Chapter 16 Probability (Additional Questions)

A random experiment is an experiment whose outcome cannot be predicted with certainty before it is performed, but the set of all possible outcomes is known.

Let's analyze the options:

(A) Calculating the area of a circle: The area of a circle is determined by its radius using the formula $A = \pi r^2$. This is a deterministic calculation, not a random experiment.

(B) Tossing a fair coin: When a fair coin is tossed, the possible outcomes are heads (H) or tails (T). We know these are the only possibilities, but we cannot predict which outcome will occur on any given toss. Therefore, tossing a fair coin is a random experiment.

(C) Determining the boiling point of water: Under standard atmospheric pressure, water boils at a specific temperature ($100^\circ$C or $212^\circ$F). This is a physical constant and its determination is not random.

(D) Measuring the length of a table: While there might be slight variations due to measurement tools or precision, the length of a table is a fixed property that can be measured. It's not about an outcome that can vary randomly from a known set of possibilities in the context of probability.

Therefore, the correct answer is (B) Tossing a fair coin.

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

When we toss three distinct coins simultaneously, we need to consider the outcome for each coin. Let 'H' represent Heads and 'T' represent Tails.

For the first coin, the outcome can be H or T.

For the second coin, the outcome can be H or T.

For the third coin, the outcome can be H or T.

To find all possible combinations, we can list them systematically:

If the first coin is H:

• Second coin H, Third coin H: HHH
• Second coin H, Third coin T: HHT
• Second coin T, Third coin H: HTH
• Second coin T, Third coin T: HTT

If the first coin is T:

• Second coin H, Third coin H: THH
• Second coin H, Third coin T: THT
• Second coin T, Third coin H: TTH
• Second coin T, Third coin T: TTT

The total number of outcomes is $2 \times 2 \times 2 = 2^3 = 8$.

Let's examine the given options:

(A) $\{HHH, TTT\}$: This lists only two outcomes, which is incomplete.

(B) $\{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$: This option lists all 8 possible unique outcomes.

(C) $\{H, T\}$: This represents the sample space for tossing a single coin.

(D) $\{3H, 2H1T, 1H2T, 3T\}$: This describes the outcomes by the number of heads and tails, not the specific sequence of each coin toss. While these categories are correct, the sample space is defined by the individual, distinguishable outcomes.

Therefore, the correct sample space for tossing three distinct coins simultaneously is the set of all unique ordered outcomes.

The correct answer is (B) $\{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$.

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

When a standard six-sided die is rolled, the possible outcomes are the numbers on its faces. These are typically the integers from 1 to 6.

The set of all possible outcomes, which is the sample space, can be represented as:

The number of elements in the sample space is the total count of these possible outcomes.

By counting the elements in the set $S$, we find there are 6 possible outcomes.

Therefore, the number of elements in the sample space when a die is rolled is 6.

The correct answer is (B) 6.

An event is a set of outcomes of an experiment to which the probability is assigned. In probability theory, an event is a subset of the sample space.

The sample space ($S$) when a fair six-sided die is rolled is $S = \{1, 2, 3, 4, 5, 6\}$.

Let's analyze each option to see if it represents a subset of $S$:

(A) Getting a number greater than 6:

The numbers in the sample space are $\{1, 2, 3, 4, 5, 6\}$. There are no numbers in this set that are greater than 6. Therefore, this is an empty set ($\emptyset$), which is a subset of any set, but it's often referred to as an impossible event in this context.

(B) Getting an odd number:

The odd numbers in the sample space are 1, 3, and 5. So, this event can be represented as the set $\{1, 3, 5\}$. Since $\{1, 3, 5\}$ is a subset of $\{1, 2, 3, 4, 5, 6\}$, this is a valid event.

(C) The outcome is 5:

This event consists of a single outcome, which is 5. This can be represented as the set $\{5\}$. Since $\{5\}$ is a subset of $\{1, 2, 3, 4, 5, 6\}$, this is also a valid event.

(D) Both (B) and (C):

Since both (B) and (C) represent valid events (subsets of the sample space), this option correctly identifies that both are events.

In summary, an event is any collection of outcomes, including a single outcome or all possible outcomes. Both "getting an odd number" and "the outcome is 5" are valid events.

Therefore, the correct answer is (D) Both (B) and (C).

The sample space ($S$) for rolling a standard six-sided die is $S = \{1, 2, 3, 4, 5, 6\}$.

An event is a subset of the sample space.

• A simple event consists of exactly one outcome.
• A compound event consists of two or more outcomes.
• An impossible event is an event that cannot occur. Its corresponding subset in the sample space is the empty set ($\emptyset$).
• A sure event is an event that is certain to occur. Its corresponding subset in the sample space is the entire sample space ($S$).

In this question, the event $E$ is defined as "getting a number greater than 6" when rolling a die.

Looking at the sample space $S = \{1, 2, 3, 4, 5, 6\}$, there are no outcomes that satisfy the condition of being greater than 6.

Therefore, the event $E$ corresponds to the empty set:

An event that cannot occur is called an impossible event.

Let's check the options:

(A) Simple event: This requires exactly one outcome, which is not the case here.

(B) Compound event: This requires two or more outcomes, which is also not the case.

(C) Impossible event: This accurately describes an event with no outcomes in the sample space.

(D) Sure event: This would be an event that includes all outcomes in the sample space, such as "getting a number less than or equal to 6".

Thus, the event of getting a number greater than 6 when rolling a die is an impossible event.

The correct answer is (C) Impossible event.

