Chapter 15 Probability (Additional Questions)

Given:

Total number of times the coin is tossed = 100

Number of times a head appears = 45

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To Find:

The empirical probability of getting a head.

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Solution:

The empirical probability of an event is calculated as the ratio of the number of times the event occurred to the total number of trials.

Empirical Probability (Event) = $\frac{\text{Number of times the event occurred}}{\text{Total number of trials}}$

Here, the event is getting a head.

Number of times a head appeared = 45

Total number of trials = 100

Therefore, the empirical probability of getting a head is:

P(Head) = $\frac{\text{Number of heads}}{\text{Total number of tosses}}$

P(Head) = $\frac{45}{100}$

Comparing this result with the given options:

(A) $\frac{45}{100}$

(B) $\frac{55}{100}$

(C) $\frac{1}{2}$

(D) 45

The calculated empirical probability matches option (A).

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Final Answer:

The empirical probability of getting a head is $\frac{45}{100}$.

The correct option is (A) $\frac{45}{100}$.

Solution:

In probability, an experiment (also called a trial or observation) is a process or action that results in one of several possible outcomes.

The outcomes of a probability experiment must be well-defined, meaning that it is clear what constitutes each outcome, and the set of all possible outcomes is known before the experiment is conducted.

Let's evaluate the given options:

(A) Only one possible outcome: This is incorrect. A probability experiment typically has two or more possible outcomes (e.g., tossing a coin can result in head or tail).

(B) Two predetermined outcomes: This is incorrect. While some experiments have exactly two outcomes (like a coin toss), many others have more (like rolling a die, which has 6 outcomes) or the outcomes are not predetermined in the sense of being fixed or certain to happen in a specific order.

(C) Some well-defined outcomes: This is correct. The key characteristic of a probability experiment is that its outcomes are known and clearly identifiable before the experiment takes place, even though the result of a single trial is uncertain.

(D) No outcome: This is incorrect. An experiment in probability must produce at least one outcome.

Thus, an experiment in probability is an action which produces some well-defined outcomes.

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Final Answer:

The correct option is (C) Some well-defined outcomes.

Given:

Total number of times the die is rolled = 200

Number of times the number 6 appears = 30

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To Find:

The empirical probability of getting a 6.

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Solution:

The empirical probability of an event is defined as the ratio of the number of times the event occurs in an experiment to the total number of trials.

Empirical Probability (Event) = $\frac{\text{Number of times the event occurred}}{\text{Total number of trials}}$

In this case, the event is getting a 6.

Number of times the number 6 appeared = 30

Total number of trials (die rolls) = 200

Therefore, the empirical probability of getting a 6 is:

P(Getting a 6) = $\frac{\text{Number of times 6 appears}}{\text{Total number of rolls}}$

P(Getting a 6) = $\frac{30}{200}$

Comparing this result with the given options:

(A) $\frac{30}{200}$

(B) $\frac{1}{6}$

(C) $\frac{170}{200}$

(D) $\frac{200}{30}$

The calculated empirical probability is $\frac{30}{200}$, which matches option (A).

Note that $\frac{30}{200}$ can be simplified to $\frac{3}{20}$, but option (A) is presented in its unsimplified form, matching the direct ratio from the problem statement.

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Final Answer:

The empirical probability of getting a 6 is $\frac{30}{200}$.

The correct option is (A) $\frac{30}{200}$.

Solution:

The empirical probability of an event is calculated based on the results of an experiment. It is given by the formula:

Empirical Probability (Event) = $\frac{\text{Number of times the event occurred}}{\text{Total number of trials}}$

Let $N$ be the total number of trials in an experiment, and let $E$ be the number of times a specific event occurs during these $N$ trials.

The number of times an event can occur ($E$) cannot be less than 0 (it cannot happen a negative number of times) and cannot be more than the total number of trials ($N$, as it cannot happen more times than the experiment was performed).

So, we have:

To find the range of the empirical probability, we divide this inequality by the total number of trials, $N$. Since $N$ is the total number of trials, it must be a positive integer ($N > 0$). Dividing by a positive number does not change the direction of the inequalities.

The ratio $\frac{E}{N}$ is the empirical probability of the event.

Therefore, the empirical probability of an event must be a value greater than or equal to 0 and less than or equal to 1.

Comparing this result with the given options:

(A) From 0 to 1 (inclusive)

(B) From 0 to infinity

(C) From -1 to 1

(D) Always greater than 1

The calculated range $0 \leq \text{Empirical Probability} \leq 1$ matches option (A).

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Final Answer:

The range of empirical probability of an event is from 0 to 1 (inclusive).

The correct option is (A) From 0 to 1 (inclusive).

Solution:

In probability theory, events are classified based on their probability of occurrence.

Let P(E) denote the probability of an event E.

If P(E) = 1, the event is called a sure or certain event. This means the event will always occur.

If P(E) = 0, the event is called an impossible event. This means the event will never occur.

If $0 < P(E) < 1$, the event is called a possible or uncertain event. This means the event may or may not occur.

A random experiment is one where the outcomes are uncertain.

The question asks for the name of an event with a probability of 0.

Based on the definitions above, an event with probability 0 is an impossible event.

Let's compare this with the given options:

(A) Sure - Incorrect (Probability is 1)

(B) Impossible - Correct (Probability is 0)

(C) Possible - Incorrect (Probability is between 0 and 1, exclusive)

(D) Random - Describes the nature of the experiment, not the event itself having probability 0.

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Final Answer:

If the probability of an event is 0, it is called an impossible event.

The correct option is (B) Impossible.

Given:

Total number of students surveyed = 200

Number of students who prefer tea = 120

Number of students who prefer coffee = 80

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To Find:

The empirical probability that a randomly chosen student prefers coffee.

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Solution:

The empirical probability of an event is calculated as the ratio of the number of times the event occurred to the total number of trials.

Empirical Probability (Event) = $\frac{\text{Number of times the event occurred}}{\text{Total number of trials}}$

Here, the event is selecting a student who prefers coffee.

Number of times the event occurred (students preferring coffee) = 80

Total number of trials (total students surveyed) = 200

Therefore, the empirical probability that a randomly chosen student prefers coffee is:

P(Student prefers coffee) = $\frac{\text{Number of students who prefer coffee}}{\text{Total number of students surveyed}}$

P(Student prefers coffee) = $\frac{80}{200}$

Comparing this result with the given options:

(A) $\frac{120}{200}$

(B) $\frac{80}{200}$

(C) $\frac{200}{80}$

(D) $\frac{120}{80}$

The calculated empirical probability is $\frac{80}{200}$, which matches option (B).

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Final Answer:

The empirical probability that a randomly chosen student prefers coffee is $\frac{80}{200}$.

The correct option is (B) $\frac{80}{200}$.

Given:

Number of red balls = 5

Number of blue balls = 3

Number of green balls = 2

A ball is drawn at random from the bag.

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To Find:

The total number of possible outcomes when drawing one ball.

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Solution:

When an experiment is performed, the set of all possible results is called the sample space, and each result is called an outcome.

In this experiment, we are drawing one ball at random from the bag.

The possible outcomes are drawing any one of the balls in the bag.

To find the total number of possible outcomes, we need to find the total number of balls in the bag.

Total number of balls = Number of red balls + Number of blue balls + Number of green balls

Total number of balls = $5 + 3 + 2$

Total number of balls = $10$

Since there are 10 balls in total, and we are drawing one ball, there are 10 distinct possible outcomes (each ball being a distinct outcome).

Therefore, the total number of possible outcomes is 10.

Comparing this result with the given options:

(A) 3

(B) 5

(C) 10

(D) Cannot be determined

The calculated total number of possible outcomes is 10, which matches option (C).

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Final Answer:

The total number of possible outcomes is 10.

The correct option is (C) 10.

Given:

From Question 7:

Number of red balls = 5

Number of blue balls = 3

Number of green balls = 2

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To Find:

The empirical probability of drawing a red ball when one ball is drawn at random.

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Solution:

First, we find the total number of possible outcomes when drawing one ball from the bag.

Total number of balls = Number of red balls + Number of blue balls + Number of green balls

Total number of balls = $5 + 3 + 2 = 10$

The number of favorable outcomes for the event "drawing a red ball" is the number of red balls in the bag, which is 5.

The empirical probability of an event is calculated as:

P(Event) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

In this case, the event is drawing a red ball.

Number of favorable outcomes (drawing a red ball) = 5

Total number of possible outcomes (drawing any ball) = 10

Therefore, the empirical probability of drawing a red ball is:

P(Drawing a red ball) = $\frac{\text{Number of red balls}}{\text{Total number of balls}}$

P(Drawing a red ball) = $\frac{5}{10}$

Comparing this result with the given options:

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

(B) $\frac{5}{10}$

(C) $\frac{3}{10}$

(D) $\frac{2}{10}$

The calculated empirical probability is $\frac{5}{10}$, which matches option (B).

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Final Answer:

The empirical probability of drawing a red ball is $\frac{5}{10}$.

The correct option is (B) $\frac{5}{10}$.

Given:

The probability of event E happening is $P(E)$.

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To Find:

The probability of event E NOT happening.

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Solution:

In probability, the event "E NOT happening" is called the complement of event E. It is often denoted by $E'$ or $\overline{E}$.

The union of an event and its complement is the entire sample space, and the intersection of an event and its complement is an empty set (they cannot happen at the same time).

For any event E, the sum of the probability of the event happening and the probability of the event not happening is always equal to 1. This is because either the event happens or it doesn't, and these are the only two possibilities covering the entire sample space.

Mathematically, this relationship is expressed as:

We are given P(E) as the probability of event E happening. We need to find the probability of event E NOT happening, which is P(E NOT happening).

Rearranging the equation to solve for P(E NOT happening):

Let's compare this result with the given options:

(A) $P(E) - 1$

(B) $1 - P(E)$

(C) $\frac{1}{P(E)}$

(D) $P(E)$

The calculated probability of event E NOT happening is $1 - P(E)$, which matches option (B).

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Final Answer:

If the probability of event E happening is $P(E)$, then the probability of event E NOT happening is $1 - P(E)$.

The correct option is (B) $1 - P(E)$.

Solution:

In probability theory, an elementary event is a single possible outcome of a random experiment.

The set of all possible elementary events constitutes the sample space of the experiment.

A fundamental property of probability distributions is that the sum of the probabilities of all the disjoint outcomes (elementary events) that make up the sample space must equal 1.

Let $E_1, E_2, \dots, E_n$ be the elementary events of an experiment. These events are mutually exclusive (they cannot occur at the same time) and exhaustive (they cover all possible outcomes).

The sum of their probabilities is:

This property reflects the fact that in any given trial of the experiment, one and only one of the elementary events must occur, and the total probability of something happening is 1 (certainty).

Comparing this result with the given options:

(A) 0

(B) 1

(C) Greater than 1

(D) Less than 1

The sum of the probabilities of all the elementary events of an experiment is always 1.

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Final Answer:

The sum of the probabilities of all the elementary events of an experiment is 1.

The correct option is (B) 1.

Solution:

Let's analyze the Assertion (A) and the Reason (R).

Assertion (A): The empirical probability of an event changes if the number of trials increases.

The empirical probability of an event is calculated as:

P(E) = $\frac{\text{Number of times the event occurred}}{\text{Total number of trials}}$

When the number of trials increases, the count of the event occurring also changes based on the new trials. Since the numerator (occurrences of the event) and the denominator (total trials) both change, the calculated ratio, which is the empirical probability, will generally change as the experiment continues with more trials. While the empirical probability tends to approach the theoretical probability as the number of trials becomes very large (Law of Large Numbers), for any increase in the number of trials, the calculated empirical probability is likely to be different from the previous calculation.

Thus, Assertion (A) is true.

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Reason (R): Empirical probability is based on the results of actual experiments, which can vary with the number of trials.

Empirical probability relies on observed frequencies from conducting an experiment. The outcome of each individual trial in a random experiment is uncertain. Therefore, the number of times a specific event occurs within a fixed number of trials can vary if the experiment is repeated with a different number of trials or even the same number of trials.

Thus, Reason (R) is true.

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Relationship between A and R:

Reason (R) states that empirical probability is based on variable experimental results. This variability in the results is precisely why the calculated empirical probability changes when the number of trials increases. As more trials are conducted, new outcomes are included in the calculation, which updates the frequency of the event and the total number of trials, leading to a potentially different empirical probability.

Therefore, Reason (R) correctly explains why Assertion (A) is true.

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Final Answer:

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

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

Solution:

Let's analyze the Assertion (A) and the Reason (R).

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Assertion (A): The probability of getting a number less than 7 when rolling a standard die is 1.

A standard die has six faces, numbered 1, 2, 3, 4, 5, and 6.

The total number of possible outcomes when rolling a standard die is 6.

The event is "getting a number less than 7". The numbers less than 7 on a standard die are 1, 2, 3, 4, 5, and 6.

The number of favorable outcomes for this event is 6.

The probability of an event is given by the ratio of the number of favorable outcomes to the total number of possible outcomes.

P(Number less than 7) = $\frac{\text{Number of outcomes less than 7}}{\text{Total number of outcomes}}$

P(Number less than 7) = $\frac{6}{6} = 1$

Thus, the probability of getting a number less than 7 when rolling a standard die is 1. Assertion (A) is true.

