possible value, which is 1.

Minimum value of $X = 1 (\text{from die 1}) + 1 (\text{from die 2}) = 2$.

This means the value 1 is not a possible sum.

Step 3: Determine the maximum possible sum.

The maximum sum occurs when both dice show the largest possible value, which is 6.

Maximum value of $X = 6 (\text{from die 1}) + 6 (\text{from die 2}) = 12$.

Step 4: Determine all possible values between the minimum and maximum.

Any integer value between 2 and 12 can be obtained as a sum. For example:

• Sum of 3: (1, 2) or (2, 1)
• Sum of 4: (1, 3), (2, 2), (3, 1)
• ... and so on up to 12.

Therefore, the set of all possible values for the random variable $X$ is $\{2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\}$.

Let's check the given options:

• (A) $\{1, 2, 3, 4, 5, 6\}$: These are the possible outcomes for a single die, not the sum of two dice.
• (B) $\{1, 2, \dots, 12\}$: This is incorrect because the minimum possible sum is 2, not 1.
• (C) $\{2, 3, \dots, 12\}$: This correctly lists all possible sums from the minimum of 2 to the maximum of 12.
• (D) $\{ (1,1), (1,2), \dots, (6,6) \}$: This represents the sample space (the set of all possible outcomes), not the numerical values of the random variable $X$.

(C) 0.5

Explanation:

The question asks for the value of the probability of success, $p$, that maximizes the variance of a binomial distribution, $X \sim B(n, p)$.

Step 1: Write down the formula for the variance.

The variance of a binomial distribution is given by:

Step 2: Analyze the function to be maximized.

For a fixed number of trials, $n$, the variance is a function of $p$. To maximize $Var(X)$, we need to maximize the term $p(1-p)$, as $n$ is just a positive constant multiplier.

Let's define a function $f(p) = p(1-p) = p - p^2$. We need to find the value of $p$ in the interval $[0, 1]$ that maximizes this function.

Step 3: Use calculus to find the maximum.

We can find the maximum by taking the derivative of $f(p)$ with respect to $p$ and setting it to zero.

First derivative:

$f'(p) = \frac{d}{dp}(p - p^2) = 1 - 2p$

Set the derivative to zero to find the critical point:

$1 - 2p = 0$

$1 = 2p$

$p = \frac{1}{2} = 0.5$

To confirm this is a maximum, we can use the second derivative test. The second derivative is:

$f''(p) = \frac{d}{dp}(1 - 2p) = -2$

Since the second derivative is negative ($-2 < 0$), the function has a maximum at the critical point $p=0.5$.

Alternative (Intuitive) Method:

The term we want to maximize is the product of two numbers, $p$ and $(1-p)$. The sum of these two numbers is constant: $p + (1-p) = 1$. A well-known mathematical principle states that the product of two numbers with a constant sum is maximized when the two numbers are equal.

So, we set the two numbers equal to each other:

$p = 1 - p$

$2p = 1$

$p = 0.5$

This confirms that the variance is maximized when the probabilities of success and failure are equal, which happens when $p=0.5$. At this point, the uncertainty of the outcome of any single trial is at its highest.

The probabilities are given as:

$P(A) = 0.8$

$P(B) = 0.5$

$P(B|A) = 0.4$

We need to find the following probabilities:

1. $P(A \cap B)$

2. $P(A|B)$

First, we will find the probability of the intersection of events A and B, i.e., $P(A \cap B)$.

The formula for the conditional probability of event B given that event A has occurred is:

By rearranging the formula, we can solve for $P(A \cap B)$:

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

Now, we substitute the given values into this formula:

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

$P(A \cap B) = 0.32$

So, the probability of the intersection of A and B is 0.32.

Next, we will find the conditional probability of event A given that event B has occurred, i.e., $P(A|B)$.

The formula for this conditional probability is:

From the previous step, we found that $P(A \cap B) = 0.32$, and we are given $P(B) = 0.5$.

Substituting these values into formula (ii):

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

$P(A|B) = 0.64$

So, the conditional probability of A given B is 0.64.

Therefore, the required probabilities are:

$P(A \cap B) = \mathbf{0.32}$

$P(A|B) = \mathbf{0.64}$

When a single die is thrown, the set of all possible outcomes, known as the sample space (S), is:

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

The total number of outcomes is $n(S) = 6$.

We are given two events:

Event E: "the number appearing is a multiple of 3".

The outcomes favorable to event E are the numbers in the sample space that are multiples of 3.

$E = \{3, 6\}$

So, the number of outcomes favorable to E is $n(E) = 2$.

Event F: "the number appearing is even".

The outcomes favorable to event F are the even numbers in the sample space.

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

So, the number of outcomes favorable to F is $n(F) = 3$.

We need to find the conditional probability $P(E|F)$, which is the probability of event E occurring given that event F has already occurred.

The formula for conditional probability is:

Alternatively, it can be expressed in terms of the number of outcomes:

First, we find the intersection of events E and F, denoted by $E \cap F$. This represents the outcomes that are common to both E and F (i.e., the number is a multiple of 3 AND is even).

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

$E \cap F = \{6\}$

The number of outcomes in the intersection is $n(E \cap F) = 1$.

Now, using formula (ii):

$P(E|F) = \frac{n(E \cap F)}{n(F)}$

We have $n(E \cap F) = 1$ and $n(F) = 3$.

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

Therefore, the probability that the number is a multiple of 3 given that it is an even number is $\frac{1}{3}$.

We can also solve this using formula (i). First, we calculate the probabilities of $P(E \cap F)$ and $P(F)$.

$P(E \cap F) = \frac{\text{Number of outcomes in } E \cap F}{\text{Total number of outcomes}} = \frac{n(E \cap F)}{n(S)} = \frac{1}{6}$

$P(F) = \frac{\text{Number of outcomes in } F}{\text{Total number of outcomes}} = \frac{n(F)}{n(S)} = \frac{3}{6} = \frac{1}{2}$

Now, substitute these values into the conditional probability formula:

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

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

Both methods yield the same result.

Given:

A bag contains:

Number of red balls = 5

Number of black balls = 4

Total number of balls = 5 + 4 = 9

Two balls are drawn one by one without replacement. This means the outcome of the second draw depends on the outcome of the first draw.

To Find:

The probability that both balls drawn are black.

Solution:

Let's define the events:

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 need to find the probability of both events occurring, which is $P(A \text{ and } B)$ or $P(A \cap B)$.

Since the balls are drawn without replacement, the events are dependent. We use the multiplication rule for dependent events:

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

where $P(B|A)$ is the probability of event B occurring given that event A has already occurred.

Step 1: Probability of the first ball being black, $P(A)$

Initially, there are 4 black balls and a total of 9 balls.