In probability theory, two events are said to be mutually exclusive if they cannot occur at the same time. This means that if one event happens, the other cannot possibly happen.

Mathematically, if events A and B are mutually exclusive, their intersection is the empty set, meaning there are no common outcomes between them.

Let's examine the given options:

(A) $A \cup B = S$: This condition means that the union of events A and B covers the entire sample space. Events satisfying this condition are called exhaustive events, but not necessarily mutually exclusive.

(B) $A \cap B = \emptyset$: This condition states that the intersection of events A and B is the empty set. This directly means that there are no common outcomes between event A and event B, which is the definition of mutually exclusive events.

(C) $A = B'$: This condition means that event A is the complement of event B. If A is the complement of B, then A and B are indeed mutually exclusive and exhaustive, but the primary definition of mutually exclusive is the absence of common outcomes.

(D) $P(A) = P(B)$: This condition states that the probabilities of events A and B are equal. This does not imply that they are mutually exclusive; events can have equal probabilities but still overlap.

The defining characteristic of mutually exclusive events is that they have no outcomes in common. This is represented by their intersection being the empty set.

Therefore, two events A and B are mutually exclusive if $A \cap B = \emptyset$.

The correct answer is (B) $A \cap B = \emptyset$.

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

Assertion (A): For any event E, $0 \leq P(E) \leq 1$.

In probability theory, the probability of an event is defined as a measure of the likelihood of that event occurring. By definition, the probability of any event must be a value between 0 and 1, inclusive. A probability of 0 means the event is impossible, and a probability of 1 means the event is certain (a sure event).

Thus, Assertion (A) is true.

Reason (R): The probability of any event is a value between 0 and 1, inclusive.

This statement is the fundamental definition and property of probability. It reiterates the range within which probabilities must fall.

Thus, Reason (R) is true.

Now, let's determine if Reason (R) correctly explains Assertion (A).

Assertion (A) states a mathematical property ($0 \leq P(E) \leq 1$). Reason (R) provides the definition or fundamental principle that underpins this property. The reason why the probability of any event E must be between 0 and 1 is precisely because the probability of any event is defined as a value within this range.

Therefore, Reason (R) is the correct explanation for Assertion (A).

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

The sample space ($S$) of a random experiment is the set of all possible outcomes.

In probability theory, the probability of an event is a measure of its likelihood. Several axioms govern probability:

1. For any event $E$, $P(E) \geq 0$.
2. For the sample space $S$, $P(S) = 1$.
3. If $E_1, E_2, \dots$ are mutually exclusive events, then $P(E_1 \cup E_2 \cup \dots) = P(E_1) + P(E_2) + \dots$.

The sample space $S$ represents all possible outcomes of an experiment. Since it is certain that one of the outcomes in the sample space will occur, its probability is 1.

Therefore, the probability of the sample space itself is always 1.

Let's review the options:

(A) 0: A probability of 0 represents an impossible event.

(B) 0.5: A probability of 0.5 represents an event that is equally likely to occur or not occur.

(C) 1: A probability of 1 represents a sure event, meaning it is certain to happen.

(D) Undefined: Probabilities are well-defined values.

Since the sample space $S$ encompasses all possible outcomes, it is a sure event that one of these outcomes will occur.

Thus, $P(S) = 1$.

The correct answer is (C) 1.

Let's define each type of event and its characteristic:

Definitions:

• A Sure Event is an event that is certain to happen. It includes all possible outcomes of an experiment, meaning it is equivalent to the entire sample space ($S$).
• An Impossible Event is an event that cannot happen under any circumstances. It has no outcomes in common with the sample space, meaning it is represented by the empty set ($\emptyset$).
• A Simple Event is an event that consists of exactly one outcome from the sample space.
• A Compound Event is an event that consists of more than one outcome from the sample space.

Now, let's match the event types with their characteristics:

(i) Sure Event: This is an event that is certain to happen and is the entire sample space. So, (i) matches with (b) "An event that is the entire sample space."

(ii) Impossible Event: This is an event that cannot happen and is represented by the empty set. So, (ii) matches with (c) "An event that is the empty set."

(iii) Simple Event: This is an event consisting of only one outcome. So, (iii) matches with (a) "An event with only one sample point."

(iv) Compound Event: This is an event consisting of more than one outcome. So, (iv) matches with (d) "An event with more than one sample point."

The correct matching is:

• (i) - (b)
• (ii) - (c)
• (iii) - (a)
• (iv) - (d)

This corresponds to option (B).

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

To find the probability of an event, we use the formula:

First, let's identify the sample space when a fair die is rolled:

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

The total number of possible outcomes is the number of elements in the sample space, which is 6.

Next, let's identify the event of "getting an odd number":

The odd numbers in the sample space are 1, 3, and 5. Let $E$ be the event of getting an odd number.

The favorable outcomes for event $E$ are $\{1, 3, 5\}$.

The number of favourable outcomes is 3.

Now, we can calculate the probability of getting an odd number:

Simplifying the fraction:

Comparing this result with the given options:

(A) 1/6

(B) 2/6 = 1/3

(C) 3/6 = 1/2

(D) 4/6 = 2/3

The calculated probability matches option (C).

The correct answer is (C) 3/6 = 1/2.

The formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ describes the probability that either event A or event B (or both) will occur.

Let's examine the provided options:

(A) Law of Complement: The Law of Complement relates the probability of an event to the probability of its complement. For an event A, $P(A') = 1 - P(A)$, where $A'$ is the complement of A. This does not match the given formula.

(B) Addition Law of Probability: This law specifically deals with the probability of the union of two events. The formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ is the general addition rule, which accounts for the possibility of overlap (intersection) between the events A and B. This matches the given formula.