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Reason (R): Getting a number less than 7 (1, 2, 3, 4, 5, or 6) is a sure event in this experiment.

A sure event (or certain event) is an event that is certain to occur. The probability of a sure event is 1.

In the experiment of rolling a standard die, the possible outcomes are {1, 2, 3, 4, 5, 6}.

The event "getting a number less than 7" includes all these possible outcomes {1, 2, 3, 4, 5, 6}.

Since the outcome of rolling a standard die must be one of these numbers, the event "getting a number less than 7" is certain to occur in every trial.

Therefore, getting a number less than 7 is a sure event. Reason (R) is true.

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Relationship between A and R:

Reason (R) states that the event "getting a number less than 7" is a sure event. The definition of a sure event is one that has a probability of 1. Assertion (A) states that the probability of this event is 1. Therefore, Reason (R) provides the correct explanation for why Assertion (A) is true.

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Final Answer:

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

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

Solution:

We need to match the probability values in Column A with the corresponding descriptions of event types in Column B.

Let's consider each probability value and its definition:

(i) 0: An event with a probability of 0 is an event that cannot happen. This is called an Impossible event. This matches option (b) in Column B.

(ii) 1: An event with a probability of 1 is an event that is certain to happen. This is called a Sure event or a Certain event. This matches option (d) in Column B.

(iii) 0.5: A probability of 0.5 (or $\frac{1}{2}$) means the event has an equal chance of happening or not happening. This matches option (c) in Column B.

(iv) 0.7: A probability of 0.7 is a value between 0.5 and 1. An event with a probability greater than 0.5 is considered more likely to happen than not. This is called a Likely event. This matches option (a) in Column B.

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Based on the analysis, the correct matching is:

(i) 0 $\to$ (b) Impossible event

(ii) 1 $\to$ (d) Sure event

(iii) 0.5 $\to$ (c) Event with equal chance of happening or not happening

(iv) 0.7 $\to$ (a) Likely event (more probable than not)

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Now let's check the given options:

(A) (i)-(b), (ii)-(d), (iii)-(c), (iv)-(a) - This matches our derived mapping.

(B) (i)-(d), (ii)-(b), (iii)-(c), (iv)-(a) - Incorrect (i) and (ii) are mismatched.

(C) (i)-(b), (ii)-(c), (iii)-(d), (iv)-(a) - Incorrect (ii) and (iii) are mismatched.

(D) (i)-(a), (ii)-(d), (iii)-(c), (iv)-(b) - Incorrect (i), (iii), and (iv) are mismatched.

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Final Answer:

The correct matching is (i)-(b), (ii)-(d), (iii)-(c), (iv)-(a).

The correct option is (A) (i)-(b), (ii)-(d), (iii)-(c), (iv)-(a).

Given:

Total number of electric bulbs inspected = 500

Number of defective bulbs found = 20

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To Find:

The empirical probability that a randomly selected bulb from this batch is defective.

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Solution:

The empirical probability of an event is determined by conducting an experiment and observing the frequency of the event.

It is calculated using the formula:

Empirical Probability (Event) = $\frac{\text{Number of times the event occurred}}{\text{Total number of trials}}$

In this case, the experiment is inspecting a batch of electric bulbs.

The total number of trials is the total number of bulbs inspected, which is 500.

The event we are interested in is selecting a defective bulb.

The number of times this event occurred in the inspection is the number of defective bulbs found, which is 20.

Therefore, the empirical probability of selecting a defective bulb is:

P(Defective bulb) = $\frac{\text{Number of defective bulbs}}{\text{Total number of bulbs inspected}}$

P(Defective bulb) = $\frac{20}{500}$

Comparing this result with the given options:

(A) $\frac{20}{480}$

(B) $\frac{20}{500}$

(C) $\frac{480}{500}$

(D) $\frac{1}{20}$

The calculated empirical probability is $\frac{20}{500}$, which matches option (B).

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Final Answer:

The empirical probability that a randomly selected bulb from this batch is defective is $\frac{20}{500}$.

The correct option is (B) $\frac{20}{500}$.

Given:

From the data in Question 14:

Total number of electric bulbs inspected = 500

Number of defective bulbs found = 20

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To Find:

The empirical probability that a randomly selected bulb is NOT defective.

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Solution:

We are looking for the empirical probability of the event "a randomly selected bulb is NOT defective". This is the complement of the event "a randomly selected bulb is defective".

First, let's find the number of non-defective bulbs.

Number of non-defective bulbs = Total number of bulbs - Number of defective bulbs

Number of non-defective bulbs = $500 - 20 = 480$

Using the empirical probability formula:

P(Not defective) = $\frac{\text{Number of non-defective bulbs}}{\text{Total number of bulbs}}$

P(Not defective) = $\frac{480}{500}$

This result matches option (B).

Alternatively, we can use the complement rule of probability. If P(E) is the probability of an event E, then the probability of the event E NOT happening is $1 - P(E)$.

Let E be the event "the bulb is defective". We found in Question 14 that $P(\text{defective}) = \frac{20}{500}$.

The event "the bulb is NOT defective" is the complement of E.

P(Not defective) = $1 - P(\text{defective})$

This expression matches option (C).

Substituting the value of P(defective):

P(Not defective) = $1 - \frac{20}{500}$

This expression matches option (A).

Let's verify if options (A) and (B) are equivalent:

Indeed, options (A) and (B) are mathematically equal and represent the correct empirical probability of a bulb not being defective. Option (C) is the general probability rule for the complement event, which is also applicable here.

Since options (A), (B), and (C) are all correct expressions or values for the empirical probability of a randomly selected bulb being not defective, the correct answer is (D).

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Final Answer:

The empirical probability that a randomly selected bulb is NOT defective can be expressed as $1 - \frac{20}{500}$, $\frac{480}{500}$, or $1 - P(\text{defective})$.

The correct option is (D) All of the above.

Solution:

A fundamental property of probability states that the probability of any event E, denoted by P(E), must be a value between 0 and 1, inclusive.

This can be expressed mathematically as:

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

(A) $0.75$:

This inequality is true. Therefore, 0.75 can be a probability of an event.

(B) $-0.5$:

This inequality is false because $-0.5 < 0$. Probability values cannot be negative. Therefore, -0.5 cannot be a probability of an event.

(C) $\frac{1}{3}$:

The value $\frac{1}{3}$ is approximately 0.333... .

This inequality is true. Therefore, $\frac{1}{3}$ can be a probability of an event.

(D) 1:

This inequality is true. A probability of 1 represents a sure event (an event that is certain to happen). Therefore, 1 can be a probability of an event.

The only value among the given options that violates the condition $0 \leq P(E) \leq 1$ is $-0.5$.

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Final Answer:

The value that cannot be the probability of an event is $-0.5$.

The correct option is (B) $-0.5$.

Given:

Total number of times the coin is tossed = 200

Number of times Heads appeared = 108

Number of times Tails appeared = 92

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To Find:

The empirical probability of getting a tail.

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Solution:

The empirical probability of an event is given by the ratio of the number of times the event occurred to the total number of trials.

Empirical Probability (Event) = $\frac{\text{Number of times the event occurred}}{\text{Total number of trials}}$

Here, the event is getting a tail.

Number of times tails appeared = 92

Total number of trials (tosses) = 200

Therefore, the empirical probability of getting a tail is:

P(Tail) = $\frac{\text{Number of tails}}{\text{Total number of tosses}}$

P(Tail) = $\frac{92}{200}$

Comparing this result with the given options:

(A) $\frac{108}{200}$ (This is the empirical probability of getting a head)

(B) $\frac{92}{200}$

(C) $\frac{1}{2}$ (This is the theoretical probability for a fair coin, but the question asks for empirical probability based on specific results)

(D) $1 - \frac{108}{200}$ (Using the complement rule, $1 - P(\text{Heads}) = P(\text{Tails})$. $1 - \frac{108}{200} = \frac{200-108}{200} = \frac{92}{200}$. This is equivalent to option B)

Option (B) directly represents the empirical probability calculated from the given data.

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Final Answer:

The empirical probability of getting a tail is $\frac{92}{200}$.

The correct option is (B) $\frac{92}{200}$.

Given:

A bag contains 10 slips of paper numbered from 1 to 10.

An experiment is performed: drawing a single slip at random.

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To Find:

The empirical probability of drawing an even number in a single draw.

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Solution:

When a slip is drawn at random from the bag, each of the 10 slips has an equal chance of being drawn.

The total number of possible outcomes in this experiment is the total number of slips in the bag.

Total number of possible outcomes = 10 (slips numbered 1, 2, 3, 4, 5, 6, 7, 8, 9, 10)

The event we are interested in is drawing an even number.

The even numbers between 1 and 10 are 2, 4, 6, 8, and 10.

The number of favorable outcomes (drawing an even number) is the count of even numbers from 1 to 10, which is 5.

The empirical (or classical) probability of an event is given by the ratio of the number of favorable outcomes to the total number of possible outcomes.

P(Event) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

For the event "drawing an even number":

P(Drawing an even number) = $\frac{\text{Number of even numbers}}{\text{Total number of slips}}$

P(Drawing an even number) = $\frac{5}{10}$

Comparing this result with the given options:

(A) $\frac{1}{10}$

(B) $\frac{5}{10}$

(C) $\frac{10}{5}$

(D) $\frac{4}{10}$

The calculated probability is $\frac{5}{10}$, which matches option (B).

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Final Answer:

The empirical probability of drawing an even number in a single draw is $\frac{5}{10}$.

The correct option is (B) $\frac{5}{10}$.

Given:

A spinner is divided into 4 equal sectors.

The colors of the sectors are Red, Blue, Green, and Yellow.

The experiment is a single spin of the spinner.

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To Find:

The empirical probability of landing on Green in a single spin.

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Solution:

Since the spinner is divided into 4 equal sectors, each sector has an equal chance of the spinner landing on it.

The total number of possible outcomes when spinning the spinner once is the total number of sectors.

Total number of possible outcomes = 4 (Red, Blue, Green, Yellow)

The event we are interested in is landing on Green.

There is only one sector colored Green.

The number of favorable outcomes for the event "landing on Green" is 1.

The empirical probability of an event in a situation with equally likely outcomes is given by the ratio of the number of favorable outcomes to the total number of possible outcomes.

P(Event) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

For the event "landing on Green":

P(Landing on Green) = $\frac{\text{Number of Green sectors}}{\text{Total number of sectors}}$

P(Landing on Green) = $\frac{1}{4}$

Comparing this result with the given options:

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

(B) $\frac{1}{3}$

(C) $\frac{3}{4}$

(D) 1

The calculated probability is $\frac{1}{4}$, which matches option (A).

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Final Answer:

The empirical probability of landing on Green in a single spin is $\frac{1}{4}$.

The correct option is (A) $\frac{1}{4}$.

Solution:

In probability, an experiment (or random experiment) is a process or action whose outcome cannot be predicted with certainty before the experiment is performed.

A trial is a single instance or performance of such an experiment.

An outcome is a possible result of a trial.

An event is a collection of one or more outcomes.

Probability is a numerical measure of the likelihood of an event occurring.

Let's consider an example: Rolling a standard die.

The action of rolling the die is the experiment.

Each time we roll the die, it is a trial.

The result of a single roll (e.g., getting a 3) is an outcome.

A set of outcomes (e.g., getting an even number, which is {2, 4, 6}) is an event.

The probability of getting an even number is the measure associated with this event.

Based on these definitions, a trial is a single performance of an experiment.

Comparing this with the given options:

(A) Outcome: Incorrect. An outcome is the result *of* a trial, not what a trial is a performance of.

(B) Event: Incorrect. An event is a collection of outcomes.

(C) Experiment: Correct. A trial is a single performance of an experiment.

(D) Probability: Incorrect. Probability is a value assigned to an event.

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Final Answer:

A trial is a single performance of an experiment.

The correct option is (C) Experiment.

Solution:

The probability of any event E, denoted by P(E), is a numerical value that lies between 0 and 1, inclusive. That is,

Let's consider what different ranges of probability imply about the likelihood of an event:

If $P(E) = 0$, the event is an impossible event (it will never happen).

If $P(E) = 1$, the event is a sure event (it will always happen).

If $0 < P(E) < 0.5$, the event is considered unlikely (it is more probable that it will not happen than it will).

If $P(E) = 0.5$, the event has an equal chance of happening or not happening.

If $0.5 < P(E) < 1$, the event is considered likely (it is more probable that it will happen than it will not).

If the probability of an event is very close to 1 (e.g., 0.9, 0.99, etc.), it falls into the category where $0.5 < P(E) < 1$. Such an event is highly probable to occur.

Among the given options:

(A) Impossible: Probability is 0.

(B) Likely: Probability is greater than 0.5. An event with probability very close to 1 fits this description.

(C) Unlikely: Probability is less than 0.5 (and greater than 0).

(D) Random: Describes an experiment whose outcome is uncertain, not the probability value itself.

An event with a probability very close to 1 is highly likely to occur.

---

Final Answer:

If the probability of an event is very close to 1, it is considered a likely event.

The correct option is (B) Likely.

Solution:

The empirical probability, also known as experimental probability, is based on the results of performing an experiment a certain number of times.

It is calculated by taking the ratio of the number of times a specific event occurs in the experiment to the total number of times the experiment is conducted (the total number of trials).