$P(A) = \frac{\text{Number of black balls}}{\text{Total number of balls}} = \frac{4}{9}$

Step 2: Probability of the second ball being black, given the first was black, $P(B|A)$

After drawing one black ball without replacing it, the bag has:

Number of remaining black balls = 4 - 1 = 3

Total number of remaining balls = 9 - 1 = 8

$P(B|A) = \frac{\text{Number of remaining black balls}}{\text{Total number of remaining balls}} = \frac{3}{8}$

Step 3: Calculate the final probability

$P(\text{both balls are black}) = P(A \cap B) = P(A) \times P(B|A)$

$P(\text{both balls are black}) = \frac{4}{9} \times \frac{3}{8}$

$P(\text{both balls are black}) = \frac{\cancel{12}^{1}}{\cancel{72}_{6}} = \frac{1}{6}$

Alternatively, simplifying before multiplication:

$P(\text{both balls are black}) = \frac{\cancel{4}^1}{9} \times \frac{3}{\cancel{8}_2} = \frac{1}{\cancel{9}_3} \times \frac{\cancel{3}^1}{2} = \frac{1 \times 1}{3 \times 2} = \frac{1}{6}$

So, the probability that both balls drawn are black is $\frac{1}{6}$.

Alternate Solution (Using Combinations)

We can also solve this by considering the number of ways to choose the balls.

Total number of possible outcomes:

This is the number of ways to choose 2 balls from the total of 9 balls, which is given by the combination formula $^nC_r = \frac{n!}{r!(n-r)!}$.

Total outcomes = $^9C_2 = \frac{9!}{2!(9-2)!} = \frac{9!}{2!7!} = \frac{9 \times 8}{2 \times 1} = 36$

Number of favorable outcomes:

This is the number of ways to choose 2 black balls from the 4 available black balls.

Favorable outcomes = $^4C_2 = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} = \frac{4 \times 3}{2 \times 1} = 6$

Calculating the probability:

Probability = $\frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

Probability = $\frac{^4C_2}{^9C_2} = \frac{6}{36} = \frac{1}{6}$

Both methods give the same result, confirming that the probability of drawing two black balls is $\frac{1}{6}$.

Definition of Independent Events

Two events, let's say A and B, associated with a random experiment are said to be independent if the probability of occurrence of one event is not affected by the occurrence or non-occurrence of the other event.

In the language of conditional probability, if A and B are independent events, then the probability of A occurring given that B has occurred is simply the probability of A. Mathematically, this is expressed as:

$P(A|B) = P(A)$, provided $P(B) \neq 0$

Similarly, the probability of B occurring given that A has occurred is the probability of B:

$P(B|A) = P(B)$, provided $P(A) \neq 0$

For example, when tossing a coin twice, the outcome of the first toss (e.g., getting a Head) is an independent event from the outcome of the second toss.

Expression for $P(A \cap B)$ for Independent Events

The probability of the simultaneous occurrence of two events A and B, denoted as $P(A \cap B)$, is given by the general multiplication rule of probability:

This rule states that the probability of both A and B occurring is the product of the probability of A and the conditional probability of B given that A has already occurred.

However, when the events A and B are independent, we know from the definition that $P(B|A) = P(B)$.

Substituting this condition into the multiplication rule, we get the expression for the intersection of two independent events:

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

This result is the fundamental property used to check for the independence of two events and to calculate the probability of their joint occurrence.

Given:

A fair coin is tossed 3 times.

X is the random variable representing the "number of heads".

To Find:

1. The possible values of the random variable X.

2. The probability distribution of X.

Solution:

Step 1: Determine the Sample Space

When a fair coin is tossed 3 times, the set of all possible outcomes (the sample space, S) is:

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

Where H represents a Head and T represents a Tail.

The total number of possible outcomes is $n(S) = 2^3 = 8$. Since the coin is fair, each of these 8 outcomes is equally likely, with a probability of $\frac{1}{8}$.

Step 2: Find the Possible Values of X

The random variable X represents the number of heads in each outcome.

• For the outcome TTT, the number of heads is 0. So, $X=0$.
• For the outcomes HTT, THT, TTH, the number of heads is 1. So, $X=1$.
• For the outcomes HHT, HTH, THH, the number of heads is 2. So, $X=2$.
• For the outcome HHH, the number of heads is 3. So, $X=3$.

Therefore, the possible values of the random variable X are {0, 1, 2, 3}.

Step 3: Calculate the Probability for Each Value of X

We now calculate the probability P(X) for each possible value of X.

P(X = 0): This means getting 0 heads (i.e., the outcome TTT).

$P(X=0) = P(\{\text{TTT}\}) = \frac{1}{8}$

P(X = 1): This means getting 1 head (i.e., the outcomes HTT, THT, TTH).

$P(X=1) = P(\{\text{HTT, THT, TTH}\}) = \frac{3}{8}$

P(X = 2): This means getting 2 heads (i.e., the outcomes HHT, HTH, THH).

$P(X=2) = P(\{\text{HHT, HTH, THH}\}) = \frac{3}{8}$

P(X = 3): This means getting 3 heads (i.e., the outcome HHH).

$P(X=3) = P(\{\text{HHH}\}) = \frac{1}{8}$

Step 4: Write the Probability Distribution

The probability distribution of a random variable X is a table or a function that lists all the possible values of X and their corresponding probabilities. We can represent this in a table:

Note that the sum of the probabilities is $\frac{1}{8} + \frac{3}{8} + \frac{3}{8} + \frac{1}{8} = \frac{8}{8} = 1$, as required for a valid probability distribution.

Alternate Solution (Using Binomial Distribution)

This experiment can be modeled as a binomial distribution because:

1. There is a fixed number of trials ($n=3$ tosses).
2. Each trial is independent.
3. Each trial has only two outcomes: success (Head) or failure (Tail).
4. The probability of success ($p$) is constant for each trial.

Here, $n=3$. Let "success" be getting a head. For a fair coin, the probability of success is $p = P(H) = \frac{1}{2}$.

The probability of failure is $q = 1 - p = 1 - \frac{1}{2} = \frac{1}{2}$.

The probability mass function for a binomial distribution is:

$P(X=k) = \binom{n}{k} p^k q^{n-k}$

where $k$ is the number of successes (heads).

$P(X=0) = \binom{3}{0} (\frac{1}{2})^0 (\frac{1}{2})^{3-0} = 1 \times 1 \times \frac{1}{8} = \frac{1}{8}$

$P(X=1) = \binom{3}{1} (\frac{1}{2})^1 (\frac{1}{2})^{3-1} = 3 \times \frac{1}{2} \times \frac{1}{4} = \frac{3}{8}$

$P(X=2) = \binom{3}{2} (\frac{1}{2})^2 (\frac{1}{2})^{3-2} = 3 \times \frac{1}{4} \times \frac{1}{2} = \frac{3}{8}$

$P(X=3) = \binom{3}{3} (\frac{1}{2})^3 (\frac{1}{2})^{3-3} = 1 \times \frac{1}{8} \times 1 = \frac{1}{8}$

These results match the first method and can be presented in the same probability distribution table.

The mean of a probability distribution is given by the formula:

We can calculate the mean by multiplying each value of X by its corresponding probability P(X) and then summing the results.

Here's the calculation:

$\mu = (0 \times 0.1) + (1 \times 0.4) + (2 \times 0.3) + (3 \times 0.2)$

$\mu = 0 + 0.4 + 0.6 + 0.6$

$\mu = 1.6$

Therefore, the mean of the given probability distribution is 1.6.