(C) Multiplication Law of Probability: The multiplication law is used to find the probability of the intersection of two events, $P(A \cap B)$. For independent events, it's $P(A \cap B) = P(A) \times P(B)$. For dependent events, it's $P(A \cap B) = P(A) \times P(B|A)$ or $P(A \cap B) = P(B) \times P(A|B)$. This does not match the given formula.

(D) Axiomatic Law: While the Addition Law is derived from the axioms of probability, "Axiomatic Law" itself is not a standard term for this specific formula.

The formula provided is the fundamental rule for calculating the probability of the union of two events, accounting for their intersection. This is known as the Addition Law of Probability.

The correct answer is (B) Addition Law of Probability.

Two events A and B are defined as mutually exclusive if they cannot occur at the same time. This means that there is no overlap between the outcomes of event A and the outcomes of event B.

In terms of set theory, if events A and B are mutually exclusive, their intersection is the empty set ($\emptyset$).

The probability of an event that is represented by the empty set is always 0.

Therefore, if A and B are mutually exclusive events, their intersection $A \cap B = \emptyset$, and the probability of this intersection is:

Let's look at the options:

(A) $P(A) + P(B)$: This would be the probability of the union of mutually exclusive events, $P(A \cup B)$, if they were mutually exclusive.

(B) $P(A)P(B)$: This is the probability of the intersection of two independent events.

(C) 0: This represents the probability of an impossible event or the intersection of mutually exclusive events.

(D) 1: This represents the probability of a sure event.

Since A and B are mutually exclusive, their intersection is empty, and thus the probability of their intersection is 0.

The correct answer is (C) 0.

The complement of an event $E$, denoted as $E'$, is the event that $E$ does not occur. The sample space $S$ can be partitioned into two mutually exclusive events: $E$ and $E'$. This means that $E \cup E' = S$ and $E \cap E' = \emptyset$.

Using the Addition Law of Probability for mutually exclusive events:

Since $E \cup E' = S$ (the sample space), we know that the probability of the sample space is 1:

Substituting these into the equation:

To find $P(E')$, we rearrange the equation:

This relationship is known as the Law of Complement.

Let's examine the options:

(A) $P(E) - 1$: This would result in a negative probability if $P(E) < 1$, which is not possible.

(B) $1 - P(E)$: This correctly represents the probability of the complement of event E.

(C) $1/P(E)$: This is related to conditional probability or odds, not the complement.

(D) $P(E \cap E)$: The intersection of an event with itself is simply the event itself, so $P(E \cap E) = P(E)$. This is not the complement.

Therefore, the probability of the complement of an event E is $1 - P(E)$.

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

To find the probability that a randomly selected student likes Football, we need to know the total number of students surveyed and the number of students who like Football.

From the case study:

• Total number of students surveyed = 100. This represents the size of the sample space.
• Number of students who like Football (F) = 60. This represents the number of favorable outcomes for the event of liking Football.
• Number of students who like Cricket (C) = 40.
• Number of students who like both Football and Cricket = 20.

The probability of an event is calculated as:

In this case, the event is "a randomly selected student likes Football".

Number of favourable outcomes (students who like Football) = 60.

Total number of outcomes (total students surveyed) = 100.

So, the probability that a randomly selected student likes Football is:

Simplifying the fraction:

Comparing this with the given options:

(A) $60/100 = 0.6$

(B) $40/100 = 0.4$

(C) $20/100 = 0.2$

(D) $80/100 = 0.8$

The calculated probability matches option (A).

The correct answer is (A) $60/100 = 0.6$.

The question asks for the probability that a randomly selected student likes both Football and Cricket. In probability theory, the word "and" typically signifies the intersection of events.

Let $F$ be the event that a student likes Football.

Let $C$ be the event that a student likes Cricket.

The phrase "likes Football and Cricket" corresponds to the intersection of these two events, denoted as $F \cap C$.

From the case study:

• Total number of students surveyed = 100.
• Number of students who like Football (F) = 60.
• Number of students who like Cricket (C) = 40.
• Number of students who like both Football and Cricket = 20. This is the number of outcomes in the intersection $F \cap C$.

The probability of the event $F \cap C$ is calculated as:

Simplifying this gives:

Now, let's evaluate the options:

(A) $P(F \cup C)$: This represents the probability that a student likes Football OR Cricket (or both), not necessarily both.

(B) $P(F \cap C)$: This notation correctly represents the probability of liking both Football and Cricket.

(C) $20/100 = 0.2$: This is the calculated probability of liking both Football and Cricket.

(D) $60/100 = 0.6$: This is the probability of liking Football only (as calculated in the previous question).

The question asks for the probability that a student likes Football AND Cricket. This is represented by $P(F \cap C)$, and its value is $20/100$ or $0.2$. Both option (B) (the notation for the probability) and option (C) (the calculated value) are correct in their own right. However, the question is asking "What is the probability...", implying a numerical answer is expected as the direct response to "What is...". Option (C) provides this direct numerical answer derived from the case study.

The correct answer is (C) $20/100 = 0.2$, as it provides the specific numerical probability based on the case study.

The question asks for the probability that a randomly selected student likes Football OR Cricket. In probability, the word "or" signifies the union of events.

Let $F$ be the event that a student likes Football.

Let $C$ be the event that a student likes Cricket.

We are looking for $P(F \cup C)$.