The formula for empirical probability is:

Empirical Probability (Event) = $\frac{\text{Number of times the event occurred}}{\text{Total number of trials}}$

In this definition, the "Number of times the event occurred" refers to the frequency of the desired outcome, which is also commonly called the "Number of favourable outcomes" based on the observed results.

Let's examine the given options:

(A) $\frac{\text{Total number of outcomes}}{\text{Number of trials}}$: Incorrect. This does not represent empirical probability.

(B) $\frac{\text{Number of trials}}{\text{Number of favourable outcomes}}$: Incorrect. This is the reciprocal of the correct definition.

(C) $\frac{\text{Number of favourable outcomes}}{\text{Total number of trials}}$: Correct. This matches the standard definition of empirical probability.

(D) $\frac{\text{Number of outcomes}}{\text{Number of events}}$: Incorrect. This ratio is not a definition of probability.

Therefore, the correct definition is given by option (C).

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Final Answer:

The empirical probability of an event is defined as the ratio of the number of favourable outcomes (the number of times the event occurred) to the total number of trials.

The correct option is (C) $\frac{\text{Number of favourable outcomes}}{\text{Total number of trials}}$.

Given:

Total number of times two coins are tossed simultaneously = 400

Number of times the outcome 'two heads' occurs = 100

---

To Find:

The empirical probability of getting two heads.

---

Solution:

The empirical probability of an event is calculated based on the observed results of an experiment.

It is given by the formula:

Empirical Probability (Event) = $\frac{\text{Number of times the event occurred}}{\text{Total number of trials}}$

Here, the experiment is tossing two coins simultaneously.

The total number of trials is the number of times the coins were tossed, which is 400.

The event we are interested in is getting 'two heads'.

The number of times this event occurred is 100.

Therefore, the empirical probability of getting two heads is:

P(Two heads) = $\frac{\text{Number of times two heads occurred}}{\text{Total number of tosses}}$

P(Two heads) = $\frac{100}{400}$

Comparing this result with the given options:

(A) $\frac{100}{100}$

(B) $\frac{100}{400}$

(C) $\frac{300}{400}$

(D) $\frac{1}{400}$

The calculated empirical probability is $\frac{100}{400}$, which matches option (B).

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Final Answer:

The empirical probability of getting two heads is $\frac{100}{400}$.

The correct option is (B) $\frac{100}{400}$.

Given:

Total number of observations = 250

Number of times the weather prediction was correct = 180

---

To Find:

The empirical probability of a correct prediction.

---

Solution:

The empirical probability of an event is calculated based on the results of an experiment or observation.

The formula for empirical probability is:

Empirical Probability (Event) = $\frac{\text{Number of times the event occurred}}{\text{Total number of trials (observations)}}$

Here, the event is a correct weather prediction.

The number of times the event occurred (correct predictions) = 180

The total number of trials (observations) = 250

Therefore, the empirical probability of a correct prediction is:

P(Correct prediction) = $\frac{\text{Number of correct predictions}}{\text{Total number of observations}}$

P(Correct prediction) = $\frac{180}{250}$

Comparing this result with the given options:

(A) $\frac{180}{250}$

(B) $\frac{70}{250}$

(C) $\frac{180}{70}$

(D) $\frac{250}{180}$

The calculated empirical probability is $\frac{180}{250}$, which matches option (A).

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Final Answer:

The empirical probability of a correct prediction is $\frac{180}{250}$.

The correct option is (A) $\frac{180}{250}$.

Solution:

Let's define each term and see its relationship to probability:

(A) Trial: A single performance of a random experiment. Experiments consist of trials, and the outcomes of these trials are used to calculate empirical probability or define theoretical probability. Thus, "Trial" is a term related to probability.

(B) Outcome: A possible result of a trial in a random experiment. Probability is fundamentally concerned with the likelihood of different outcomes occurring. Thus, "Outcome" is a term related to probability.

(C) Event: A collection of one or more outcomes of a random experiment. Probability is assigned to events to quantify their likelihood of occurrence. Thus, "Event" is a term related to probability.

(D) Sample Space: The set of all possible outcomes of a random experiment. The sample space is the foundation upon which events are defined and probabilities are calculated. Thus, "Sample Space" is a term related to probability.

All the given terms (Trial, Outcome, Event, and Sample Space) are fundamental concepts used in the study and application of probability.

---

Final Answer:

All the listed terms are related to probability.

The correct options are (A) Trial, (B) Outcome, (C) Event, and (D) Sample Space.

Given:

A box contains 100 cards numbered from 1 to 100.

A card is drawn at random.

---

To Find:

The probability of drawing a card with a number divisible by 5.

---

Solution:

When a card is drawn at random from the box, each of the 100 cards has an equal chance of being drawn.

The total number of possible outcomes is the total number of cards in the box.

Total number of possible outcomes = 100

The event we are interested in is drawing a card with a number divisible by 5.

The numbers between 1 and 100 (inclusive) that are divisible by 5 are the multiples of 5 up to 100. These numbers are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, and 100.

To count the number of multiples of 5 up to 100, we can divide 100 by 5.

So, there are 20 numbers from 1 to 100 that are divisible by 5.

The probability of an event is calculated as the ratio of the number of favorable outcomes to the total number of possible outcomes.

P(Event) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

For the event "drawing a number divisible by 5":

P(Divisible by 5) = $\frac{\text{Number of cards divisible by 5}}{\text{Total number of cards}}$

P(Divisible by 5) = $\frac{20}{100}$

This fraction can be simplified:

So, the probability is $\frac{20}{100}$ or $\frac{1}{5}$.

Comparing this result with the given options:

(A) $\frac{1}{5}$

(B) $\frac{20}{100}$

(C) $\frac{80}{100}$

(D) Both A and B are correct.

Both option (A) and option (B) represent the correct probability, with (A) being the simplified form of (B).

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Final Answer:

The probability of drawing a card with a number divisible by 5 is $\frac{20}{100}$ or $\frac{1}{5}$.

Since both forms are given as options (A) and (B), the correct option is (D).

The correct option is (D) Both A and B are correct.

Solution:

In probability theory, an event that cannot occur in any trial of an experiment is called an impossible event.

The probability of an event quantifies how likely it is to occur.

A probability of 0 indicates that the event will never occur, i.e., it is an impossible event.

A probability of 1 indicates that the event is certain to occur, i.e., it is a sure event.

Probabilities between 0 and 1 indicate that the event may or may not occur.

The range of probability for any event E is given by:

Specifically, for an impossible event, the number of favorable outcomes is 0, regardless of the total number of possible outcomes (as long as there is at least one possible outcome in the sample space). If we consider the empirical probability formula, P(Impossible Event) = $\frac{\text{Number of times the impossible event occurred}}{\text{Total number of trials}} = \frac{0}{\text{Total number of trials}} = 0$.

Let's examine the given options:

(A) 1: This is the probability of a sure event, not an impossible event.

(B) 0: This is the probability assigned to an impossible event.

(C) Greater than 0: This is the probability for any event that is possible, but not necessarily certain.

(D) Less than 0: Probability values cannot be negative.

Therefore, the probability of an event which is impossible to happen is 0.

---

Final Answer:

The probability of an event which is impossible to happen is 0.

The correct option is (B) 0.

Given:

The empirical probability of an event E is $P(E) = 0.4$.

---

To Find:

The empirical probability of the complementary event (E NOT happening).

---

Solution:

The complementary event to event E is the event that E does not occur. It is denoted by $E'$ or $\overline{E}$.

A fundamental property of probability states that the sum of the probability of an event and the probability of its complementary event is always 1.

This applies to both theoretical and empirical probability.

P(E) + P(E') = 1

We are given $P(E) = 0.4$. We need to find $P(E')$.

Substituting the given value into the equation:

Subtracting 0.4 from both sides:

Therefore, the empirical probability of the complementary event is 0.6.

Comparing this result with the given options:

(A) 0.4

(B) 0.6

(C) 1

(D) Cannot be determined

The calculated probability is 0.6, which matches option (B).

---

Final Answer:

If the empirical probability of an event is $0.4$, the empirical probability of the complementary event is $0.6$.

The correct option is (B) 0.6.

Given:

Total number of students surveyed = 30

Number of students born in March = 4

---

To Find:

The empirical probability that a randomly selected student was born in March.

---

Solution:

The empirical probability of an event is calculated based on the observed frequency of the event in an experiment.

Empirical Probability (Event) = $\frac{\text{Number of times the event occurred}}{\text{Total number of trials}}$

In this case, the experiment is surveying the students about their birth month. The total number of trials is the total number of students surveyed, which is 30.

The event we are interested in is a student being born in March.

The number of times this event occurred is the number of students born in March, which is 4.

Therefore, the empirical probability that a student was born in March is:

P(Born in March) = $\frac{\text{Number of students born in March}}{\text{Total number of students surveyed}}$

P(Born in March) = $\frac{4}{30}$

Comparing this result with the given options:

(A) $\frac{4}{12}$

(B) $\frac{4}{30}$

(C) $\frac{3}{30}$

(D) $\frac{1}{30}$

The calculated empirical probability is $\frac{4}{30}$, which matches option (B).

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Final Answer:

The empirical probability that a student was born in March is $\frac{4}{30}$.

The correct option is (B) $\frac{4}{30}$.

Given:

Total number of students surveyed = 30

Number of students born in each month:

Jan (3), Feb (2), Mar (4), Apr (2), May (3), Jun (1), Jul (2), Aug (3), Sep (2), Oct (3), Nov (1), Dec (2)

Months with 31 days: Jan, Mar, May, Jul, Aug, Oct, Dec.

---

To Find:

The empirical probability that a randomly selected student was born in a month with 31 days.

---

Solution:

The empirical probability of an event is the ratio of the number of times the event occurred to the total number of trials.

Empirical Probability (Event) = $\frac{\text{Number of times the event occurred}}{\text{Total number of trials}}$

The total number of trials is the total number of students surveyed, which is 30.

The event we are interested in is a student being born in a month with 31 days.

The months with 31 days are Jan, Mar, May, Jul, Aug, Oct, and Dec.

To find the number of students born in a month with 31 days, we sum the number of students born in each of these months:

Number of students born in months with 31 days = (Students in Jan) + (Students in Mar) + (Students in May) + (Students in Jul) + (Students in Aug) + (Students in Oct) + (Students in Dec)

Number of students born in months with 31 days = $3 + 4 + 3 + 2 + 3 + 3 + 2$

The total number of students is 30.

Therefore, the empirical probability that a student was born in a month with 31 days is:

P(Born in month with 31 days) = $\frac{\text{Number of students born in months with 31 days}}{\text{Total number of students surveyed}}$

P(Born in month with 31 days) = $\frac{20}{30}$

Comparing this result with the given options:

(A) $\frac{3+4+3+2+3+3+2}{30} = \frac{20}{30}$

(B) $\frac{7}{30}$

(C) $\frac{20}{12}$

(D) $\frac{21}{30}$

Option (A) shows the correct sum of favorable outcomes divided by the total number of trials, and the result matches our calculation.

---

Final Answer:

The empirical probability that a student was born in a month with 31 days is $\frac{20}{30}$.

The correct option is (A) $\frac{3+4+3+2+3+3+2}{30} = \frac{20}{30}$.

Given:

A bag contains only red and blue balls.

The probability of drawing a red ball is $P(\text{Red}) = \frac{2}{5}$.

---

To Find:

The probability of drawing a blue ball, $P(\text{Blue})$.

---

Solution:

Since the bag contains only red and blue balls, drawing a red ball and drawing a blue ball are the only two possible outcomes when drawing a single ball. These two events are mutually exclusive and exhaustive.

The event "drawing a blue ball" is the complementary event to "drawing a red ball".

For any event and its complementary event, the sum of their probabilities is 1.

P(Event) + P(Complement of Event) = 1

In this context:

We are given P(Drawing a red ball) = $\frac{2}{5}$.

Substitute the given probability into the equation:

To find $P(\text{Blue})$, subtract $\frac{2}{5}$ from 1:

Perform the subtraction:

Thus, the probability of drawing a blue ball is $\frac{3}{5}$.

Comparing this result with the given options:

(A) $\frac{2}{5}$

(B) $\frac{3}{5}$

(C) 1

(D) 0

The calculated probability matches option (B).

---

Final Answer:

The probability of drawing a blue ball is $\frac{3}{5}$.

The correct option is (B) $\frac{3}{5}$.

Given:

Total number of students in the class = 40

Number of girls in the class = 15

Number of boys in the class = 25

A student is selected at random.

---

To Find:

The probability that the selected student is a girl.

---

Solution:

When a student is selected at random from the class, each student has an equal chance of being selected.

The total number of possible outcomes is the total number of students in the class, which is 40.

The event we are interested in is selecting a girl.

The number of favorable outcomes for this event is the number of girls in the class, which is 15.

The probability of an event is given by the ratio of the number of favorable outcomes to the total number of possible outcomes:

P(Event) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

For the event "selecting a girl":

P(Selecting a girl) = $\frac{\text{Number of girls}}{\text{Total number of students}}$

P(Selecting a girl) = $\frac{15}{40}$

Comparing this result with the given options:

(A) $\frac{15}{25}$

(B) $\frac{25}{40}$

(C) $\frac{15}{40}$

(D) $\frac{40}{15}$

The calculated probability is $\frac{15}{40}$, which matches option (C).