In a binomial distribution, the probability of getting exactly $k$ successes in $n$ trials is given by:

Where:

• $n$ is the number of trials
• $k$ is the number of successes we want
• $p$ is the probability of success on a single trial
• ${n \choose k}$ is the binomial coefficient, which can be calculated as $\frac{n!}{k!(n-k)!}$

In this case:

• $n = 4$ (number of coin tosses)
• $k = 2$ (number of heads we want)
• $p = 0.5$ (probability of getting a head on a single toss, assuming a fair coin)

Plugging these values into the formula:

$P(X = 2) = {4 \choose 2} (0.5)^2 (1-0.5)^{4-2}$

$P(X = 2) = {4 \choose 2} (0.5)^2 (0.5)^2$

Now, we need to calculate the binomial coefficient ${4 \choose 2}$:

${4 \choose 2} = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} = \frac{4 \times 3 \times 2 \times 1}{(2 \times 1)(2 \times 1)} = \frac{24}{4} = 6$

Substituting this back into the probability equation:

$P(X = 2) = 6 \times (0.5)^2 \times (0.5)^2$

$P(X = 2) = 6 \times 0.25 \times 0.25$

$P(X = 2) = 6 \times 0.0625$

$P(X = 2) = 0.375$

Therefore, the probability of getting exactly 2 heads when a coin is tossed 4 times is 0.375.

The probability mass function for a binomial distribution is given by:

Where:

• $n$ is the number of trials
• $k$ is the number of successes
• $p$ is the probability of success on a single trial
• ${n \choose k}$ is the binomial coefficient, calculated as $\frac{n!}{k!(n-k)!}$

In this problem, we have $n = 5$, $p = 0.3$, and we want to find $P(X = 0)$. This means we want to find the probability of getting exactly 0 successes in 5 trials.

Plugging in the values:

$P(X = 0) = {5 \choose 0} (0.3)^0 (1-0.3)^{5-0}$

$P(X = 0) = {5 \choose 0} (0.3)^0 (0.7)^5$

We need to calculate ${5 \choose 0}$:

${5 \choose 0} = \frac{5!}{0!(5-0)!} = \frac{5!}{0!5!} = \frac{1}{1} = 1$

Note: 0! is defined as 1.

Now, substitute this back into the equation:

$P(X = 0) = 1 \times (0.3)^0 \times (0.7)^5$

$P(X = 0) = 1 \times 1 \times (0.7)^5$

$P(X = 0) = (0.7)^5$

Calculating $(0.7)^5$:

$(0.7)^5 = 0.7 \times 0.7 \times 0.7 \times 0.7 \times 0.7 = 0.16807$

Therefore, $P(X = 0) = 0.16807$

The probability of the union of two events, $A$ and $B$, is given by:

Since A and B are independent events, the probability of their intersection is given by:

We are given $P(A) = 0.3$ and $P(B) = 0.6$. Let's first find $P(A \cap B)$:

$P(A \cap B) = 0.3 \times 0.6 = 0.18$

Now, we can find $P(A \cup B)$ using the formula:

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

$P(A \cup B) = 0.3 + 0.6 - 0.18$

$P(A \cup B) = 0.9 - 0.18$

$P(A \cup B) = 0.72$

Therefore, the probability of $A \cup B$ is 0.72.

Part 1: Probability of drawing a black ball initially

Total number of balls = 10 (red) + 12 (black) = 22

Number of black balls = 12

Probability of drawing a black ball, $P(\text{Black}) = \frac{\text{Number of black balls}}{\text{Total number of balls}}$

$P(\text{Black}) = \frac{12}{22} = \frac{6}{11}$

Part 2: Probability of drawing a black ball after a red ball is drawn and not replaced

If a red ball is drawn and not replaced:

New total number of balls = 22 - 1 = 21

Number of red balls remaining = 10 - 1 = 9

Number of black balls remains the same = 12

Probability of drawing a black ball next, $P(\text{Black after Red}) = \frac{\text{Number of black balls}}{\text{New total number of balls}}$

$P(\text{Black after Red}) = \frac{12}{21} = \frac{4}{7}$

Therefore:

The probability of drawing a black ball initially is $\frac{6}{11}$.

If a red ball is drawn and not replaced, the probability of drawing a black ball next is $\frac{4}{7}$.

The variance of a probability distribution is a measure of how spread out the distribution is. It quantifies the expected squared deviation of a random variable from its mean.

In simpler terms, the variance tells us how much the individual values in the distribution tend to differ from the average value (mean).

Formula:

For a discrete probability distribution, the variance (denoted as $\sigma^2$ or $Var(X)$) is calculated as:

Where:

• $x_i$ represents each individual value of the random variable $X$
• $\mu$ represents the mean (expected value) of the distribution, calculated as $\mu = \sum x_i P(x_i)$
• $P(x_i)$ is the probability of the value $x_i$
• $\sum$ denotes the sum over all possible values of $x_i$

Alternatively, the variance can also be calculated using the following formula:

Where:

• $E(X^2)$ is the expected value of $X^2$, calculated as $\sum x_i^2 P(x_i)$
• $E(X)$ is the expected value of $X$, calculated as $\sum x_i P(x_i)$ which is the mean $\mu$

Let's define the sample space for the possible genders of two children. We'll use B for boy and G for girl. The sample space is:

S = {BB, BG, GB, GG}

Each outcome is equally likely with a probability of $\frac{1}{4}$.

We want to find the probability that both children are boys (BB), given that at least one of them is a boy. Let's define the events:

• A: Both children are boys (BB)
• B: At least one child is a boy (BB, BG, GB)

We want to find $P(A|B)$, which is the probability of A given B. The formula for conditional probability is:

First, let's find $P(B)$. The event B (at least one boy) includes the outcomes BB, BG, and GB. Therefore:

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

Next, we need to find $P(A \cap B)$. This is the probability that both children are boys (A) and at least one child is a boy (B). Since A (BB) is a subset of B, $A \cap B$ is simply A (BB).

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

Now, we can calculate $P(A|B)$:

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

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

Let's define the events:

• A: The card drawn is a face card (Jack, Queen, or King)
• B: The card drawn is red (Heart or Diamond)

We want to find the conditional probability $P(A|B)$, which is the probability that the card is a face card, given that it is red.

First, let's find $P(B)$, the probability that the card drawn is red.

There are 26 red cards in a deck of 52 cards (13 hearts and 13 diamonds). Therefore:

$P(B) = \frac{26}{52} = \frac{1}{2}$

Next, we need to find $P(A \cap B)$, the probability that the card is a face card and red. This means the card is either a Jack, Queen, or King of Hearts or Diamonds.

There are 3 face cards (Jack, Queen, King) in each suit. So there are 3 red face cards in hearts and 3 red face cards in diamonds, for a total of 6 red face cards.

Therefore, $P(A \cap B) = \frac{6}{52} = \frac{3}{26}$

Now, we can calculate $P(A|B)$:

$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{3}{26}}{\frac{1}{2}} = \frac{3}{26} \times \frac{2}{1} = \frac{6}{26} = \frac{3}{13}$

Therefore, the probability that the card is a face card, given that it is red, is $\frac{3}{13}$.

X represents the number of red balls drawn. Since we are drawing two balls, X can take the values 0, 1, or 2.