From the case study and previous questions, we have:

• Total students = 100
• $P(F) = 60/100 = 0.6$
• $P(C) = 40/100 = 0.4$
• $P(F \cap C) = 20/100 = 0.2$ (Probability of liking both)

We use the Addition Law of Probability:

Substituting the known values:

Let's analyze the options:

(A) $P(F) + P(C) - P(F \cap C) = 0.6 + 0.4 - 0.2 = 0.8$: This option correctly states the Addition Law and applies it to the given values, arriving at the correct probability.

(B) $0.6 + 0.4 = 1.0$: This incorrectly adds $P(F)$ and $P(C)$ without subtracting the intersection, which is only valid if F and C were mutually exclusive (which they are not, as 20 students like both).

(C) $0.8$: This is the correct numerical value of the probability.

(D) $0.2$: This is the probability of liking both Football and Cricket ($P(F \cap C)$).

Option (A) correctly shows the formula and calculation leading to the probability of liking Football or Cricket. Option (C) provides the final numerical answer. Since option (A) explains how the answer is obtained using the correct formula and also provides the final result, it is the most comprehensive correct answer. The question asks "What is the probability...", and option (A) provides the method and the result.

The correct answer is (A) $P(F) + P(C) - P(F \cap C) = 0.6 + 0.4 - 0.2 = 0.8$.

The event "a student likes neither Football nor Cricket" is the complement of the event "a student likes Football or Cricket".

Let $F$ be the event that a student likes Football.

Let $C$ be the event that a student likes Cricket.

The event "a student likes Football or Cricket" is denoted by $F \cup C$.

The event "a student likes neither Football nor Cricket" is the complement of $F \cup C$, which is denoted as $(F \cup C)'$.

We know from the previous question that $P(F \cup C) = 0.8$.

The probability of the complement of an event is given by the Law of Complement:

Applying this to our event $(F \cup C)'$:

Substituting the value of $P(F \cup C)$:

Now let's examine the options:

(A) $P((F \cup C)')$: This is the correct notation for the event "neither Football nor Cricket", but it's not the direct calculation.

(B) $1 - P(F \cup C) = 1 - 0.8 = 0.2$: This option correctly states the formula using the complement rule and shows the calculation with the previously found probability of $P(F \cup C)$.

(C) $1 - 0.8 = 0.2$: This is the correct numerical calculation, but it assumes the reader knows $P(F \cup C)$ is 0.8.

(D) $0.6 + 0.4 - 0.2 = 0.8$: This is the calculation for $P(F \cup C)$, not for the complement of that event.

Option (B) provides both the correct formula relating to the complement and the correct calculation using the previously determined value, making it the most complete and accurate answer.

The correct answer is (B) $1 - P(F \cup C) = 1 - 0.8 = 0.2$.

To calculate the probability of drawing a red card from a well-shuffled deck of 52 playing cards, we need to determine the total number of cards and the number of red cards.

Understanding a Deck of Cards:

A standard deck of 52 playing cards has four suits:

• Hearts (♥) - Red
• Diamonds (♦) - Red
• Clubs (♣) - Black
• Spades (♠) - Black

Each suit has 13 cards (Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King).

Determining the Numbers:

Total number of cards in the deck = 52. This is the total number of possible outcomes.

Number of red cards: There are two red suits, Hearts and Diamonds. Each suit has 13 cards.

Number of red cards = (Number of Hearts) + (Number of Diamonds) = 13 + 13 = 26.

This is the number of favorable outcomes for drawing a red card.

Calculating the Probability:

The probability of an event is given by:

Here, the event is "drawing a red card".

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

Simplifying the fraction:

Matching with Options:

(A) 1/52: Probability of drawing a specific card.

(B) 13/52 = 1/4: Probability of drawing a card of a specific suit (e.g., Hearts or Spades).

(C) 26/52 = 1/2: Probability of drawing a red card.

(D) 39/52 = 3/4: Probability of drawing a card that is not black (i.e., red) or a face card from a specific suit, or some other combination.

The calculated probability matches option (C).

The correct answer is (C) 26/52 = 1/2.

To find the probability of getting a sum of 7 when rolling two fair dice, we first need to determine the total number of possible outcomes and then the number of outcomes that result in a sum of 7.

1. Total Number of Possible Outcomes:

When rolling one fair die, there are 6 possible outcomes: {1, 2, 3, 4, 5, 6}.

When rolling two fair dice, the total number of possible outcomes is the product of the number of outcomes for each die:

The sample space consists of 36 ordered pairs, e.g., (1,1), (1,2), ..., (6,6).

2. Number of Favorable Outcomes (Sum of 7):

We need to find the combinations of two dice rolls that add up to 7. Let the outcome of the first die be $d_1$ and the outcome of the second die be $d_2$. We are looking for pairs $(d_1, d_2)$ such that $d_1 + d_2 = 7$.

The possible pairs are:

• (1, 6)
• (2, 5)
• (3, 4)
• (4, 3)
• (5, 2)
• (6, 1)

There are 6 such pairs.

3. Calculating the Probability:

The probability of getting a sum of 7 is:

Simplifying the fraction:

Matching with Options:

(A) 1/36: Probability of a single specific outcome, e.g., (1,1).

(B) 6/36 = 1/6: This matches our calculated probability.

(C) 7/36: This is not directly a standard probability calculation for dice rolls.

(D) 12/36 = 1/3: This might relate to sums like 6 or 8, which have 5 favorable outcomes each.

The correct answer is (B) 6/36 = 1/6.

We are asked to find the probability of the union of two events, $P(A \cup B)$, given the probabilities of the individual events and their intersection.