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Final Answer:

The probability that the student selected at random is a girl is $\frac{15}{40}$.

The correct option is (C) $\frac{15}{40}$.

Given:

A standard die is rolled.

---

To Find:

The probability of getting an odd number.

---

Solution:

When a standard die is rolled, the possible outcomes are the numbers on its faces.

The sample space (set of all possible outcomes) is $\{1, 2, 3, 4, 5, 6\}$.

The total number of possible outcomes is 6.

We assume the die is fair, so each outcome is equally likely.

The event we are interested in is getting an odd number.

The odd numbers in the sample space are $\{1, 3, 5\}$.

The number of favorable outcomes for this event is 3.

The probability of an event in a situation with equally likely outcomes is calculated as:

P(Event) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

For the event "getting an odd number":

P(Getting an odd number) = $\frac{\text{Number of odd numbers}}{\text{Total number of outcomes}}$

P(Getting an odd number) = $\frac{3}{6}$

Comparing this result with the given options:

(A) $\frac{1}{6}$

(B) $\frac{3}{6}$

(C) $\frac{2}{6}$

(D) 1

The calculated probability is $\frac{3}{6}$, which matches option (B).

Note that $\frac{3}{6}$ can be simplified to $\frac{1}{2}$, but option (B) is given in the unsimplified form.

---

Final Answer:

The probability of getting an odd number when rolling a die is $\frac{3}{6}$.

The correct option is (B) $\frac{3}{6}$.

Given:

A standard deck of playing cards contains 52 cards.

A single card is drawn at random.

---

To Find:

The probability of drawing a red card.

---

Solution:

A standard deck of 52 playing cards consists of four suits: Hearts, Diamonds, Clubs, and Spades.

Each suit has 13 cards.

The suits Hearts and Diamonds are red.

The suits Clubs and Spades are black.

The number of red cards in the deck is the sum of the number of cards in the red suits:

Number of red cards = Number of Hearts + Number of Diamonds

Number of red cards = $13 + 13 = 26$

The total number of possible outcomes when drawing a single card is the total number of cards in the deck, which is 52.

The event we are interested in is drawing a red card.

The number of favorable outcomes for this event is the number of red cards, which is 26.

The probability of an event is calculated as the ratio of the number of favorable outcomes to the total number of possible outcomes, assuming all outcomes are equally likely (which is the case when drawing a card at random from a standard deck).

P(Event) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

For the event "drawing a red card":

P(Drawing a red card) = $\frac{\text{Number of red cards}}{\text{Total number of cards}}$

P(Drawing a red card) = $\frac{26}{52}$

Comparing this result with the given options:

(A) $\frac{1}{52}$

(B) $\frac{26}{52}$

(C) $\frac{13}{52}$

(D) $\frac{52}{26}$

The calculated probability is $\frac{26}{52}$, which matches option (B).

Note that $\frac{26}{52}$ can be simplified to $\frac{1}{2}$, but this option is not provided explicitly.

---

Final Answer:

The probability of drawing a red card from a deck of 52 playing cards is $\frac{26}{52}$.

The correct option is (B) $\frac{26}{52}$.

Solution:

When a standard coin is tossed, in the most common probability model, there are two possible results: Heads (H) or Tails (T).

These are the outcomes of the experiment. The set of all possible outcomes is called the sample space, which is {Head, Tail}.

Let's consider each option provided:

(A) Head: Getting a Head is a possible outcome of tossing a single coin. This is an element of the sample space.

(B) Tail: Getting a Tail is also a possible outcome of tossing a single coin. This is an element of the sample space.

(C) Both Head and Tail simultaneously: When you toss a single coin, the outcome is either Head or Tail, but it cannot be both at the exact same time in a single trial. These are mutually exclusive outcomes. Therefore, "Both Head and Tail simultaneously" is not a possible outcome.

(D) Neither Head nor Tail (assuming it doesn't land on the edge): In the standard model, where we assume the coin does not land on its edge or disappear, the outcome must be either Head or Tail. The phrase "(assuming it doesn't land on the edge)" reinforces this standard assumption. If the outcome must be either Head or Tail, then getting "Neither Head nor Tail" is not possible under this assumption. Thus, this is also not a possible outcome.

---

Both options (C) and (D) describe events that are impossible in the standard model of a single coin toss (with the assumption in D). However, option (C) directly contradicts the nature of an elementary outcome in this experiment – a single toss yields a single result, not multiple results at once.

Option (D)'s impossibility relies on the explicit assumption stated. Option (C) is an impossibility based on the definition of distinct outcomes from a single trial.

In the context of a multiple choice question asking for something that is NOT a possible outcome, both (C) and (D) fit the description of not being elements of the standard sample space {Head, Tail}. However, "Both Head and Tail simultaneously" is a more fundamental conceptual impossibility related to the nature of the outcomes themselves in a single trial.

Considering the standard understanding in introductory probability, "Both Head and Tail simultaneously" from a single coin toss is the most direct example of something that is not a possible outcome.

---

Final Answer:

The value that cannot be a possible outcome when tossing a single coin is "Both Head and Tail simultaneously".

The correct option is (C) Both Head and Tail simultaneously.

Solution:

In probability theory, an event that is certain to occur is called a sure event or a certain event.

The probability of an event is a measure of its likelihood of happening. The value of probability ranges from 0 to 1.

Let E be an event. The probability of E, denoted as P(E), satisfies:

Specific values of probability correspond to specific types of events:

- If P(E) = 0, the event is an impossible event (it will never happen).

- If P(E) = 1, the event is a sure event (it will always happen).

- If $0 < P(E) < 1$, the event is a possible event (it may or may not happen).

The question asks for the probability of an event that is sure to happen. By definition, a sure event has a probability of 1.

Let's compare this with the given options:

(A) 0: This is the probability of an impossible event.

(B) 1: This is the probability of a sure event.

(C) Between 0 and 1: This is the probability for events that are possible but not certain or impossible.

(D) Greater than 1: Probability values cannot be greater than 1.

Thus, the probability of an event which is sure to happen is 1.

---

Final Answer:

If an event is sure to happen, its probability is 1.

The correct option is (B) 1.

Solution:

The empirical probability of an event is calculated based on the actual results of an experiment:

Empirical Probability (Event) = $\frac{\text{Number of times the event occurred}}{\text{Total number of trials}}$

The theoretical probability of an event is based on the possible outcomes of an experiment, assuming they are equally likely:

Theoretical Probability (Event) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

The relationship between empirical probability and theoretical probability is described by the Law of Large Numbers.

The Law of Large Numbers states that as the number of trials in a random experiment increases, the empirical probability (observed frequency) of an event will tend to get closer and closer to the theoretical probability (expected proportion) of that event.

For example, when tossing a fair coin, the theoretical probability of getting a head is $\frac{1}{2}$. In a small number of trials (e.g., 10 tosses), the empirical probability might be quite different from 0.5 (e.g., $\frac{7}{10}$ or $\frac{3}{10}$). However, if the coin is tossed a very large number of times (e.g., 1000 or 10000 times), the empirical probability will likely be very close to 0.5.

Therefore, the empirical probability of an event approaches the theoretical probability as the number of trials becomes very large.

Let's consider the options:

(A) Small: A small number of trials typically leads to empirical probability that can vary significantly from the theoretical probability.

(B) Large: A very large number of trials causes the empirical probability to converge towards the theoretical probability.

(C) Equal to the number of outcomes: This is not a general relationship between the number of trials and the convergence of probabilities.

(D) Zero: The number of trials must be greater than zero to calculate empirical probability.

---

Final Answer:

The empirical probability of an event approaches the theoretical probability as the number of trials becomes very large.

The correct option is (B) Large.

Given:

Total number of families surveyed = 100

Number of families owning a scooter = 60

Number of families owning a car = 40

Number of families owning both a scooter and a car = 20

---

To Find:

The empirical probability that a randomly chosen family owns only a scooter.

---

Solution:

The empirical probability of an event is the ratio of the number of times the event occurred to the total number of trials.

Empirical Probability (Event) = $\frac{\text{Number of times the event occurred}}{\text{Total number of trials}}$

The total number of trials is the total number of families surveyed, which is 100.

The event we are interested in is that a family owns only a scooter.

We are given the number of families that own a scooter (which includes those who own both) and the number of families who own both.

To find the number of families who own only a scooter, we subtract the number of families who own both from the total number of families who own a scooter.

Number of families owning only a scooter = (Number of families owning a scooter) - (Number of families owning both)

Number of families owning only a scooter = $60 - 20 = 40$

The number of favorable outcomes for the event "owns only a scooter" is 40.

Therefore, the empirical probability that a family owns only a scooter is:

P(Owns only a scooter) = $\frac{\text{Number of families owning only a scooter}}{\text{Total number of families surveyed}}$

P(Owns only a scooter) = $\frac{40}{100}$

Comparing this result with the given options:

(A) $\frac{60}{100}$

(B) $\frac{40}{100}$

(C) $\frac{20}{100}$

(D) $\frac{60-20}{100} = \frac{40}{100}$

Option (B) directly matches our calculated probability. Option (D) shows the calculation that leads to the correct probability and matches the result.

Since both (B) and (D) represent the same correct probability, and (D) explicitly shows the calculation for "only a scooter", both are valid representations of the answer.

In a single-choice context, option (D) provides more detail about how the specific event "only a scooter" was derived from the given data about "owns a scooter" and "owns both". However, option (B) is also numerically correct based on the calculated number of families who own only a scooter.

Assuming the question expects the simplest form or the direct numerical ratio, (B) is sufficient. If the intention is to show the derivation, (D) is better.

Let's re-examine the options. Option (D) is presented as a calculation and a resulting value. Both the calculation and the result are correct for the probability of owning only a scooter.

Option (B) is just the result of that calculation.

Often, in such questions, the option that best represents the derived value or process is preferred.

---

Final Answer:

The empirical probability that a randomly selected family owns only a scooter is $\frac{40}{100}$.

The correct option is (D) $\frac{60-20}{100} = \frac{40}{100}$.

Given:

Total number of times the experiment (drawing a slip) is repeated = 50

Number of times the number 3 is drawn = 12

---

To Find:

The empirical probability of drawing the number 3.

---

Solution:

The empirical probability of an event is calculated based on the observed frequency of the event in a series of trials.

Empirical Probability (Event) = $\frac{\text{Number of times the event occurred}}{\text{Total number of trials}}$

Here, the experiment is drawing a slip from the bag, and the experiment was performed 50 times. So, the total number of trials is 50.

The event we are interested in is drawing the number 3.

The number of times this event occurred is given as 12.

Therefore, the empirical probability of drawing the number 3 is:

P(Drawing the number 3) = $\frac{\text{Number of times 3 was drawn}}{\text{Total number of trials}}$

P(Drawing the number 3) = $\frac{12}{50}$

Comparing this result with the given options:

(A) $\frac{1}{5}$ (This is the theoretical probability if each number were equally likely, but the question asks for empirical probability based on the given results)

(B) $\frac{12}{50}$

(C) $\frac{38}{50}$ (This is the empirical probability of NOT drawing the number 3, i.e., $50-12=38$ times other numbers were drawn)

(D) $\frac{50}{12}$

The calculated empirical probability is $\frac{12}{50}$, which matches option (B).

---

Final Answer:

The empirical probability of drawing the number 3 is $\frac{12}{50}$.

The correct option is (B) $\frac{12}{50}$.

Given:

Total number of vehicles recorded = 100

Number of speeding vehicles = 30

Number of vehicles within the limit = 70

---

To Find:

The empirical probability that a randomly selected vehicle was speeding.

---

Solution:

The empirical probability of an event is based on the observed results of an experiment or survey. It is calculated as the ratio of the number of times the event occurred to the total number of trials.

Empirical Probability (Event) = $\frac{\text{Number of times the event occurred}}{\text{Total number of trials}}$

In this case, the experiment is observing the speed of vehicles.

The total number of trials is the total number of vehicles recorded, which is 100.

The event we are interested in is a vehicle being speeding.

The number of times this event occurred is the number of speeding vehicles recorded, which is 30.

Therefore, the empirical probability that a randomly selected vehicle was speeding is:

P(Vehicle was speeding) = $\frac{\text{Number of speeding vehicles}}{\text{Total number of vehicles recorded}}$

P(Vehicle was speeding) = $\frac{30}{100}$

Comparing this result with the given options:

(A) $\frac{100}{30}$

(B) $\frac{70}{100}$

(C) $\frac{30}{100}$

(D) $\frac{1}{2}$

The calculated empirical probability is $\frac{30}{100}$, which matches option (C).

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Final Answer:

The empirical probability that a randomly selected vehicle was speeding is $\frac{30}{100}$.

The correct option is (C) $\frac{30}{100}$.

In the context of probability, an experiment is an operation which can produce some well-defined outcomes.

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For example, tossing a fair coin is an experiment because it can produce one of two well-defined outcomes: Heads or Tails.

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Another example is rolling a standard six-sided die, which can produce one of six well-defined outcomes: 1, 2, 3, 4, 5, or 6.

An outcome is a possible result of a probabilistic experiment.

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When a standard six-sided die is thrown, the possible outcomes are the numbers that can appear on the top face. These are:

1, 2, 3, 4, 5, 6.