• X = 0: No red balls are drawn (both are black)
• X = 1: One red ball and one black ball are drawn
• X = 2: Two red balls are drawn

We need to calculate the probability for each value of X.

Total number of balls in the bag = 3 (Red) + 5 (Black) = 8

We will use combinations to calculate the probabilities since the order in which the balls are drawn doesn't matter.

The total number of ways to draw 2 balls from 8 is ${8 \choose 2} = \frac{8!}{2!6!} = \frac{8 \times 7}{2 \times 1} = 28$

1. Probability of X = 0 (Both balls are black)

Number of ways to draw 2 black balls from 5 black balls is ${5 \choose 2} = \frac{5!}{2!3!} = \frac{5 \times 4}{2 \times 1} = 10$

$P(X = 0) = \frac{\text{Number of ways to draw 2 black balls}}{\text{Total number of ways to draw 2 balls}} = \frac{10}{28} = \frac{5}{14}$

2. Probability of X = 1 (One red and one black ball)

Number of ways to draw 1 red ball from 3 red balls is ${3 \choose 1} = \frac{3!}{1!2!} = 3$

Number of ways to draw 1 black ball from 5 black balls is ${5 \choose 1} = \frac{5!}{1!4!} = 5$

Number of ways to draw 1 red and 1 black ball is ${3 \choose 1} \times {5 \choose 1} = 3 \times 5 = 15$

$P(X = 1) = \frac{\text{Number of ways to draw 1 red and 1 black ball}}{\text{Total number of ways to draw 2 balls}} = \frac{15}{28}$

3. Probability of X = 2 (Both balls are red)

Number of ways to draw 2 red balls from 3 red balls is ${3 \choose 2} = \frac{3!}{2!1!} = \frac{3 \times 2}{2 \times 1} = 3$

$P(X = 2) = \frac{\text{Number of ways to draw 2 red balls}}{\text{Total number of ways to draw 2 balls}} = \frac{3}{28}$

Probability Distribution of X

The probability distribution of X is:

We can verify that the probabilities sum up to 1:

$\frac{5}{14} + \frac{15}{28} + \frac{3}{28} = \frac{10}{28} + \frac{15}{28} + \frac{3}{28} = \frac{28}{28} = 1$

For a binomial distribution with parameters $n$ (number of trials) and $p$ (probability of success on a single trial):

• Mean ($\mu$) = $np$
• Variance ($\sigma^2$) = $np(1-p)$

In this case, $n = 4$ and $p = 0.5$.

1. Mean:

$\mu = np = 4 \times 0.5 = 2$

2. Variance:

$\sigma^2 = np(1-p) = 4 \times 0.5 \times (1-0.5) = 4 \times 0.5 \times 0.5 = 1$

Therefore:

• Mean ($\mu$) = 2
• Variance ($\sigma^2$) = 1

For a probability distribution, the sum of all probabilities must equal 1.

Therefore, we have:

$P(X=0) + P(X=1) + P(X=2) + P(X=3) = 1$

Substituting the given probabilities:

$k + 2k + 3k + 4k = 1$

Combining the terms:

$10k = 1$

Solving for $k$:

$k = \frac{1}{10} = 0.1$

Therefore, the value of $k$ is 0.1.

Let $P(T)$ be the probability of getting a tail and $P(H)$ be the probability of getting a head.

Given that the head is 3 times as likely to occur as tail, we have $P(H) = 3P(T)$.

Since the sum of probabilities must be 1, we have $P(H) + P(T) = 1$.

Substituting $P(H) = 3P(T)$ into the equation $P(H) + P(T) = 1$:

$3P(T) + P(T) = 1$

$4P(T) = 1$

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

Therefore, $P(H) = 3P(T) = 3 \times \frac{1}{4} = \frac{3}{4}$

We are tossing the coin twice. Let X be the number of tails. X can take the values 0, 1, or 2.

• X = 0: No tails (HH)
• X = 1: One tail (HT or TH)
• X = 2: Two tails (TT)

1. Probability of X = 0 (HH):

$P(X = 0) = P(HH) = P(H) \times P(H) = \frac{3}{4} \times \frac{3}{4} = \frac{9}{16}$

2. Probability of X = 1 (HT or TH):

$P(X = 1) = P(HT) + P(TH) = P(H) \times P(T) + P(T) \times P(H) = \frac{3}{4} \times \frac{1}{4} + \frac{1}{4} \times \frac{3}{4} = \frac{3}{16} + \frac{3}{16} = \frac{6}{16} = \frac{3}{8}$

3. Probability of X = 2 (TT):

$P(X = 2) = P(TT) = P(T) \times P(T) = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16}$

Probability Distribution of X

The probability distribution of X is:

We can verify that the probabilities sum up to 1:

$\frac{9}{16} + \frac{3}{8} + \frac{1}{16} = \frac{9}{16} + \frac{6}{16} + \frac{1}{16} = \frac{16}{16} = 1$

We are given $P(A) = 0.4$, $P(B) = 0.8$, and $P(B|A) = 0.6$. We want to find $P(A \cup B)$.

We know that:

We also know the conditional probability formula:

We can rearrange the conditional probability formula to find $P(A \cap B)$:

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

Substituting the given values:

$P(A \cap B) = 0.6 \times 0.4 = 0.24$

Now we can substitute the values of $P(A)$, $P(B)$, and $P(A \cap B)$ into the formula for $P(A \cup B)$:

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

$P(A \cup B) = 0.4 + 0.8 - 0.24$

$P(A \cup B) = 1.2 - 0.24$

$P(A \cup B) = 0.96$

Therefore, $P(A \cup B) = 0.96$

We want to find the probability of drawing two aces simultaneously from a deck of 52 cards.

There are 4 aces in a standard deck of 52 cards.

Method 1: Using Combinations

The total number of ways to choose 2 cards from 52 is ${52 \choose 2} = \frac{52!}{2!50!} = \frac{52 \times 51}{2 \times 1} = 1326$

The number of ways to choose 2 aces from 4 aces is ${4 \choose 2} = \frac{4!}{2!2!} = \frac{4 \times 3}{2 \times 1} = 6$

The probability of drawing two aces is:

$P(\text{Two Aces}) = \frac{\text{Number of ways to choose 2 aces}}{\text{Total number of ways to choose 2 cards}} = \frac{6}{1326} = \frac{1}{221}$

Method 2: Step-by-Step Probability

The probability that the first card drawn is an ace is $\frac{4}{52} = \frac{1}{13}$

Given that the first card drawn is an ace, there are now 3 aces left in a deck of 51 cards.

The probability that the second card drawn is also an ace, given that the first card was an ace, is $\frac{3}{51} = \frac{1}{17}$

The probability of drawing two aces in a row is:

$P(\text{Two Aces}) = P(\text{First card is an ace}) \times P(\text{Second card is an ace | First card is an ace})$

$P(\text{Two Aces}) = \frac{1}{13} \times \frac{1}{17} = \frac{1}{221}$

Therefore, the probability that both cards are aces is $\frac{1}{221}$.

We are given $P(A) = \frac{1}{2}$, $P(B) = \frac{7}{12}$, and $P(\text{not } A \text{ or not } B) = P(A' \cup B') = \frac{1}{4}$. We need to determine if A and B are independent.