We are given:

• $P(A) = 0.5$
• $P(B) = 0.3$
• $P(A \cap B) = 0.1$

We use the Addition Law of Probability:

Substituting the given values into the formula:

First, add $P(A)$ and $P(B)$:

Then, subtract $P(A \cap B)$ from the sum:

So, $P(A \cup B) = 0.7$.

Let's examine the options:

(A) $0.5 + 0.3 + 0.1 = 0.9$: This incorrectly adds the intersection instead of subtracting it.

(B) $0.5 + 0.3 - 0.1 = 0.7$: This option correctly states the calculation using the Addition Law of Probability and arrives at the correct answer.

(C) 0.7: This is the correct numerical value of the probability.

(D) 0.8: This is the sum of $P(A)$ and $P(B)$ without subtracting the intersection, which would only be correct if A and B were mutually exclusive.

Option (B) provides the full calculation using the correct formula, which leads to the result shown in option (C). Since option (B) shows the derivation, it is the most complete correct answer.

The correct answer is (B) $0.5 + 0.3 - 0.1 = 0.7$.

The fundamental rule of probability states that the probability of any event must be a value between 0 and 1, inclusive. Mathematically, for any event E, $0 \leq P(E) \leq 1$.

Let's analyze each option based on this rule:

(A) 0.7: This value is between 0 and 1, so it CAN be a probability of an event.

(B) 1.5: This value is greater than 1. Probabilities cannot exceed 1. Therefore, 1.5 CANNOT be a probability of an event.

(C) 0: This value is between 0 and 1, representing an impossible event. So, it CAN be a probability of an event.

(D) 1: This value is between 0 and 1, representing a sure event. So, it CAN be a probability of an event.

The question asks which value CANNOT be a probability of an event.

The only value that violates the fundamental rule of probability is 1.5.

The correct answer is (B) 1.5.

In probability theory, specific terms are used to define different concepts related to random experiments.

Definitions:

• An outcome is a single result of a random experiment.
• An event is a set of one or more outcomes.
• Probability is a measure of the likelihood of an event occurring, expressed as a number between 0 and 1.
• The sample space is the set of all possible outcomes of a random experiment.

The question asks for the term that describes "The set of all possible outcomes of a random experiment".

Based on the definitions, this is precisely the definition of the sample space.

Let's look at the options:

(A) event: An event is a subset of the sample space, not the set of all possible outcomes itself.

(B) probability: Probability is a numerical measure of likelihood, not a set of outcomes.

(C) sample space: This is the definition of the set of all possible outcomes.

(D) outcome: An outcome is a single result, not the set of all possible results.

Therefore, the correct term is sample space.

The correct answer is (C) sample space.

We need to find the probability of the union of two events, A and B, given that they are mutually exclusive.

Definition of Mutually Exclusive Events:

Two events A and B are mutually exclusive if they cannot occur at the same time. This means their intersection is empty, i.e., $A \cap B = \emptyset$.

Consequently, the probability of their intersection is zero: $P(A \cap B) = P(\emptyset) = 0$.

The Addition Law of Probability:

The general formula for the probability of the union of two events is:

Applying the Condition of Mutual Exclusivity:

Since A and B are mutually exclusive, we know that $P(A \cap B) = 0$. Substituting this into the Addition Law:

Matching with Options:

(A) $P(A) + P(B)$: This is the correct formula for the union of mutually exclusive events.

(B) $P(A)P(B)$: This is the probability of the intersection of independent events.

(C) $P(A) + P(B) - P(A \cap B)$: This is the general Addition Law of Probability, which holds true for all events, but when $P(A \cap B)=0$, it simplifies to option (A).

(D) $P(A \cap B)$: This is the probability of the intersection of events.

The question asks what $P(A \cup B)$ is *equal to* given that A and B are mutually exclusive. While option (C) is always true, option (A) is the simplified and specific form for mutually exclusive events, which directly results from the given condition.

The correct answer is (A) $P(A) + P(B)$.

We need to find the probability of drawing a spade OR a King from a standard deck of 52 cards. This involves the union of two events.

Let S be the event of drawing a spade.

Let K be the event of drawing a King.

We want to find $P(S \cup K)$.

Understanding the Deck:

Total number of cards = 52.

Number of spades = 13 (since there are 13 cards in each suit).

Number of Kings = 4 (one King in each suit: King of Spades, King of Hearts, King of Diamonds, King of Clubs).

The events "drawing a spade" and "drawing a King" are not mutually exclusive because there is one card that is both a spade and a King: the King of Spades.

Calculating Probabilities:

Probability of drawing a spade:

Probability of drawing a King:

Probability of drawing the King of Spades (which is both a spade and a King):

Using the Addition Law of Probability:

$P(S \cup K) = P(S) + P(K) - P(S \cap K)$

Combine the numerators:

Simplify the fraction:

Evaluating the Options:

(A) $13/52 + 4/52 = 17/52$: This is incorrect because it assumes the events are mutually exclusive and does not subtract the intersection.

(B) $P(\text{Spade}) + P(\text{King}) - P(\text{King of Spade}) = 13/52 + 4/52 - 1/52 = 16/52 = 4/13$: This option correctly states the Addition Law, shows the calculation, and provides the simplified answer.

(C) $16/52 = 4/13$: This is the correct simplified probability value.

(D) $1/52$: This is the probability of drawing a specific card, like the King of Spades.

Option (B) correctly shows the application of the Addition Law of Probability for non-mutually exclusive events, including the calculation. Option (C) provides the final simplified answer. Since option (B) demonstrates the reasoning and calculation, it is the most complete correct answer.