The sample space of an experiment is the set of all possible outcomes of the experiment.

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When two coins are tossed simultaneously, each coin can land either Heads (H) or Tails (T).

The possible outcomes for the pair of coins are:

First coin is H and second coin is H (HH)

First coin is H and second coin is T (HT)

First coin is T and second coin is H (TH)

First coin is T and second coin is T (TT)

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Therefore, the sample space for tossing two coins simultaneously is the set containing these outcomes:

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

An event is a subset of the sample space of an experiment. It is a collection of one or more outcomes of the experiment.

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When a standard six-sided die is thrown, the sample space is $S = \{1, 2, 3, 4, 5, 6\}$.

An example of an event could be "getting an even number". This event corresponds to the subset of outcomes $\{2, 4, 6\}$ from the sample space.

Another example of an event could be "getting a number greater than 4". This event corresponds to the subset of outcomes $\{5, 6\}$ from the sample space.

Definition of Empirical Probability

Empirical probability, also known as experimental probability or relative frequency, is a probability measure that is based on the results of an experiment or observation. It is calculated by observing the frequency of an event in a finite sequence of trials.

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Calculation of Empirical Probability

Empirical probability is calculated as the ratio of the number of times a specific event occurs to the total number of trials performed. The formula is:

$P(\text{Event}) = \frac{\text{Number of times the event occurred}}{\text{Total number of trials}}$

For example, if a coin is tossed 100 times and heads appears 45 times, the empirical probability of getting heads is $\frac{45}{100} = 0.45$.

When a fair coin is tossed, there are two possible outcomes:

• Head (H)
• Tail (T)

These two outcomes are equally likely.

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The total number of possible outcomes is 2.

The favourable outcome is getting a 'Tail'.

The number of favourable outcomes is 1.

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The probability of an event is calculated using the formula:

$P(\text{Event}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

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Using this formula for getting a 'Tail':

$P(\text{Getting a Tail}) = \frac{\text{Number of times Tail can occur}}{\text{Total number of possible outcomes}}$

$P(\text{Getting a Tail}) = \frac{1}{2}$

---

Therefore, the probability of getting a 'Tail' when a coin is tossed is $\frac{1}{2}$ or 0.5.

When a standard six-sided die is thrown, there are six possible outcomes:

• 1
• 2
• 3
• 4
• 5
• 6

These outcomes are equally likely.

---

The total number of possible outcomes is 6.

The favourable outcome is getting the number '4'.

The number of favourable outcomes is 1 (since '4' appears only once on the die).

---

The probability of an event is calculated using the formula:

$P(\text{Event}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

---

Using this formula for getting a '4':

$P(\text{Getting a 4}) = \frac{\text{Number of times 4 can occur}}{\text{Total number of possible outcomes}}$

$P(\text{Getting a 4}) = \frac{1}{6}$

---

Therefore, the probability of getting a number '4' when a die is thrown is $\frac{1}{6}$.

Given Information

Number of red marbles = 5

Number of blue marbles = 3

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To Find

The probability of drawing a red marble.

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Solution

The total number of marbles in the bag is the sum of the number of red marbles and the number of blue marbles.

Total number of marbles = Number of red marbles + Number of blue marbles

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

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The favourable outcome is drawing a red marble.

The number of favourable outcomes is the number of red marbles, which is 5.

---

The probability of an event is given by the formula:

$P(\text{Event}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

---

In this case, the event is drawing a red marble. So,

$P(\text{Drawing a red marble}) = \frac{\text{Number of red marbles}}{\text{Total number of marbles}}$

$P(\text{Drawing a red marble}) = \frac{5}{8}$

---

Therefore, the probability of drawing a red marble is $\frac{5}{8}$.

Given Information

Number of red marbles = 5

Number of blue marbles = 3

---

To Find

The probability of drawing a marble that is not red.

---

Solution

The total number of marbles in the bag is the sum of the number of red marbles and the number of blue marbles.

Total number of marbles = Number of red marbles + Number of blue marbles

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

---

The event is drawing a marble that is not red.

In this bag, marbles that are not red are the blue marbles.

The number of favourable outcomes (drawing a marble that is not red) is the number of blue marbles, which is 3.

---

The probability of an event is given by the formula:

$P(\text{Event}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

---

In this case, the event is drawing a marble that is not red. So,

$P(\text{Drawing a marble that is not red}) = \frac{\text{Number of marbles that are not red}}{\text{Total number of marbles}}$

$P(\text{Drawing a marble that is not red}) = \frac{3}{8}$

---

Therefore, the probability of drawing a marble that is not red is $\frac{3}{8}$.

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Alternate Solution

We know the probability of drawing a red marble from the previous question is $\frac{5}{8}$.

The event of drawing a marble that is not red is the complement of the event of drawing a red marble.

The sum of the probability of an event and the probability of its complement is 1.

$P(\text{Event}) + P(\text{Not Event}) = 1$

---

Let E be the event of drawing a red marble.

$P(E) = P(\text{Drawing a red marble}) = \frac{5}{8}$

Let E' be the event of drawing a marble that is not red.

$P(E') = P(\text{Drawing a marble that is not red})$

$P(E) + P(E') = 1$

$\frac{5}{8} + P(E') = 1$

$P(E') = 1 - \frac{5}{8}$

$P(E') = \frac{8}{8} - \frac{5}{8}$

$P(E') = \frac{8-5}{8} = \frac{3}{8}$

---

The probability of drawing a marble that is not red is $\frac{3}{8}$.

Definition of a Sure Event

A sure event (or certain event) is an event that is certain to happen in any given trial of an experiment. In the context of sample space, a sure event is the entire sample space itself.

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Probability of a Sure Event

The probability of a sure event is always 1.

This is because the number of favourable outcomes for a sure event is equal to the total number of possible outcomes (the entire sample space).

Using the probability formula:

$P(\text{Sure Event}) = \frac{\text{Number of outcomes in the event}}{\text{Total number of outcomes}}$

$P(\text{Sure Event}) = \frac{\text{Total number of outcomes}}{\text{Total number of outcomes}} = 1$

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Example of a Sure Event

Consider the experiment of rolling a standard six-sided die.

The sample space is $S = \{1, 2, 3, 4, 5, 6\}$. The total number of outcomes is 6.

Consider the event E: "Getting a number less than 7".

The outcomes for this event are $\{1, 2, 3, 4, 5, 6\}$.

The number of outcomes favourable to event E is 6.

---

The probability of event E is:

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

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

---

Since the probability is 1, the event "Getting a number less than 7" when rolling a standard six-sided die is a sure event.

Definition of an Impossible Event

An impossible event is an event that cannot happen under any circumstances in a given trial of an experiment. In the context of sample space, an impossible event is represented by the empty set ($\emptyset$).

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Probability of an Impossible Event

The probability of an impossible event is always 0.

This is because there are no outcomes favourable to an impossible event.

Using the probability formula:

$P(\text{Impossible Event}) = \frac{\text{Number of outcomes in the event}}{\text{Total number of outcomes}}$

$P(\text{Impossible Event}) = \frac{0}{\text{Total number of outcomes}} = 0$

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Example of an Impossible Event

Consider the experiment of rolling a standard six-sided die.

The sample space is $S = \{1, 2, 3, 4, 5, 6\}$. The total number of outcomes is 6.

Consider the event E: "Getting a number 7".

The standard six-sided die has faces numbered 1 through 6. It is not possible to get a 7 when rolling this die.

The set of outcomes favourable to event E is empty, i.e., $E = \emptyset$. The number of favourable outcomes is 0.

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The probability of event E is:

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

$P(E) = \frac{0}{6} = 0$

---

Since the probability is 0, the event "Getting a number 7" when rolling a standard six-sided die is an impossible event.

Can the probability of an event be a negative number?

No, the probability of an event cannot be a negative number.

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Explanation:

The probability of an event is calculated using the formula:

$P(\text{Event}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

---

Both the number of favourable outcomes and the total number of outcomes are counts of occurrences or possibilities. Counts are non-negative numbers (they can be zero or positive integers).

Since the number of favourable outcomes can be zero (for an impossible event) but cannot be negative, and the total number of outcomes is always positive (for a valid experiment), the ratio $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$ will always be non-negative.

Therefore, $P(\text{Event}) \geq 0$.

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Can the probability of an event be greater than 1?

No, the probability of an event cannot be greater than 1.

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Explanation:

The number of favourable outcomes for any event in an experiment can never exceed the total number of possible outcomes for that experiment.

The most favourable outcomes an event can have is when the event includes all possible outcomes, which is the case of a sure event. In this case, the number of favourable outcomes is equal to the total number of outcomes.

---

Using the probability formula:

$P(\text{Event}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

---

Since the number of favourable outcomes is always less than or equal to the total number of outcomes, the ratio $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$ will always be less than or equal to 1.

Therefore, $P(\text{Event}) \leq 1$.

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Conclusion

Based on the definition and calculation of probability, the probability of any event must lie between 0 and 1, inclusive.

$0 \leq P(\text{Event}) \leq 1$

A probability of 0 indicates an impossible event, and a probability of 1 indicates a sure event.

Definition of an Elementary Event

An elementary event (or simple event) is an event that consists of only one single outcome of a random experiment.

When we consider the sample space of an experiment, each individual element (outcome) in the sample space corresponds to an elementary event.

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Example of an Elementary Event

Consider the experiment of rolling a standard six-sided die.

The sample space for this experiment is $S = \{1, 2, 3, 4, 5, 6\}$. The outcomes are the numbers that can appear on the top face.

Each of the following is an elementary event:

• Event E$_1$: Getting a 1. This event is the set $\{1\}$. It contains only one outcome.
• Event E$_2$: Getting a 2. This event is the set $\{2\}$. It contains only one outcome.
• Event E$_3$: Getting a 3. This event is the set $\{3\}$. It contains only one outcome.
• Event E$_4$: Getting a 4. This event is the set $\{4\}$. It contains only one outcome.
• Event E$_5$: Getting a 5. This event is the set $\{5\}$. It contains only one outcome.
• Event E$_6$: Getting a 6. This event is the set $\{6\}$. It contains only one outcome.

In this example, $\{1\}, \{2\}, \{3\}, \{4\}, \{5\},$ and $\{6\}$ are the elementary events.

Definition of a Compound Event

A compound event is an event that consists of more than one outcome of a random experiment.

It is formed by combining two or more elementary events from the sample space.

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Example of a Compound Event

Consider the experiment of rolling a standard six-sided die.

The sample space, which is the set of all possible outcomes, is $S = \{1, 2, 3, 4, 5, 6\}$. Each individual outcome (1, 2, 3, etc.) corresponds to an elementary event.

Consider the event E: "Getting an even number".

The outcomes from the sample space that satisfy this event are 2, 4, and 6.

So, the event E can be represented as the set of outcomes $\{2, 4, 6\}$.

Since this event $\{2, 4, 6\}$ contains more than one outcome (it contains three outcomes), it is a compound event.

Another example of a compound event in the same experiment could be "Getting a number greater than 3", which corresponds to the outcomes $\{4, 5, 6\}$.

Given Information

The box contains slips numbered from 1 to 10.

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To Find

The probability of drawing a slip with an even number.

---

Solution

The possible outcomes when drawing a slip are the numbers from 1 to 10.

The sample space is $S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$.

The total number of possible outcomes is 10.

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The favourable outcome is drawing a slip with an even number.

The even numbers between 1 and 10 are 2, 4, 6, 8, and 10.

The set of favourable outcomes is $\{2, 4, 6, 8, 10\}$.

The number of favourable outcomes is 5.

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The probability of an event is calculated using the formula:

$P(\text{Event}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

---

In this case, the event is drawing a slip with an even number. So,

$P(\text{Drawing an even number}) = \frac{\text{Number of even numbers}}{\text{Total number of slips}}$

$P(\text{Drawing an even number}) = \frac{5}{10}$

$P(\text{Drawing an even number}) = \frac{1}{2}$

---

Therefore, the probability of drawing a slip with an even number on it is $\frac{1}{2}$ or 0.5.

Given Information

Total number of times the die is thrown (Total trials) = 200

Frequency of getting a '6' (Number of times the event occurred) = 35

---

To Find

The empirical probability of getting a '6'.

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Solution

The empirical probability of an event is calculated using the formula:

$P(\text{Event}) = \frac{\text{Number of times the event occurred}}{\text{Total number of trials}}$

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In this case, the event is getting a '6'.

Number of times the event occurred (getting a '6') = 35

Total number of trials = 200

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Substitute these values into the formula:

$P(\text{Getting a 6}) = \frac{35}{200}$

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We can simplify this fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 5.

$P(\text{Getting a 6}) = \frac{\cancel{35}^{7}}{\cancel{200}_{40}}$

$P(\text{Getting a 6}) = \frac{7}{40}$

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Alternatively, we can express the probability as a decimal:

$P(\text{Getting a 6}) = \frac{35}{200} = \frac{35 \times 5}{200 \times 5} = \frac{175}{1000} = 0.175$

---

Therefore, the empirical probability of getting a '6' is $\frac{7}{40}$ or 0.175.

Given Information

Total number of times the coins are tossed (Total trials) = 150

Frequency of getting 'Two Heads' (Number of times the event occurred) = 30

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To Find

The empirical probability of getting 'Two Heads'.