By De Morgan's Law:

So, $P(A \cap B)' = \frac{1}{4}$. This means the probability of the complement of ($A \cap B$) is $\frac{1}{4}$.

Therefore:

For A and B to be independent, the following condition must be true:

Let's check if this condition holds:

$P(A) \times P(B) = \frac{1}{2} \times \frac{7}{12} = \frac{7}{24}$

We found that $P(A \cap B) = \frac{3}{4}$ and $P(A) \times P(B) = \frac{7}{24}$.

Since $\frac{3}{4} \neq \frac{7}{24}$, the condition for independence is not met.

Therefore, A and B are not independent events.

Since we are drawing cards with replacement, this is a binomial distribution problem.

Let X be the number of spades drawn. We want to find the probability of getting exactly 3 spades in 5 draws, so we want to find $P(X = 3)$.

The binomial probability formula is:

Where:

• $n$ is the number of trials (draws)
• $k$ is the number of successes (spades) we want
• $p$ is the probability of success on a single trial (probability of drawing a spade)

In this problem:

• $n = 5$ (five cards are drawn)
• $k = 3$ (exactly three spades are desired)
• $p = \frac{13}{52} = \frac{1}{4}$ (probability of drawing a spade in a single draw, since there are 13 spades in a deck of 52 cards)

Plugging these values into the formula:

$P(X = 3) = {5 \choose 3} \left(\frac{1}{4}\right)^3 \left(1 - \frac{1}{4}\right)^{5-3}$

$P(X = 3) = {5 \choose 3} \left(\frac{1}{4}\right)^3 \left(\frac{3}{4}\right)^2$

Now, we calculate the binomial coefficient ${5 \choose 3}$:

${5 \choose 3} = \frac{5!}{3!(5-3)!} = \frac{5!}{3!2!} = \frac{5 \times 4}{2 \times 1} = 10$

Substituting this back into the equation:

$P(X = 3) = 10 \times \left(\frac{1}{4}\right)^3 \times \left(\frac{3}{4}\right)^2$

$P(X = 3) = 10 \times \frac{1}{64} \times \frac{9}{16}$

$P(X = 3) = 10 \times \frac{9}{1024}$

$P(X = 3) = \frac{90}{1024} = \frac{45}{512}$

Therefore, the probability of getting exactly three spades when drawing five cards with replacement is $\frac{45}{512}$.

Let E be the event that the student passes the English examination, and H be the event that the student passes the Hindi examination.

We are given:

• $P(E \cap H) = 0.5$ (probability of passing both English and Hindi)
• $P(\text{neither E nor H}) = P(E' \cap H') = 0.1$ (probability of passing neither)
• $P(E) = 0.75$ (probability of passing English)

We want to find $P(H)$ (probability of passing Hindi).

By De Morgan's Law, $P(E' \cap H') = P(E \cup H)'$. So,

Therefore,

We also know that:

Substituting the known values:

$0.9 = 0.75 + P(H) - 0.5$

Now, we can solve for $P(H)$:

$0.9 = 0.25 + P(H)$

$P(H) = 0.9 - 0.25$

$P(H) = 0.65$

Therefore, the probability of passing the Hindi examination is 0.65.

Given:

Let's define the following events:

$E_1$: The event of selecting the first bag.

$E_2$: The event of selecting the second bag.

$A$: The event of drawing a red ball.

Bag 1:

Number of red balls = 4

Number of black balls = 4

Total number of balls = $4 + 4 = 8$

Bag 2:

Number of red balls = 2

Number of black balls = 6

Total number of balls = $2 + 6 = 8$

To Find:

The probability that the ball is drawn from the first bag, given that the ball drawn is red. This is represented as $P(E_1|A)$.

Solution:

Since one of the two bags is selected at random, the probability of selecting each bag is:

$P(E_1) = \frac{1}{2}$

$P(E_2) = \frac{1}{2}$

Now, we find the probability of drawing a red ball from each bag.

The probability of drawing a red ball given that the first bag is selected is $P(A|E_1)$.

$P(A|E_1) = \frac{\text{Number of red balls in Bag 1}}{\text{Total balls in Bag 1}} = \frac{4}{8} = \frac{1}{2}$

The probability of drawing a red ball given that the second bag is selected is $P(A|E_2)$.

$P(A|E_2) = \frac{\text{Number of red balls in Bag 2}}{\text{Total balls in Bag 2}} = \frac{2}{8} = \frac{1}{4}$

We need to find the probability that the ball was drawn from the first bag, given that it is red, which is $P(E_1|A)$.

Using Bayes' Theorem:

$P(E_1|A) = \frac{P(E_1) \cdot P(A|E_1)}{P(E_1) \cdot P(A|E_1) + P(E_2) \cdot P(A|E_2)}$

Let's substitute the values we have calculated:

$P(E_1|A) = \frac{\frac{1}{2} \times \frac{1}{2}}{(\frac{1}{2} \times \frac{1}{2}) + (\frac{1}{2} \times \frac{1}{4})}$

$P(E_1|A) = \frac{\frac{1}{4}}{\frac{1}{4} + \frac{1}{8}}$

To add the fractions in the denominator, we find a common denominator, which is 8:

$P(E_1|A) = \frac{\frac{1}{4}}{\frac{2}{8} + \frac{1}{8}}$

$P(E_1|A) = \frac{\frac{1}{4}}{\frac{3}{8}}$

Now, we can simplify the expression:

$P(E_1|A) = \frac{1}{4} \times \frac{8}{3}$

$P(E_1|A) = \frac{\cancel{8}^2}{\cancel{12}_3}$

$P(E_1|A) = \frac{2}{3}$

Therefore, the probability that the ball is drawn from the first bag is $\frac{2}{3}$.

Given:

The probability distribution of a random variable X is:

To Find:

1. The value of k.

2. $P(X \leq 2)$

3. $P(X > 2)$

4. $P(1 \leq X \leq 3)$

5. The mean and variance of X.

Solution:

1. Finding the value of k

We know that for a probability distribution, the sum of all probabilities must be equal to 1.

$\sum P(X_i) = 1$

$P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) = 1$

Substituting the given values:

$k + 2k + 3k + 2k + k = 1$

$9k = 1$

$k = \frac{1}{9}$

So, the value of k is $\frac{1}{9}$.

2. Calculating Probabilities

Now we can find the specific probabilities:

(i) $P(X \leq 2)$

$P(X \leq 2) = P(X=0) + P(X=1) + P(X=2)$

$P(X \leq 2) = k + 2k + 3k = 6k$

Substituting $k = \frac{1}{9}$:

$P(X \leq 2) = 6 \times \frac{1}{9} = \frac{6}{9} = \frac{2}{3}$

Thus, $P(X \leq 2) = \frac{2}{3}$.

(ii) $P(X > 2)$

$P(X > 2) = P(X=3) + P(X=4)$

$P(X > 2) = 2k + k = 3k$

Substituting $k = \frac{1}{9}$:

$P(X > 2) = 3 \times \frac{1}{9} = \frac{3}{9} = \frac{1}{3}$

Thus, $P(X > 2) = \frac{1}{3}$.