The correct answer is (B) $P(\text{Spade}) + P(\text{King}) - P(\text{King of Spade}) = 13/52 + 4/52 - 1/52 = 16/52 = 4/13$.

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

Assertion (A): The probability of getting a sum of 1 when rolling two dice is 0.

When rolling two fair dice, the minimum possible outcome for each die is 1. Therefore, the minimum possible sum when rolling two dice is $1 + 1 = 2$.

It is impossible to get a sum of 1 by rolling two dice. An event that cannot occur is called an impossible event, and its probability is 0.

Thus, Assertion (A) is true.

Reason (R): The minimum sum when rolling two dice is $1+1=2$. An event with no possible outcomes has a probability of 0.

The first part of the reason correctly states that the minimum sum is 2.

The second part of the reason states that an event with no possible outcomes has a probability of 0. This is a fundamental principle in probability theory: an impossible event has a probability of 0.

Thus, Reason (R) is true.

Relationship between Assertion (A) and Reason (R):

Assertion (A) states that the probability of getting a sum of 1 is 0. Reason (R) explains why this is the case: the minimum sum possible is 2, meaning the event of getting a sum of 1 has no possible outcomes, and events with no possible outcomes have a probability of 0.

Therefore, Reason (R) correctly explains Assertion (A).

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

To find the probability of getting a doublet when throwing two dice, we need to identify the total number of possible outcomes and the number of outcomes that constitute a doublet.

1. Total Number of Possible Outcomes:

As established when rolling two dice, the total number of possible outcomes is $6 \times 6 = 36$.

2. Number of Favorable Outcomes (Doublets):

A doublet occurs when both dice show the same number. The possible doublets are:

• (1, 1)
• (2, 2)
• (3, 3)
• (4, 4)
• (5, 5)
• (6, 6)

There are 6 such outcomes.

3. Calculating the Probability:

The probability of getting a doublet is the ratio of the number of doublets to the total number of outcomes:

Simplifying the fraction:

Matching with Options:

(A) 6/36 = 1/6: This matches our calculated probability.

(B) 1/36: This is the probability of a specific single outcome, like (1,1).

(C) 12/36 = 1/3: This might be the probability of getting a sum of 5 or 9.

(D) 1/2: This is the probability of getting a head when flipping a coin, or drawing a red card.

The correct answer is (A) 6/36 = 1/6.

To find the probability of drawing a blue ball from the bag, we need to know the total number of balls and the number of blue balls.

1. Total Number of Balls:

The bag contains:

• 5 red balls
• 3 blue balls
• 2 green balls

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

This is the total number of possible outcomes when drawing one ball.

2. Number of Favorable Outcomes (Blue Balls):

The number of blue balls in the bag is 3. This is the number of favorable outcomes for drawing a blue ball.

3. Calculating the Probability:

The probability of drawing a blue ball is given by:

Matching with Options:

(A) 3/10: This matches our calculated probability.

(B) 5/10: This is the probability of drawing a red ball.

(C) 2/10: This is the probability of drawing a green ball.

(D) 1/10: This is the probability of drawing a specific color if there was only one ball of that color.

The correct answer is (A) 3/10.

We need to find the probability of drawing a ball that is not red from the bag. This can be approached in two ways: using the complement rule or by directly counting the non-red balls.

Method 1: Using the Complement Rule

First, identify the total number of balls and the number of red balls.

• Total balls = 5 red + 3 blue + 2 green = 10.
• Number of red balls = 5.

The probability of drawing a red ball is:

The event "not red" is the complement of the event "red". Using the Law of Complement, $P(\text{not red}) = 1 - P(\text{red})$.

Simplifying this gives $\frac{1}{2}$.

Method 2: Directly Counting Non-Red Balls

The balls that are not red are the blue balls and the green balls.

• Number of blue balls = 3.
• Number of green balls = 2.

Number of non-red balls = Number of blue balls + Number of green balls = $3 + 2 = 5$.

The probability of drawing a non-red ball is:

Simplifying this gives $\frac{1}{2}$.

Evaluating the Options:

(A) $1 - P(\text{red}) = 1 - 5/10 = 5/10 = 1/2$: This correctly uses the complement rule and shows the calculation. It's a valid explanation and result.

(B) 5/10 = 1/2: This states the simplified numerical answer correctly.

(C) 5/10: This states the unsimplified numerical answer correctly.

(D) 3/10 + 2/10 = 5/10 = 1/2: This correctly adds the probabilities of drawing a blue ball and a green ball (the non-red balls) and shows the calculation.

All options (A), (B), (C), and (D) correctly arrive at the probability of drawing a non-red ball, which is 5/10 or 1/2. However, option (A) and (D) provide a more complete explanation of how the probability is derived using probability rules, whereas (B) and (C) just state the answer. Option (A) demonstrates the complement rule, and option (D) demonstrates the direct counting of favorable outcomes. Both are valid methods.

Given the structure of the question and options, options that show the reasoning are generally preferred. Option (A) uses the complement rule and option (D) uses direct counting of favorable events. Both are correct. However, if we have to pick the *most* descriptive answer that shows the process, both A and D are strong candidates. Option A directly relates to the complement rule which is a key probability concept.

Let's re-evaluate the question asking "what is the probability". Options B, C, and D provide the numerical answer. Option A provides the calculation for that answer. Typically, a numerical answer is expected. Between B, C, and D, B and D show the simplification. D explicitly shows the sum of the probabilities of the constituent non-red events.