---

Solution

The empirical probability of an event is calculated using the formula:

$P(\text{Event}) = \frac{\text{Number of times the event occurred}}{\text{Total number of trials}}$

---

In this case, the event is getting 'Two Heads'.

Number of times the event occurred (getting 'Two Heads') = 30

Total number of trials = 150

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Substitute these values into the formula:

$P(\text{Getting Two Heads}) = \frac{30}{150}$

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We can simplify this fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 30.

$P(\text{Getting Two Heads}) = \frac{\cancel{30}^{1}}{\cancel{150}_{5}}$

$P(\text{Getting Two Heads}) = \frac{1}{5}$

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Alternatively, we can express the probability as a decimal:

$P(\text{Getting Two Heads}) = \frac{30}{150} = \frac{3}{15} = 0.2$

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Therefore, the empirical probability of getting 'Two Heads' is $\frac{1}{5}$ or 0.2.

Given Information

Total number of students surveyed (Total trials) = 100

Number of students who like tea = 60

Number of students who like coffee (Number of times the event occurred) = 40

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To Find

The empirical probability that a randomly chosen student likes coffee.

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Solution

The empirical probability of an event is calculated using the formula:

$P(\text{Event}) = \frac{\text{Number of times the event occurred}}{\text{Total number of trials}}$

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In this case, the event is selecting a student who likes coffee.

Number of times the event occurred (students who like coffee) = 40

Total number of trials (total students) = 100

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Substitute these values into the formula:

$P(\text{Student likes coffee}) = \frac{40}{100}$

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We can simplify this fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 20.

$P(\text{Student likes coffee}) = \frac{\cancel{40}^{2}}{\cancel{100}_{5}}$

$P(\text{Student likes coffee}) = \frac{2}{5}$

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Alternatively, we can express the probability as a decimal:

$P(\text{Student likes coffee}) = \frac{40}{100} = 0.40 = 0.4$

---

Therefore, the empirical probability that the student likes coffee is $\frac{2}{5}$ or 0.4.

An elementary event is an event that consists of only one single outcome of a random experiment.

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The sum of the probabilities of all the elementary events of an experiment is always 1.

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Explanation:

Let the sample space of an experiment be $S = \{e_1, e_2, ..., e_n\}$, where $e_1, e_2, ..., e_n$ are the individual outcomes.

Each outcome $e_i$ corresponds to an elementary event $\{e_i\}$.

The probability of an elementary event $\{e_i\}$ is given by:

$P(\{e_i\}) = \frac{\text{Number of outcomes in } \{e_i\}}{\text{Total number of outcomes in } S} = \frac{1}{n}$

(Assuming all outcomes are equally likely. If not equally likely, the sum still equals 1 based on the definition of probability axioms).

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The sum of the probabilities of all elementary events is:

Sum = $P(\{e_1\}) + P(\{e_2\}) + \dots + P(\{e_n\})$

Sum = $\frac{1}{n} + \frac{1}{n} + \dots + \frac{1}{n}$ (n times)

Sum = $n \times \frac{1}{n}$

Sum = 1

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This is because the collection of all elementary events covers the entire sample space, and exactly one of the elementary events must occur when the experiment is performed.

Given Information

Total number of cards in a standard deck = 52

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To Find

The probability of drawing a King of Spades.

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Solution

The total number of possible outcomes is the total number of cards in the deck.

Total number of outcomes = 52

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The favourable outcome is drawing a King of Spades.

In a standard deck of 52 cards, there is only one King of Spades.

The number of favourable outcomes is 1.

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The probability of an event is calculated using the formula:

$P(\text{Event}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

---

In this case, the event is drawing a King of Spades.

$P(\text{Drawing King of Spades}) = \frac{\text{Number of King of Spades}}{\text{Total number of cards}}$

$P(\text{Drawing King of Spades}) = \frac{1}{52}$

---

Therefore, the probability of drawing a King of Spades from a deck of 52 cards is $\frac{1}{52}$.

Given Information

The experiment is throwing a standard six-sided die.

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To Find

The probability of getting a number less than 7.

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Solution

When a standard six-sided die is thrown, the possible outcomes are the numbers on its faces.

The sample space is $S = \{1, 2, 3, 4, 5, 6\}$.

The total number of possible outcomes is 6.

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The favourable outcome is getting a number less than 7.

We check which numbers in the sample space are less than 7. These are 1, 2, 3, 4, 5, and 6.

The set of favourable outcomes is $\{1, 2, 3, 4, 5, 6\}$.

The number of favourable outcomes is 6.

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The probability of an event is calculated using the formula:

$P(\text{Event}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

---

In this case, the event is getting a number less than 7. So,

$P(\text{Getting a number less than 7}) = \frac{\text{Number of outcomes less than 7}}{\text{Total number of outcomes}}$

$P(\text{Getting a number less than 7}) = \frac{6}{6}$

$P(\text{Getting a number less than 7}) = 1$

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This is a sure event, as it is certain to happen when throwing a standard die.

Therefore, the probability of getting a number less than 7 when a die is thrown is 1.

Given Information

The probability of an event happening, let's call it Event E, is $P(E) = 0.45$.

---

To Find

The probability of the event not happening, which is the probability of the complement of Event E, denoted as $P(\text{not E})$ or $P(E')$.

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Solution

We know that the sum of the probability of an event and the probability of its complement (the event not happening) is always 1.

This can be expressed by the formula:

$P(\text{Event}) + P(\text{Not Event}) = 1$

or

$P(E) + P(E') = 1$

---

We are given $P(E) = 0.45$. We need to find $P(E')$.

Substitute the given value into the formula:

$0.45 + P(E') = 1$

---

Now, solve for $P(E')$:

$P(E') = 1 - 0.45$

$P(E') = 0.55$

---

Therefore, the probability of the event not happening is 0.55.

Given Information

The spinner has 4 equal sectors with the following colours: Blue, Green, Red, Yellow.

The sectors are equal, which implies each colour is equally likely to be landed upon.

---

To Find

The probability of landing on a colour that starts with the letter 'R'.

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Solution

The total number of possible outcomes is the total number of distinct sectors on the spinner.

Total number of outcomes = 4 (representing the colours Blue, Green, Red, Yellow)

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The favourable outcome is landing on a colour that starts with the letter 'R'.

We examine the given list of colours: Blue, Green, Red, Yellow.

Only one colour in this list starts with the letter 'R', which is 'Red'.

The number of favourable outcomes is 1 (the colour Red).

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The probability of an event is calculated using the basic probability formula:

$P(\text{Event}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

---

In this case, the event is landing on a colour that starts with 'R'. So,

$P(\text{Landing on colour starting with R}) = \frac{\text{Number of colours starting with R}}{\text{Total number of colours}}$

$P(\text{Landing on colour starting with R}) = \frac{1}{4}$

---

Therefore, the probability of landing on a colour that starts with the letter 'R' is $\frac{1}{4}$ or 0.25.

Equally Likely Outcomes

Outcomes of a random experiment are said to be equally likely if each outcome has the same chance of occurring. In such cases, the probability of each elementary event is the same.

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Example of Equally Likely Outcomes:

Consider the experiment of tossing a fair coin.

The possible outcomes are Head (H) and Tail (T).

Because the coin is fair, there is no reason to expect a Head more often than a Tail, or vice versa.

The probability of getting a Head is $\frac{1}{2}$, and the probability of getting a Tail is $\frac{1}{2}$.

$P(H) = P(T) = \frac{1}{2}$

Since $P(H) = P(T)$, the outcomes Head and Tail are equally likely.

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Not Equally Likely Outcomes (or Unequally Likely Outcomes)

Outcomes of a random experiment are said to be not equally likely if each outcome does not have the same chance of occurring. The probability of the elementary events are different.

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Example of Not Equally Likely Outcomes:

Consider a bag containing 5 red marbles and 1 blue marble.

The possible outcomes when drawing a marble from the bag are drawing a Red marble (R) or drawing a Blue marble (B).

The total number of marbles is $5 + 1 = 6$.

The probability of drawing a Red marble is:

$P(R) = \frac{\text{Number of red marbles}}{\text{Total number of marbles}} = \frac{5}{6}$

The probability of drawing a Blue marble is:

$P(B) = \frac{\text{Number of blue marbles}}{\text{Total number of marbles}} = \frac{1}{6}$

Since $P(R) = \frac{5}{6}$ and $P(B) = \frac{1}{6}$, and $\frac{5}{6} \neq \frac{1}{6}$, the outcomes drawing a Red marble and drawing a Blue marble are not equally likely. It is five times more likely to draw a red marble than a blue marble.

Given Information

Probability of event A, $P(A) = 0.6$

Probability of event B, $P(B) = 0.3$

---

To Check

Whether A and B can be the only two possible outcomes of the experiment.

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Solution/Explanation

A fundamental principle of probability states that the sum of the probabilities of all possible outcomes in a random experiment must be equal to 1.

If A and B were the only two possible outcomes of the experiment, then the sum of their probabilities must be 1.

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Let's calculate the sum of the given probabilities:

Sum of probabilities = $P(A) + P(B)$

Sum of probabilities = $0.6 + 0.3$

Sum of probabilities = $0.9$

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We observe that the sum of the probabilities ($0.9$) is not equal to 1.

$0.9 \neq 1$

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Conclusion

Since the sum of the probabilities of events A and B is 0.9, which is less than 1, A and B cannot be the only two possible outcomes of the experiment.

There must be at least one other possible outcome (or set of outcomes) in this experiment, the probability of which is $1 - 0.9 = 0.1$, to make the total probability sum up to 1.

Let's explain the concepts using the example of throwing a standard six-sided die once.

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1. Experiment

An experiment is an action or process that results in one of several possible outcomes. It must be repeatable under essentially the same conditions.

In this example, the experiment is the act of throwing a die once.

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2. Sample Space

The sample space ($S$) of an experiment is the set of all possible outcomes of that experiment. It is usually denoted by the letter 'S'.

When a standard six-sided die is thrown once, the possible outcomes are the numbers that can appear on the uppermost face.

The possible outcomes are: 1, 2, 3, 4, 5, 6.

The sample space for this experiment is $S = \{1, 2, 3, 4, 5, 6\}$.

The total number of outcomes in the sample space is $|S| = 6$.

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3. Event

An event is any subset of the sample space. It is a collection of one or more outcomes of the experiment.

An event occurs if any one of its outcomes occurs.

An event can be an elementary event (a single outcome) or a compound event (more than one outcome).

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4. Elementary Event

An elementary event is an event that consists of just one single outcome from the sample space.

For the experiment of throwing a die, the elementary events are the individual outcomes:

• Getting a 1: $\{1\}$
• Getting a 2: $\{2\}$
• Getting a 3: $\{3\}$
• Getting a 4: $\{4\}$
• Getting a 5: $\{5\}$
• Getting a 6: $\{6\}$

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Example of a Compound Event and its Elementary Events

Consider the event E: "Getting an even number" when throwing a die.

This event consists of the outcomes where the number is even. Looking at the sample space $S = \{1, 2, 3, 4, 5, 6\}$, the even numbers are 2, 4, and 6.

So, the event E is the set of outcomes $\{2, 4, 6\}$.

Since the event E contains more than one outcome, it is a compound event.

The elementary events that make up the compound event E = $\{2, 4, 6\}$ are the individual outcomes within this set:

• The elementary event $\{2\}$
• The elementary event $\{4\}$
• The elementary event $\{6\}$

The event "Getting an even number" occurs if any one of these elementary events occurs (i.e., if the die shows a 2, a 4, or a 6).

Given Information

Total number of times two coins are tossed (Total trials) = 200

Frequency of 'Two Heads' (HH) = 45

Frequency of 'One Head' (HT or TH) = 110

Frequency of 'No Heads' (TT) = 45

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To Find

The empirical probability of:

(a) Exactly one Head

(b) At least one Head

(c) No Heads

Verify that the sum of probabilities of all outcomes is 1.

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Solution

The empirical probability of an event is given by the formula:

$P(\text{Event}) = \frac{\text{Number of times the event occurred}}{\text{Total number of trials}}$

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(a) Empirical probability of getting Exactly one Head:

The event is 'Exactly one Head' (HT or TH).

Number of times this event occurred = 110

$P(\text{Exactly one Head}) = \frac{110}{200}$

Simplifying the fraction by dividing the numerator and denominator by 10:

$P(\text{Exactly one Head}) = \frac{\cancel{110}^{11}}{\cancel{200}_{20}} = \frac{11}{20}$

---

(b) Empirical probability of getting At least one Head:

The event 'At least one Head' means getting either one Head (HT or TH) or two Heads (HH).

Number of times 'At least one Head' occurred = Frequency(One Head) + Frequency(Two Heads)

Number of times 'At least one Head' occurred = $110 + 45 = 155$

$P(\text{At least one Head}) = \frac{155}{200}$

Simplifying the fraction by dividing the numerator and denominator by 5:

$P(\text{At least one Head}) = \frac{\cancel{155}^{31}}{\cancel{200}_{40}} = \frac{31}{40}$

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(c) Empirical probability of getting No Heads:

The event is 'No Heads' (TT).