(iii) $P(1 \leq X \leq 3)$

$P(1 \leq X \leq 3) = P(X=1) + P(X=2) + P(X=3)$

$P(1 \leq X \leq 3) = 2k + 3k + 2k = 7k$

Substituting $k = \frac{1}{9}$:

$P(1 \leq X \leq 3) = 7 \times \frac{1}{9} = \frac{7}{9}$

Thus, $P(1 \leq X \leq 3) = \frac{7}{9}$.

3. Finding the Mean and Variance

To find the mean and variance, let's create a calculation table using the value $k = 1/9$.

Mean of X:

The mean (or expected value) $E(X)$ is given by the formula:

$Mean (\mu) = E(X) = \sum X_i P(X_i)$

From the table, the sum of the $X_i P(X_i)$ column is:

$\mu = \frac{18}{9} = 2$

The Mean of X is 2.

Variance of X:

The variance $Var(X)$ is given by the formula:

$Variance (\sigma^2) = E(X^2) - [E(X)]^2 = \sum X_i^2 P(X_i) - \mu^2$

From the table, $\sum X_i^2 P(X_i) = \frac{48}{9} = \frac{16}{3}$.

We already calculated the mean, $\mu = 2$.

Substituting these values into the variance formula:

$\sigma^2 = \frac{16}{3} - (2)^2$

$\sigma^2 = \frac{16}{3} - 4$

$\sigma^2 = \frac{16 - 12}{3}$

$\sigma^2 = \frac{4}{3}$

The Variance of X is $\frac{4}{3}$.

Let $A$, $B$, and $C$ be the events that a product is produced by machine A, machine B, and machine C respectively. Let $D$ be the event that a product is defective.

We are given the following probabilities:

$P(A) = 0.30$

$P(B) = 0.25$

$P(C) = 0.45$

$P(D|A) = 0.01$

$P(D|B) = 0.012$

$P(D|C) = 0.02$

We want to find $P(A|D)$, the probability that the product was produced by machine A given that it is defective. We can use Bayes' Theorem:

$P(A|D) = \frac{P(D|A)P(A)}{P(D)}$

First, we need to find $P(D)$, the probability that a product is defective. We can use the law of total probability:

$P(D) = P(D|A)P(A) + P(D|B)P(B) + P(D|C)P(C)$

$P(D) = (0.01)(0.30) + (0.012)(0.25) + (0.02)(0.45)$

$P(D) = 0.003 + 0.003 + 0.009 = 0.015$

Now we can find $P(A|D)$:

$P(A|D) = \frac{P(D|A)P(A)}{P(D)} = \frac{(0.01)(0.30)}{0.015} = \frac{0.003}{0.015} = \frac{3}{15} = \frac{1}{5} = 0.2$

Therefore, the probability that the defective product was produced by machine A is 0.2 or 20%.

Let $X$ be the number of throws required until a six comes up. The probability of getting a six on any throw is $\frac{1}{6}$, and the probability of not getting a six is $\frac{5}{6}$.

**(i) Probability that the six comes up on the third throw:**

For the six to come up on the third throw, we must not get a six on the first two throws and then get a six on the third throw. The probability of this is:

$P(X=3) = \left(\frac{5}{6}\right) \left(\frac{5}{6}\right) \left(\frac{1}{6}\right) = \frac{25}{216}$

**(ii) Probability that the six comes up at most on the third throw:**

This means the six can come up on the first, second, or third throw. We can calculate the probability for each of these cases and add them together:

$P(X \leq 3) = P(X=1) + P(X=2) + P(X=3)$

$P(X=1) = \frac{1}{6}$

$P(X=2) = \left(\frac{5}{6}\right)\left(\frac{1}{6}\right) = \frac{5}{36}$

$P(X=3) = \left(\frac{5}{6}\right)^2\left(\frac{1}{6}\right) = \frac{25}{216}$

So, $P(X \leq 3) = \frac{1}{6} + \frac{5}{36} + \frac{25}{216} = \frac{36}{216} + \frac{30}{216} + \frac{25}{216} = \frac{36+30+25}{216} = \frac{91}{216}$

Therefore, the probability that the six comes up at most on the third throw is $\frac{91}{216}$.

Alternatively for (ii):

The probability of the six appearing at most on the third throw is the complement of the probability of the six not appearing in the first three throws.

$P(X \leq 3) = 1 - P(\text{no six in first 3 throws})$

$P(\text{no six in first 3 throws}) = \left(\frac{5}{6}\right)^3 = \frac{125}{216}$

$P(X \leq 3) = 1 - \frac{125}{216} = \frac{216 - 125}{216} = \frac{91}{216}$

Let $p$ be the probability that a student will graduate, so $p = 0.4$. Let $q$ be the probability that a student will not graduate, so $q = 1 - p = 1 - 0.4 = 0.6$. We have 5 students, so $n = 5$. This is a binomial distribution problem.

**(i) Probability that none will graduate:**

This means all 5 students will not graduate. The probability is:

$P(\text{none graduate}) = P(X=0) = \binom{5}{0} p^0 q^5 = \binom{5}{0} (0.4)^0 (0.6)^5 = 1 \cdot 1 \cdot (0.6)^5 = 0.07776$

**(ii) Probability that exactly one will graduate:**

This means exactly one student will graduate. The probability is:

$P(\text{exactly one graduates}) = P(X=1) = \binom{5}{1} p^1 q^4 = \binom{5}{1} (0.4)^1 (0.6)^4 = 5 \cdot (0.4) \cdot (0.6)^4 = 5 \cdot 0.4 \cdot 0.1296 = 2 \cdot 0.1296 = 0.2592$

**(iii) Probability that at least one will graduate:**

This is the complement of the probability that none will graduate. Therefore:

$P(\text{at least one graduates}) = 1 - P(\text{none graduate}) = 1 - P(X=0) = 1 - 0.07776 = 0.92224$

Let $T_A$ be the event that A speaks the truth, and $T_B$ be the event that B speaks the truth. Let $F_A$ be the event that A lies, and $F_B$ be the event that B lies.

We are given:

$P(T_A) = 0.60$, so $P(F_A) = 1 - 0.60 = 0.40$

$P(T_B) = 0.90$, so $P(F_B) = 1 - 0.90 = 0.10$

They contradict each other if one speaks the truth and the other lies. This can happen in two ways:

1. A speaks the truth and B lies: $P(T_A \text{ and } F_B) = P(T_A) \cdot P(F_B) = (0.60)(0.10) = 0.06$

2. A lies and B speaks the truth: $P(F_A \text{ and } T_B) = P(F_A) \cdot P(T_B) = (0.40)(0.90) = 0.36$

The probability that they contradict each other is the sum of these probabilities:

$P(\text{contradiction}) = P(T_A \text{ and } F_B) + P(F_A \text{ and } T_B) = 0.06 + 0.36 = 0.42$

Therefore, in 42% of the cases, they are likely to contradict each other.

When three coins are tossed, the possible outcomes are {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}. The random variable $X$ represents the number of heads. Therefore, $X$ can take values 0, 1, 2, or 3.