Considering the provided options, Option (A) and (D) show the derivation. Option (B) and (C) show the answer. Since the question asks "what is the probability", a numerical answer is primarily sought. Option (D) shows the sum of probabilities of disjoint events (blue or green) which is a direct way to calculate "not red". Option (A) uses the complement.

Let's assume the question seeks the most direct method of calculating "not red" by summing the probabilities of the component events.

The correct answer is (D) 3/10 + 2/10 = 5/10 = 1/2.

We are given two events, A and B, with their probabilities, and we are told they are mutually exclusive. We need to find the probability of their union, $P(A \cup B)$.

Given Information:

• $P(A) = 0.6$
• $P(B) = 0.4$
• A and B are mutually exclusive events.

Understanding Mutually Exclusive Events:

Mutually exclusive events are events that cannot occur at the same time. This means their intersection is empty, so $A \cap B = \emptyset$. Consequently, the probability of their intersection is zero: $P(A \cap B) = 0$.

Applying the Addition Law of Probability:

The general formula for the probability of the union of two events is:

Since A and B are mutually exclusive, $P(A \cap B) = 0$. Substituting this into the formula:

Calculation:

Now, substitute the given probabilities:

Evaluating the Options:

(A) $0.6 + 0.4 = 1.0$: This option correctly shows the calculation for mutually exclusive events and the result.

(B) 1.0: This is the correct numerical answer.

(C) $0.6 \times 0.4 = 0.24$: This would be the probability of the intersection if the events were independent, not the union of mutually exclusive events.

(D) 0: This is the probability of the intersection of mutually exclusive events.

Option (A) provides the calculation process, leading to the numerical answer. Option (B) provides the numerical answer directly. Since option (A) shows the correct application of the rule for mutually exclusive events, it is the most complete correct answer.

The correct answer is (A) $0.6 + 0.4 = 1.0$.

To find the probability of getting at least one tail when tossing two fair coins, we first need to determine the sample space and then identify the outcomes that satisfy the condition "at least one tail".

1. Sample Space:

When tossing two fair coins, the possible outcomes are:

• HH (Head on first coin, Head on second coin)
• HT (Head on first coin, Tail on second coin)
• TH (Tail on first coin, Head on second coin)
• TT (Tail on first coin, Tail on second coin)

The sample space is $S = \{HH, HT, TH, TT\}$. The total number of possible outcomes is 4.

2. Outcomes with at Least One Tail:

The event "at least one tail" means that we are interested in outcomes that have one tail or two tails.

The outcomes with at least one tail are:

• HT (one tail)
• TH (one tail)
• TT (two tails)

There are 3 such outcomes.

3. Calculating the Probability:

The probability of getting at least one tail is the ratio of the number of favorable outcomes to the total number of outcomes:

Alternative Method (Using Complement):

The complement of "at least one tail" is "no tails", which means getting all heads (HH).

The probability of getting no tails (HH) is $P(HH) = 1/4$.

Using the complement rule, $P(\text{at least one tail}) = 1 - P(\text{no tails}) = 1 - P(HH)$.

Evaluating the Options:

(A) Sample space = $\{HH, HT, TH, TT\}$. At least one tail = $\{HT, TH, TT\}$. Probability = 3/4: This option correctly describes the sample space, the favorable outcomes, and the calculated probability.

(B) 1/4: This is the probability of getting no tails (HH).

(C) 2/4 = 1/2: This might be the probability of getting exactly one tail (HT or TH).

(D) 3/4: This is the correct numerical probability.

Option (A) provides a complete explanation, including the sample space, the event of interest, and the final probability. Option (D) provides just the numerical answer.

The correct answer is (A) Sample space = $\{HH, HT, TH, TT\}$. At least one tail = $\{HT, TH, TT\}$. Probability = 3/4.

The question asks for the sample space for selecting a vowel from the English alphabet. The sample space is the set of all possible outcomes of an experiment.

Understanding the English Alphabet and Vowels:

The English alphabet consists of 26 letters: $\{a, b, c, ..., z\}$.

The vowels in the English alphabet are generally considered to be 'a', 'e', 'i', 'o', and 'u'. The letter 'y' can sometimes function as a vowel, but in the standard definition of vowels for such problems, it is usually excluded unless specified.

Analyzing the Options:

(A) $\{a, e, i, o, u\}$: This set represents the standard vowels in the English alphabet. If the experiment is "selecting a vowel", then this set lists all the possible outcomes.

(B) $\{a, b, c, ..., z\}$: This is the sample space for selecting any letter from the English alphabet, not specifically a vowel.

(C) $\{a, e, i, o, u, \text{sometimes } y\}$: While 'y' can act as a vowel, the term "sample space" usually refers to a definitive set of outcomes. Including "sometimes y" makes it ambiguous and not a standard representation of a sample space unless the experiment's rules explicitly include it.

(D) 5: This is the number of standard vowels, not the set of vowels itself. A sample space is a set of outcomes.

The question asks for the sample space for selecting a vowel. This implies the outcomes are the vowels themselves. Option (A) correctly lists the standard vowels as the set of possible outcomes.

The correct answer is (A) $\{a, e, i, o, u\}$.

We are given the probabilities of two events A and B, and the probability of their union, and we need to find the probability of their intersection.

Given Information:

• $P(A) = 0.7$
• $P(B) = 0.5$
• $P(A \cup B) = 0.8$

We need to find $P(A \cap B)$.

Using the Addition Law of Probability:

The Addition Law of Probability states:

We need to rearrange this formula to solve for $P(A \cap B)$:

Calculation:

Substitute the given values into the rearranged formula:

First, add $P(A)$ and $P(B)$:

Then, subtract $P(A \cup B)$:

So, $P(A \cap B) = 0.4$.