Number of times this event occurred = 45

$P(\text{No Heads}) = \frac{45}{200}$

Simplifying the fraction by dividing the numerator and denominator by 5:

$P(\text{No Heads}) = \frac{\cancel{45}^{9}}{\cancel{200}_{40}} = \frac{9}{40}$

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Verification of the sum of probabilities

The possible outcomes in this experiment are Two Heads (HH), One Head (HT or TH), and No Heads (TT). The sum of their empirical probabilities should be 1.

$P(\text{HH}) = \frac{45}{200} = \frac{9}{40}$

$P(\text{One Head}) = \frac{110}{200} = \frac{11}{20} = \frac{22}{40}$

$P(\text{No Heads}) = \frac{45}{200} = \frac{9}{40}$

---

Sum of probabilities = $P(\text{HH}) + P(\text{One Head}) + P(\text{No Heads})$

Sum of probabilities = $\frac{9}{40} + \frac{22}{40} + \frac{9}{40}$

Sum of probabilities = $\frac{9 + 22 + 9}{40}$

Sum of probabilities = $\frac{40}{40}$

Sum of probabilities = $1$

---

Since the sum of the probabilities of all possible outcomes (HH, One Head, TT) is 1, the result is verified.

Given Information

Total number of times the die is thrown (Total trials) = 300

Frequencies of the outcomes are given in the table:

Frequency of 1 = 40

Frequency of 2 = 52

Frequency of 3 = 48

Frequency of 4 = 50

Frequency of 5 = 60

Frequency of 6 = 50

---

To Find

The empirical probability of getting:

(a) The number 5.

(b) An even number.

(c) A number less than 3.

(d) A number greater than or equal to 4.

---

Solution

The empirical probability of an event is calculated using the formula:

$P(\text{Event}) = \frac{\text{Number of times the event occurred}}{\text{Total number of trials}}$

---

(a) Empirical probability of getting the number 5:

The event is getting the number 5.

Number of times the event occurred (Frequency of 5) = 60

$P(\text{Getting 5}) = \frac{\text{Frequency of 5}}{\text{Total number of trials}}$

$P(\text{Getting 5}) = \frac{60}{300}$

Simplifying the fraction:

$P(\text{Getting 5}) = \frac{\cancel{60}^{1}}{\cancel{300}_{5}} = \frac{1}{5}$

---

(b) Empirical probability of getting an even number:

The event is getting an even number. The even numbers on a die are 2, 4, and 6.

Number of times the event occurred = Frequency(2) + Frequency(4) + Frequency(6)

Number of times the event occurred = $52 + 50 + 50 = 152$

$P(\text{Getting an even number}) = \frac{\text{Frequency(2) + Frequency(4) + Frequency(6)}}{\text{Total number of trials}}$

$P(\text{Getting an even number}) = \frac{152}{300}$

Simplifying the fraction by dividing the numerator and denominator by 4:

$152 \div 4 = 38$

$300 \div 4 = 75$

$P(\text{Getting an even number}) = \frac{38}{75}$

---

(c) Empirical probability of getting a number less than 3:

The event is getting a number less than 3. The numbers less than 3 on a die are 1 and 2.

Number of times the event occurred = Frequency(1) + Frequency(2)

Number of times the event occurred = $40 + 52 = 92$

$P(\text{Getting a number less than 3}) = \frac{\text{Frequency(1) + Frequency(2)}}{\text{Total number of trials}}$

$P(\text{Getting a number less than 3}) = \frac{92}{300}$

Simplifying the fraction by dividing the numerator and denominator by 4:

$92 \div 4 = 23$

$300 \div 4 = 75$

$P(\text{Getting a number less than 3}) = \frac{23}{75}$

---

(d) Empirical probability of getting a number greater than or equal to 4:

The event is getting a number greater than or equal to 4. The numbers $\geq 4$ on a die are 4, 5, and 6.

Number of times the event occurred = Frequency(4) + Frequency(5) + Frequency(6)

Number of times the event occurred = $50 + 60 + 50 = 160$

$P(\text{Getting a number } \geq 4) = \frac{\text{Frequency(4) + Frequency(5) + Frequency(6)}}{\text{Total number of trials}}$

$P(\text{Getting a number } \geq 4) = \frac{160}{300}$

Simplifying the fraction by dividing the numerator and denominator by 20:

$160 \div 20 = 8$

$300 \div 20 = 15$

$P(\text{Getting a number } \geq 4) = \frac{8}{15}$

Concept of Equally Likely Outcomes

In a random experiment, outcomes are considered equally likely if each outcome has the exact same chance of occurring as any other outcome in the sample space. There is no inherent bias or factor that makes one outcome more or less probable than another.

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Relation to Probability Calculation

When the outcomes of an experiment are equally likely, the probability of any specific event can be calculated easily using the classical definition of probability:

$P(\text{Event}) = \frac{\text{Number of favourable outcomes for the event}}{\text{Total number of possible outcomes in the sample space}}$

In this formula, the denominator represents the size of the sample space, and the numerator represents the number of outcomes within that sample space that satisfy the conditions of the event. This formula is valid only when all outcomes are equally likely.

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Examples of Experiments with Equally Likely Outcomes

1. Tossing a Fair Coin:

The possible outcomes are Head (H) and Tail (T).

Sample Space $S = \{H, T\}$. Total outcomes = 2.

Assuming the coin is fair, the probability of getting a Head is $\frac{1}{2}$, and the probability of getting a Tail is $\frac{1}{2}$. Since $P(H) = P(T) = \frac{1}{2}$, the outcomes H and T are equally likely.

---

2. Rolling a Fair Six-Sided Die:

The possible outcomes are the numbers 1, 2, 3, 4, 5, and 6.

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

Assuming the die is fair, the probability of getting any specific number (e.g., 1, 2, 3, etc.) is $\frac{1}{6}$. Since the probability of each individual outcome is the same ($\frac{1}{6}$), the outcomes are equally likely.

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Examples of Experiments with Outcomes that are Not Equally Likely

1. Drawing a Marble from a Bag with Different Quantities:

Consider a bag containing 5 red marbles and 1 blue marble.

The possible outcomes when drawing one marble are getting a Red marble (R) or getting a Blue marble (B).

Total number of marbles = $5 + 1 = 6$.

Probability of drawing Red: $P(R) = \frac{\text{Number of red marbles}}{\text{Total marbles}} = \frac{5}{6}$.

Probability of drawing Blue: $P(B) = \frac{\text{Number of blue marbles}}{\text{Total marbles}} = \frac{1}{6}$.

Since $P(R) = \frac{5}{6}$ and $P(B) = \frac{1}{6}$, and $\frac{5}{6} \neq \frac{1}{6}$, the outcomes are not equally likely. Drawing a red marble is more likely than drawing a blue marble.

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2. Tossing a Biased Coin:

Consider a coin that is weighted such that Heads appears more often than Tails.

The possible outcomes are Head (H) and Tail (T).

If the coin is biased to land on Heads, the probability of getting a Head might be, say, $P(H) = 0.7$, and the probability of getting a Tail would be $P(T) = 1 - 0.7 = 0.3$.

Since $P(H) = 0.7$ and $P(T) = 0.3$, and $0.7 \neq 0.3$, the outcomes Heads and Tails are not equally likely.

Given Information

Number of red balls = 15

Number of blue balls = 10

Number of green balls = 5

A ball is drawn from the bag at random, meaning each ball has an equal chance of being drawn.

---

To Find

The probability of drawing:

(a) A red ball.

(b) A blue ball.

(c) A green ball.

(d) A ball that is not green.

(e) A yellow ball.

---

Solution

First, calculate the total number of balls in the bag.

Total number of balls = Number of red balls + Number of blue balls + Number of green balls

Total number of balls = $15 + 10 + 5 = 30$

The total number of possible outcomes is 30.

---

The probability of an event is given by the formula:

$P(\text{Event}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

---

(a) Probability of drawing a red ball:

The event is drawing a red ball.

Number of favourable outcomes = Number of red balls = 15

$P(\text{Drawing a red ball}) = \frac{\text{Number of red balls}}{\text{Total number of balls}}$

$P(\text{Drawing a red ball}) = \frac{15}{30}$

Simplifying the fraction:

$P(\text{Drawing a red ball}) = \frac{\cancel{15}^{1}}{\cancel{30}_{2}} = \frac{1}{2}$

---

(b) Probability of drawing a blue ball:

The event is drawing a blue ball.

Number of favourable outcomes = Number of blue balls = 10

$P(\text{Drawing a blue ball}) = \frac{\text{Number of blue balls}}{\text{Total number of balls}}$

$P(\text{Drawing a blue ball}) = \frac{10}{30}$

Simplifying the fraction:

$P(\text{Drawing a blue ball}) = \frac{\cancel{10}^{1}}{\cancel{30}_{3}} = \frac{1}{3}$

---

(c) Probability of drawing a green ball:

The event is drawing a green ball.

Number of favourable outcomes = Number of green balls = 5

$P(\text{Drawing a green ball}) = \frac{\text{Number of green balls}}{\text{Total number of balls}}$

$P(\text{Drawing a green ball}) = \frac{5}{30}$

Simplifying the fraction:

$P(\text{Drawing a green ball}) = \frac{\cancel{5}^{1}}{\cancel{30}_{6}} = \frac{1}{6}$

---

(d) Probability of drawing a ball that is not green:

The event is drawing a ball that is not green.

The balls that are not green are the red balls and the blue balls.

Number of favourable outcomes = Number of red balls + Number of blue balls = $15 + 10 = 25$

$P(\text{Drawing a ball that is not green}) = \frac{\text{Number of balls that are not green}}{\text{Total number of balls}}$

$P(\text{Drawing a ball that is not green}) = \frac{25}{30}$

Simplifying the fraction by dividing numerator and denominator by 5:

$P(\text{Drawing a ball that is not green}) = \frac{\cancel{25}^{5}}{\cancel{30}_{6}} = \frac{5}{6}$

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Alternate method for (d): Using the complement rule

The event 'drawing a ball that is not green' is the complement of the event 'drawing a green ball'.

$P(\text{Not Green}) = 1 - P(\text{Green})$

We found $P(\text{Green}) = \frac{1}{6}$ in part (c).

$P(\text{Not Green}) = 1 - \frac{1}{6} = \frac{6}{6} - \frac{1}{6} = \frac{5}{6}$

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(e) Probability of drawing a yellow ball:

The event is drawing a yellow ball.

There are no yellow balls in the bag.

Number of favourable outcomes = Number of yellow balls = 0

$P(\text{Drawing a yellow ball}) = \frac{\text{Number of yellow balls}}{\text{Total number of balls}}$

$P(\text{Drawing a yellow ball}) = \frac{0}{30}$

$P(\text{Drawing a yellow ball}) = 0$

This is an impossible event.

Given Information

Total number of students surveyed = 50

Number of students who like Cricket = 30

Number of students who like Football = 20

Number of students who like both Cricket and Football = 10

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To Find

The empirical probability that a randomly selected student:

(a) Likes Cricket only.

(b) Likes Football only.

(c) Likes neither Cricket nor Football.

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Solution

The total number of possible outcomes is the total number of students surveyed, which is 50.

The empirical probability of an event is given by the formula:

$P(\text{Event}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of trials}}$

---

(a) Probability that the student likes Cricket only:

Students who like Cricket only are those who like Cricket but do not like Football.

Number of students who like Cricket only = (Number of students who like Cricket) - (Number of students who like both)

Number of students who like Cricket only = $30 - 10 = 20$

$P(\text{Likes Cricket only}) = \frac{\text{Number of students who like Cricket only}}{\text{Total number of students}}$

$P(\text{Likes Cricket only}) = \frac{20}{50}$

Simplifying the fraction:

$P(\text{Likes Cricket only}) = \frac{\cancel{20}^{2}}{\cancel{50}_{5}} = \frac{2}{5}$

---

(b) Probability that the student likes Football only:

Students who like Football only are those who like Football but do not like Cricket.

Number of students who like Football only = (Number of students who like Football) - (Number of students who like both)

Number of students who like Football only = $20 - 10 = 10$

$P(\text{Likes Football only}) = \frac{\text{Number of students who like Football only}}{\text{Total number of students}}$

$P(\text{Likes Football only}) = \frac{10}{50}$

Simplifying the fraction:

$P(\text{Likes Football only}) = \frac{\cancel{10}^{1}}{\cancel{50}_{5}} = \frac{1}{5}$

---

(c) Probability that the student likes neither Cricket nor Football:

First, find the number of students who like at least one of the sports. This is the number of students who like Cricket only, Football only, or both.

Number of students who like at least one sport = (Like Cricket only) + (Like Football only) + (Like both)

Number of students who like at least one sport = $20 + 10 + 10 = 40$

Alternatively, using set theory principle (though not strictly needed with 'only' counts):

$n(C \cup F) = n(C) + n(F) - n(C \cap F)$

$n(C \cup F) = 30 + 20 - 10 = 40$

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Students who like neither sport are the total students minus those who like at least one sport.

Number of students who like neither = Total students - Number of students who like at least one sport

Number of students who like neither = $50 - 40 = 10$

$P(\text{Likes neither}) = \frac{\text{Number of students who like neither}}{\text{Total number of students}}$

$P(\text{Likes neither}) = \frac{10}{50}$

Simplifying the fraction:

$P(\text{Likes neither}) = \frac{\cancel{10}^{1}}{\cancel{50}_{5}} = \frac{1}{5}$

Given Information

Number of cars = 120

Number of bikes = 80

Number of trucks = 50

Number of buses = 30

A vehicle is chosen at random, meaning each vehicle has an equal chance of being selected from the surveyed data.