The probability distribution of $X$ is as follows:

$P(X=0) = P(\text{TTT}) = \frac{1}{8}$

$P(X=1) = P(\text{HTT, THT, TTH}) = \frac{3}{8}$

$P(X=2) = P(\text{HHT, HTH, THH}) = \frac{3}{8}$

$P(X=3) = P(\text{HHH}) = \frac{1}{8}$

We can summarize this in a table:

Now, we calculate the mean of $X$, denoted by $\mu$ or $E(X)$:

$E(X) = \sum_{i} x_i P(x_i) = 0 \cdot \frac{1}{8} + 1 \cdot \frac{3}{8} + 2 \cdot \frac{3}{8} + 3 \cdot \frac{1}{8} = 0 + \frac{3}{8} + \frac{6}{8} + \frac{3}{8} = \frac{12}{8} = \frac{3}{2} = 1.5$

Next, we calculate the variance of $X$, denoted by $Var(X)$ or $\sigma^2$:

$Var(X) = E(X^2) - [E(X)]^2$

First, we find $E(X^2)$:

$E(X^2) = \sum_{i} x_i^2 P(x_i) = 0^2 \cdot \frac{1}{8} + 1^2 \cdot \frac{3}{8} + 2^2 \cdot \frac{3}{8} + 3^2 \cdot \frac{1}{8} = 0 + \frac{3}{8} + \frac{12}{8} + \frac{9}{8} = \frac{24}{8} = 3$

Then, $Var(X) = E(X^2) - [E(X)]^2 = 3 - \left(\frac{3}{2}\right)^2 = 3 - \frac{9}{4} = \frac{12 - 9}{4} = \frac{3}{4} = 0.75$

Therefore, the mean of $X$ is 1.5 and the variance of $X$ is 0.75.

Let $A$, $B$, and $C$ be the events that the item was produced by machine operators A, B, and C, respectively. Let $D$ be the event that the item is defective.

We are given the following probabilities:

$P(A) = 0.50$

$P(B) = 0.30$

$P(C) = 0.20$

$P(D|A) = 0.01$

$P(D|B) = 0.05$

$P(D|C) = 0.07$

We want to find $P(A|D)$, the probability that the item was produced by operator A given that it is defective. We can use Bayes' Theorem:

$P(A|D) = \frac{P(D|A)P(A)}{P(D)}$

First, we need to find $P(D)$, the probability that an item is defective. We can use the law of total probability:

$P(D) = P(D|A)P(A) + P(D|B)P(B) + P(D|C)P(C)$

$P(D) = (0.01)(0.50) + (0.05)(0.30) + (0.07)(0.20)$

$P(D) = 0.005 + 0.015 + 0.014 = 0.034$

Now we can find $P(A|D)$:

$P(A|D) = \frac{P(D|A)P(A)}{P(D)} = \frac{(0.01)(0.50)}{0.034} = \frac{0.005}{0.034} = \frac{5}{34} \approx 0.1471$

Therefore, the probability that the defective item was produced by machine operator A is approximately 0.1471 or 14.71%.

Let $X$ be the random variable representing the number of defective bulbs in the sample of 4. Since the bulbs are drawn with replacement, this is a binomial distribution.

The probability of drawing a defective bulb is $p = \frac{6}{30} = \frac{1}{5} = 0.2$.

The probability of drawing a non-defective bulb is $q = 1 - p = 1 - \frac{1}{5} = \frac{4}{5} = 0.8$.

We have a sample of $n = 4$ bulbs.

The probability distribution of $X$ is given by $P(X=k) = \binom{n}{k} p^k q^{n-k}$, where $k$ can be 0, 1, 2, 3, or 4.

$P(X=0) = \binom{4}{0} (0.2)^0 (0.8)^4 = 1 \cdot 1 \cdot (0.8)^4 = 0.4096$

$P(X=1) = \binom{4}{1} (0.2)^1 (0.8)^3 = 4 \cdot (0.2) \cdot (0.8)^3 = 4 \cdot 0.2 \cdot 0.512 = 0.4096$

$P(X=2) = \binom{4}{2} (0.2)^2 (0.8)^2 = 6 \cdot (0.2)^2 \cdot (0.8)^2 = 6 \cdot 0.04 \cdot 0.64 = 0.1536$

$P(X=3) = \binom{4}{3} (0.2)^3 (0.8)^1 = 4 \cdot (0.2)^3 \cdot (0.8)^1 = 4 \cdot 0.008 \cdot 0.8 = 0.0256$

$P(X=4) = \binom{4}{4} (0.2)^4 (0.8)^0 = 1 \cdot (0.2)^4 \cdot 1 = 0.0016$

The probability distribution is:

The mean of this distribution is given by $\mu = E(X) = np = 4 \cdot (0.2) = 0.8$.

Therefore, the mean of the distribution is 0.8.

Let $D$ be the event that the lost card was a diamond. Let $ND$ be the event that the lost card was not a diamond. Let $A$ be the event that two cards drawn from the remaining cards are both diamonds.

We want to find $P(D|A)$, the probability that the lost card was a diamond given that two cards drawn are diamonds.

We have $P(D) = \frac{13}{52} = \frac{1}{4}$ (since there are 13 diamonds in a deck of 52 cards).

And $P(ND) = 1 - P(D) = 1 - \frac{1}{4} = \frac{3}{4}$.

We need to find $P(A|D)$ and $P(A|ND)$.

If the lost card was a diamond, then there are 12 diamonds left in the deck of 51 cards. The probability of drawing two diamonds is:

$P(A|D) = \frac{\binom{12}{2}}{\binom{51}{2}} = \frac{\frac{12 \cdot 11}{2}}{\frac{51 \cdot 50}{2}} = \frac{12 \cdot 11}{51 \cdot 50} = \frac{132}{2550} = \frac{22}{425}$

If the lost card was not a diamond, then there are 13 diamonds left in the deck of 51 cards. The probability of drawing two diamonds is:

$P(A|ND) = \frac{\binom{13}{2}}{\binom{51}{2}} = \frac{\frac{13 \cdot 12}{2}}{\frac{51 \cdot 50}{2}} = \frac{13 \cdot 12}{51 \cdot 50} = \frac{156}{2550} = \frac{26}{425}$

By Bayes' Theorem, we have:

$P(D|A) = \frac{P(A|D)P(D)}{P(A|D)P(D) + P(A|ND)P(ND)}$

$P(D|A) = \frac{\frac{22}{425} \cdot \frac{1}{4}}{\frac{22}{425} \cdot \frac{1}{4} + \frac{26}{425} \cdot \frac{3}{4}} = \frac{\frac{22}{425 \cdot 4}}{\frac{22}{425 \cdot 4} + \frac{78}{425 \cdot 4}} = \frac{22}{22 + 78} = \frac{22}{100} = \frac{11}{50} = 0.22$

Therefore, the probability that the lost card was a diamond is $\frac{11}{50}$ or 0.22.

Given:

A die is thrown 6 times. So, the number of trials is $n=6$.

A 'success' is defined as getting an odd number.

When a fair die is thrown, the possible outcomes are $\{1, 2, 3, 4, 5, 6\}$.

The odd numbers are $\{1, 3, 5\}$. There are 3 odd numbers.

The probability of success (getting an odd number) in a single throw is $p = \frac{\text{Number of odd outcomes}}{\text{Total outcomes}} = \frac{3}{6} = \frac{1}{2}$.

The probability of failure (not getting an odd number) is $q = 1 - p = 1 - \frac{1}{2} = \frac{1}{2}$.

Let $X$ be the random variable representing the number of successes in 6 trials.

$X$ follows a binomial distribution with parameters $n=6$ and $p=\frac{1}{2}$.

The probability mass function (PMF) for a binomial distribution is given by:

In this case, substituting $n=6$, $p=\frac{1}{2}$, and $q=\frac{1}{2}$ into equation (1):

$P(X=k) = \binom{6}{k} \left(\frac{1}{2}\right)^k \left(\frac{1}{2}\right)^{6-k} = \binom{6}{k} \left(\frac{1}{2}\right)^{k+6-k} = \binom{6}{k} \left(\frac{1}{2}\right)^6$

We know that $\left(\frac{1}{2}\right)^6 = \frac{1}{2^6} = \frac{1}{64}$. So,

$P(X=k) = \binom{6}{k} \frac{1}{64}$

To Find:

(i) Probability of 5 successes, i.e., $P(X=5)$.

(ii) Probability of at least 5 successes, i.e., $P(X \ge 5)$.

(iii) Probability of at most 5 successes, i.e., $P(X \le 5)$.

Solution:

(i) Probability of 5 successes

We need to find $P(X=5)$. Using the formula $P(X=k) = \binom{6}{k} \frac{1}{64}$ with $k=5$:

$P(X=5) = \binom{6}{5} \frac{1}{64}$

First, calculate the binomial coefficient $\binom{6}{5}$:

$\binom{6}{5} = \frac{6!}{5!(6-5)!} = \frac{6!}{5!1!} = \frac{6 \times 5!}{5! \times 1} = 6$.

So, $P(X=5) = 6 \times \frac{1}{64} = \frac{6}{64}$.

This can be simplified to $P(X=5) = \frac{3}{32}$.

Thus, the probability of 5 successes is $\frac{3}{32}$.

(ii) Probability of at least 5 successes

At least 5 successes means $X=5$ or $X=6$. So we need to find $P(X \ge 5) = P(X=5) + P(X=6)$.

We already found $P(X=5) = \frac{6}{64}$.

Now, we calculate $P(X=6)$. Using the formula $P(X=k) = \binom{6}{k} \frac{1}{64}$ with $k=6$:

$P(X=6) = \binom{6}{6} \frac{1}{64}$

Calculate the binomial coefficient $\binom{6}{6}$:

$\binom{6}{6} = \frac{6!}{6!(6-6)!} = \frac{6!}{6!0!} = 1$ (since $0!=1$).

So, $P(X=6) = 1 \times \frac{1}{64} = \frac{1}{64}$.

Now, $P(X \ge 5) = P(X=5) + P(X=6) = \frac{6}{64} + \frac{1}{64} = \frac{6+1}{64} = \frac{7}{64}$.

Thus, the probability of at least 5 successes is $\frac{7}{64}$.

(iii) Probability of at most 5 successes

At most 5 successes means $X \le 5$. This means the number of successes can be $0, 1, 2, 3, 4,$ or $5$.

This can be calculated as $P(X \le 5) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5)$.

Alternatively, it is easier to calculate this using the complement rule: $P(X \le 5) = 1 - P(X > 5)$.

The event $X > 5$ means $X=6$ (since the maximum number of trials is $n=6$).

So, $P(X \le 5) = 1 - P(X=6)$.

We calculated $P(X=6) = \frac{1}{64}$ in part (ii).

$P(X \le 5) = 1 - \frac{1}{64} = \frac{64}{64} - \frac{1}{64} = \frac{64-1}{64} = \frac{63}{64}$.

Thus, the probability of at most 5 successes is $\frac{63}{64}$.

Given:

There are two bags, Bag I and Bag II.

Bag I contains: 5 white (W) balls and 7 black (B) balls. Total balls in Bag I = $5 + 7 = 12$.

Bag II contains: 3 white (W) balls and 5 black (B) balls. Total balls in Bag II = $3 + 5 = 8$.

Process:

1. One ball is drawn from Bag I and transferred to Bag II.

2. Then a ball is drawn from Bag II.

To Find:

The probability that the ball drawn from Bag II is white.

Solution:

Let $E_1$ be the event that a white ball is transferred from Bag I to Bag II.

Let $E_2$ be the event that a black ball is transferred from Bag I to Bag II.

Let $A$ be the event that the ball drawn from Bag II is white.

We need to find $P(A)$.

First, let's calculate the probabilities of events $E_1$ and $E_2$:

The probability of drawing a white ball from Bag I is:

$P(E_1) = \frac{\text{Number of white balls in Bag I}}{\text{Total balls in Bag I}} = \frac{5}{12}$

The probability of drawing a black ball from Bag I is:

$P(E_2) = \frac{\text{Number of black balls in Bag I}}{\text{Total balls in Bag I}} = \frac{7}{12}$

Note that $P(E_1) + P(E_2) = \frac{5}{12} + \frac{7}{12} = \frac{12}{12} = 1$.

Now, let's consider the composition of Bag II after the transfer, and the probability of drawing a white ball from Bag II in each case.

Case 1: A white ball is transferred from Bag I to Bag II (event $E_1$ occurs).

If a white ball is transferred from Bag I to Bag II:

Bag II will now contain $(3+1)$ white balls and $5$ black balls.

So, Bag II has $4$ white balls and $5$ black balls. Total balls in Bag II = $4 + 5 = 9$.

The probability of drawing a white ball from Bag II, given that a white ball was transferred (event $A|E_1$), is:

$P(A|E_1) = \frac{\text{Number of white balls in Bag II}}{\text{Total balls in Bag II}} = \frac{4}{9}$

Case 2: A black ball is transferred from Bag I to Bag II (event $E_2$ occurs).

If a black ball is transferred from Bag I to Bag II:

Bag II will now contain $3$ white balls and $(5+1)$ black balls.

So, Bag II has $3$ white balls and $6$ black balls. Total balls in Bag II = $3 + 6 = 9$.

The probability of drawing a white ball from Bag II, given that a black ball was transferred (event $A|E_2$), is:

$P(A|E_2) = \frac{\text{Number of white balls in Bag II}}{\text{Total balls in Bag II}} = \frac{3}{9} = \frac{1}{3}$

Now, we can find the total probability of drawing a white ball from Bag II, $P(A)$, using the law of total probability:

Substituting the values we calculated:

$P(A) = \left(\frac{4}{9}\right) \left(\frac{5}{12}\right) + \left(\frac{3}{9}\right) \left(\frac{7}{12}\right)$

$P(A) = \frac{4 \times 5}{9 \times 12} + \frac{3 \times 7}{9 \times 12}$

$P(A) = \frac{20}{108} + \frac{21}{108}$

$P(A) = \frac{20 + 21}{108}$

$P(A) = \frac{41}{108}$

The number 41 is a prime number, and 108 is not divisible by 41. So, the fraction is in its simplest form.

Thus, the probability that the ball drawn from Bag II is white is $\frac{41}{108}$.