---

To Find

The empirical probability that the vehicle is:

(a) A car.

(b) A bike.

(c) Not a bus.

---

Solution

First, calculate the total number of vehicles surveyed.

Total number of vehicles = Number of cars + Number of bikes + Number of trucks + Number of buses

Total number of vehicles = $120 + 80 + 50 + 30 = 280$

The total number of possible outcomes is 280.

---

The empirical probability of an event is calculated using the formula:

$P(\text{Event}) = \frac{\text{Number of times the event occurred}}{\text{Total number of trials}}$

---

(a) Empirical probability that the vehicle is a car:

The event is selecting a car.

Number of times the event occurred (Number of cars) = 120

$P(\text{Vehicle is a car}) = \frac{\text{Number of cars}}{\text{Total number of vehicles}}$

$P(\text{Vehicle is a car}) = \frac{120}{280}$

Simplifying the fraction by dividing numerator and denominator by 40:

$P(\text{Vehicle is a car}) = \frac{\cancel{120}^{3}}{\cancel{280}_{7}} = \frac{3}{7}$

---

(b) Empirical probability that the vehicle is a bike:

The event is selecting a bike.

Number of times the event occurred (Number of bikes) = 80

$P(\text{Vehicle is a bike}) = \frac{\text{Number of bikes}}{\text{Total number of vehicles}}$

$P(\text{Vehicle is a bike}) = \frac{80}{280}$

Simplifying the fraction by dividing numerator and denominator by 40:

$P(\text{Vehicle is a bike}) = \frac{\cancel{80}^{2}}{\cancel{280}_{7}} = \frac{2}{7}$

---

(c) Empirical probability that the vehicle is not a bus:

The event is selecting a vehicle that is not a bus.

The vehicles that are not buses are cars, bikes, and trucks.

Number of times the event occurred (Number of non-bus vehicles) = Number of cars + Number of bikes + Number of trucks

Number of non-bus vehicles = $120 + 80 + 50 = 250$

$P(\text{Vehicle is not a bus}) = \frac{\text{Number of vehicles that are not buses}}{\text{Total number of vehicles}}$

$P(\text{Vehicle is not a bus}) = \frac{250}{280}$

Simplifying the fraction by dividing numerator and denominator by 10:

$P(\text{Vehicle is not a bus}) = \frac{\cancel{250}^{25}}{\cancel{280}_{28}} = \frac{25}{28}$

Relationship between $P(E)$ and $P(\text{not } E)$

Let E be any event in a random experiment.

The event 'not E' (also called the complement of E, often denoted as E') is the event that occurs if and only if event E does not occur.

The sample space ($S$) of an experiment is made up of outcomes where event E occurs and outcomes where event E does not occur.

These two events, E and 'not E', are mutually exclusive (they cannot happen at the same time) and exhaustive (one of them must happen).

---

The fundamental relationship between the probability of an event E and the probability of the event 'not E' is that their sum is always equal to 1.

Mathematically, this relationship is expressed as:

$P(E) + P(\text{not } E) = 1$

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This formula can be rearranged to find the probability of 'not E' if the probability of E is known:

$P(\text{not } E) = 1 - P(E)$

---

Example 1: Probability of Rain

Given Information:

The probability of rain tomorrow, $P(\text{Rain}) = 0.7$.

To Find:

The probability that it will not rain tomorrow, $P(\text{Not Rain})$.

Solution:

Using the relationship $P(E) + P(\text{not } E) = 1$:

$P(\text{Rain}) + P(\text{Not Rain}) = 1$

$0.7 + P(\text{Not Rain}) = 1$

$P(\text{Not Rain}) = 1 - 0.7$

$P(\text{Not Rain}) = 0.3$

The probability that it will not rain tomorrow is 0.3.

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Example 2: Probability of Winning a Game

Given Information:

The probability of winning a game, $P(\text{Winning}) = \frac{3}{8}$.

To Find:

The probability of not winning the game, $P(\text{Not Winning})$.

Solution:

Using the relationship $P(E) + P(\text{not } E) = 1$:

$P(\text{Winning}) + P(\text{Not Winning}) = 1$

$\frac{3}{8} + P(\text{Not Winning}) = 1$

$P(\text{Not Winning}) = 1 - \frac{3}{8}$

$P(\text{Not Winning}) = \frac{8}{8} - \frac{3}{8}$

$P(\text{Not Winning}) = \frac{8-3}{8} = \frac{5}{8}$

The probability of not winning the game is $\frac{5}{8}$.

Given Information

Total number of eggs in the carton = 144

Number of rotten eggs = 12

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Calculation of Good Eggs

Number of good eggs = Total number of eggs - Number of rotten eggs

Number of good eggs = $144 - 12 = 132$

---

To Find

The probability that a randomly selected egg is:

(a) A rotten egg.

(b) A good egg.

Verify that the sum of these probabilities is 1.

---

Solution

The total number of possible outcomes is the total number of eggs in the carton, which is 144.

The probability of an event is given by the formula:

$P(\text{Event}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

---

(a) Probability of drawing a rotten egg:

The event is selecting a rotten egg.

Number of favourable outcomes = Number of rotten eggs = 12

$P(\text{Rotten egg}) = \frac{\text{Number of rotten eggs}}{\text{Total number of eggs}}$

$P(\text{Rotten egg}) = \frac{12}{144}$

Simplifying the fraction by dividing numerator and denominator by 12:

$P(\text{Rotten egg}) = \frac{\cancel{12}^{1}}{\cancel{144}_{12}} = \frac{1}{12}$

---

(b) Probability of drawing a good egg:

The event is selecting a good egg.

Number of favourable outcomes = Number of good eggs = 132

$P(\text{Good egg}) = \frac{\text{Number of good eggs}}{\text{Total number of eggs}}$

$P(\text{Good egg}) = \frac{132}{144}$

Simplifying the fraction by dividing numerator and denominator by 12:

$132 \div 12 = 11$

$144 \div 12 = 12$

$P(\text{Good egg}) = \frac{11}{12}$

---

Verification of the sum of probabilities

Let E be the event of selecting a rotten egg, and E' be the event of selecting a good egg.

These two events are complements of each other (an egg is either rotten or good, and cannot be both).

The sum of the probabilities of an event and its complement should be 1.

$P(\text{Rotten egg}) + P(\text{Good egg}) = \frac{1}{12} + \frac{11}{12}$

$P(\text{Rotten egg}) + P(\text{Good egg}) = \frac{1+11}{12} = \frac{12}{12} = 1$

---

Since the sum of the probabilities is 1, the result is verified.

Given Information

Probability that the student gives a correctly answered question, $P(\text{Correct}) = 0.75$

Probability that the student gives an incorrectly answered question, $P(\text{Incorrect}) = 0.20$

---

Analysis

For a set of outcomes to be the only possible outcomes of an experiment, the sum of their probabilities must be equal to 1.

---

Let's calculate the sum of the given probabilities:

Sum of probabilities = $P(\text{Correct}) + P(\text{Incorrect})$

Sum of probabilities = $0.75 + 0.20$

Sum of probabilities = $0.95$

---

We compare the sum of the probabilities to 1:

Sum $= 0.95$

$1 = 1.00$

Since $0.95 \neq 1$, the sum of the probabilities of the given outcomes is not equal to 1.

---

Conclusion

These are not the only two possible outcomes of the experiment.

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Other Possible Outcome(s)

If the student writes an answer, possible outcomes are typically 'Correct', 'Incorrect', or perhaps outcomes that fall into neither category from a scoring perspective, such as 'Partially Correct', or even 'No Answer' (though the question states the student writes an answer, ruling out a blank paper entirely, a partially correct answer might still be distinct from fully correct or fully incorrect depending on the scoring criteria). Given the context of 'Correct' and 'Incorrect', the missing probability accounts for any other scenario not covered by these two, often implicitly representing outcomes like 'Partially Correct' or an answer that cannot be strictly classified as fully correct or incorrect.

Let O be the event representing any other possible outcome(s).

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Probability of the Other Outcome(s)

The sum of the probabilities of all possible outcomes must equal 1.

$P(\text{Correct}) + P(\text{Incorrect}) + P(\text{Other Outcome(s)}) = 1$

Substitute the given probabilities:

$0.75 + 0.20 + P(\text{Other Outcome(s)}) = 1$

$0.95 + P(\text{Other Outcome(s)}) = 1$

$P(\text{Other Outcome(s)}) = 1 - 0.95$

$P(\text{Other Outcome(s)}) = 0.05$

---

The probability of the other possible outcome(s) (e.g., partially correct, unclassifiable answer) is 0.05.

Given Information

Total number of women surveyed = 300

Number of women suffering due to Lack of proper diet = 80

Number of women suffering due to Unsafe drinking water = 70

Number of women suffering due to Lack of medical facilities = 60

Number of women suffering due to Others = 90

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Verification of Total Count

Sum of frequencies = $80 + 70 + 60 + 90 = 300$

This matches the total number of women surveyed.

---

To Find

The empirical probability that a randomly selected woman is suffering due to:

(a) Lack of proper diet.

(b) Unsafe drinking water.

(c) Lack of medical facilities or unsafe drinking water.

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Solution

The total number of possible outcomes is the total number of women surveyed, which is 300.

The empirical probability of an event is given by the formula:

$P(\text{Event}) = \frac{\text{Number of times the event occurred}}{\text{Total number of trials}}$

---

(a) Empirical probability that the woman is suffering due to Lack of proper diet:

The event is suffering due to 'Lack of proper diet'.

Number of favourable outcomes = Number of women suffering due to Lack of proper diet = 80

$P(\text{Lack of proper diet}) = \frac{\text{Number of women suffering due to Lack of proper diet}}{\text{Total number of women}}$

$P(\text{Lack of proper diet}) = \frac{80}{300}$

Simplifying the fraction by dividing numerator and denominator by 20:

$P(\text{Lack of proper diet}) = \frac{\cancel{80}^{4}}{\cancel{300}_{15}} = \frac{4}{15}$

---

(b) Empirical probability that the woman is suffering due to Unsafe drinking water:

The event is suffering due to 'Unsafe drinking water'.

Number of favourable outcomes = Number of women suffering due to Unsafe drinking water = 70

$P(\text{Unsafe drinking water}) = \frac{\text{Number of women suffering due to Unsafe drinking water}}{\text{Total number of women}}$

$P(\text{Unsafe drinking water}) = \frac{70}{300}$

Simplifying the fraction by dividing numerator and denominator by 10:

$P(\text{Unsafe drinking water}) = \frac{\cancel{70}^{7}}{\cancel{300}_{30}} = \frac{7}{30}$

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(c) Empirical probability that the woman is suffering due to Lack of medical facilities or unsafe drinking water:

The event is suffering due to 'Lack of medical facilities' or 'Unsafe drinking water'. Since these are presented as distinct categories in the survey data, we assume they are mutually exclusive for the purpose of counting.

Number of favourable outcomes = (Number suffering due to Lack of medical facilities) + (Number suffering due to Unsafe drinking water)

Number of favourable outcomes = $60 + 70 = 130$

$P(\text{Lack of medical facilities or Unsafe drinking water}) = \frac{\text{Number suffering due to L.M.F. or U.D.W.}}{\text{Total number of women}}$

$P(\text{Lack of medical facilities or Unsafe drinking water}) = \frac{130}{300}$

Simplifying the fraction by dividing numerator and denominator by 10:

$P(\text{Lack of medical facilities or Unsafe drinking water}) = \frac{\cancel{130}^{13}}{\cancel{300}_{30}} = \frac{13}{30}$

Empirical Probability

Empirical probability is determined by conducting an experiment or observing real-world events and recording the outcomes. It is calculated based on the actual frequency of an event occurring in a series of trials.

Formula:

$P(\text{Event}) = \frac{\text{Number of times the event occurred in the experiment}}{\text{Total number of trials in the experiment}}$

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Theoretical Probability

Theoretical probability is determined by analyzing the possible outcomes of an experiment based on mathematical theory and the assumption that all elementary outcomes are equally likely. It represents the expected likelihood of an event occurring in an idealized situation.

Formula (for equally likely outcomes):

$P(\text{Event}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

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Key Difference

The main difference lies in their basis:

• Empirical probability is based on observation and actual results from experiments that have been performed. It describes "what did happen".
• Theoretical probability is based on calculation and reasoning about equally likely possibilities in a theoretical model. It describes "what should happen" in an ideal scenario.

As the number of trials in an experiment increases, the empirical probability of an event generally gets closer to its theoretical probability (Law of Large Numbers).

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When Empirical Probability is Usually Used

We usually use empirical probability in the following situations:

• When the outcomes of an experiment are not equally likely, making the theoretical probability formula (based on equally likely outcomes) inappropriate.
• When it is difficult or impossible to determine the total number of possible outcomes or the number of favourable outcomes through theoretical analysis alone (e.g., complex systems, weather prediction).
• When we want to find the probability of an event in a real-world context based on past data or experience (e.g., insurance risk assessment, quality control in manufacturing, predicting customer behaviour).
• To verify or test theoretical probability models.

In essence, empirical probability is used when dealing with real-world data or situations where a theoretical model is not available or does not accurately reflect the probabilities.