Evaluating the Options:

(A) $P(A) + P(B) - P(A \cup B) = 0.7 + 0.5 - 0.8 = 1.2 - 0.8 = 0.4$: This option correctly shows the formula rearrangement, the substitution of values, and the calculation leading to the correct answer.

(B) 0.4: This is the correct numerical answer.

(C) 1.2: This is the sum of $P(A)$ and $P(B)$ before subtracting $P(A \cup B)$.

(D) 0.2: This is an incorrect calculation.

Option (A) provides the complete derivation of the answer using the correct formula, while option (B) provides just the numerical answer.

The correct answer is (A) $P(A) + P(B) - P(A \cup B) = 0.7 + 0.5 - 0.8 = 1.2 - 0.8 = 0.4$.

To find the probability of drawing a face card from a deck of 52 cards, we need to determine the total number of cards and the number of face cards.

Understanding a Deck of Cards:

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

In each suit, there are 13 cards: Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King.

Identifying Face Cards:

Face cards are typically defined as Kings, Queens, and Jacks.

Number of Kings in the deck = 4 (one for each suit).

Number of Queens in the deck = 4 (one for each suit).

Number of Jacks in the deck = 4 (one for each suit).

Total number of face cards = (Number of Kings) + (Number of Queens) + (Number of Jacks)

So, there are 12 face cards in a deck of 52 cards. These are the favorable outcomes.

Calculating the Probability:

The probability of drawing a face card is the ratio of the number of face cards to the total number of cards:

Simplify the fraction:

Divide both the numerator and the denominator by their greatest common divisor, which is 4:

Matching with Options:

(A) 3/52: This would be the probability of drawing a specific face card from one suit (e.g., King of Hearts).

(B) 12/52 = 3/13: This matches our calculated probability.

(C) 16/52 = 4/13: This number doesn't directly correspond to common card probabilities in this context.

(D) 1/4: This is the probability of drawing a card of a specific suit.

The correct answer is (B) 12/52 = 3/13.

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

Assertion (A): If A and B are mutually exclusive and exhaustive events, then $P(A) + P(B) = 1$.

Mutually exclusive events means they cannot happen at the same time, so $A \cap B = \emptyset$, which implies $P(A \cap B) = 0$.

Exhaustive events means that their union covers the entire sample space, so $A \cup B = S$, which implies $P(A \cup B) = P(S) = 1$.

Using the Addition Law of Probability for any two events:

Substituting the conditions for mutually exclusive and exhaustive events:

Thus, Assertion (A) is true.

Reason (R): Mutually exclusive means $A \cap B = \emptyset$, so $P(A \cup B) = P(A) + P(B)$. Exhaustive means $A \cup B = S$, so $P(A \cup B) = P(S) = 1$. Thus, $P(A) + P(B) = 1$.

The reason correctly breaks down the definitions and their implications for probability:

• "Mutually exclusive means $A \cap B = \emptyset$, so $P(A \cup B) = P(A) + P(B)$." This part is correct because when $P(A \cap B) = 0$, the Addition Law simplifies to $P(A \cup B) = P(A) + P(B)$.
• "Exhaustive means $A \cup B = S$, so $P(A \cup B) = P(S) = 1$." This part is also correct.
• "Thus, $P(A) + P(B) = 1$." This conclusion logically follows from the preceding statements.

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

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

We are given a biased coin where the probability of getting a head is twice the probability of getting a tail. We need to find the probability of getting a tail.

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

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

We are given the condition:

For any coin toss, there are only two possible outcomes: head or tail. These events are mutually exclusive and exhaustive. Therefore, the sum of their probabilities must be 1:

Now we can substitute the first equation into the second equation:

$(2 \times P(T)) + P(T) = 1$

Combine the terms involving $P(T)$:

Solve for $P(T)$:

If needed, we can also find $P(H)$: $P(H) = 2 \times P(T) = 2 \times \frac{1}{3} = \frac{2}{3}$.

Matching with Options:

(A) 1/3: This matches our calculated probability of getting a tail.

(B) 2/3: This is the probability of getting a head.

(C) 1/2: This is the probability for a fair coin.

(D) 1/4: This is an incorrect probability value.

The correct answer is (A) 1/3.

The odds in favor of an event are defined as the ratio of the probability that the event will occur to the probability that the event will not occur.

Let $P(A)$ be the probability of event A occurring.

The probability that event A will not occur (the complement of A) is denoted by $P(A')$, and it is equal to $1 - P(A)$.

The odds in favor of event A are expressed as the ratio of $P(A)$ to $P(A')$:

Since $P(A') = 1 - P(A)$, we can also express the odds in favor of A as:

Evaluating the Options:

(A) $P(A) : P(A')$: This is the definition of odds in favor of A.

(B) $P(A') : P(A)$: This represents the odds *against* A.

(C) $P(A) : 1 - P(A)$: This is an alternative way of expressing the odds in favor of A, using $1 - P(A)$ instead of $P(A')$.

(D) Both (A) and (C): Since both (A) and (C) correctly define the odds in favor of A, this option is the most comprehensive.

Both expressions $P(A) : P(A')$ and $P(A) : 1 - P(A)$ correctly represent the odds in favor of event A.

The correct answer is (D) Both (A) and (C).

We need to identify which of the given statements are always true for any event E in the context of probability theory.

Analyzing each statement:

(A) $0 \leq P(E) \leq 1$: This is one of the fundamental axioms of probability. The probability of any event must always be between 0 and 1, inclusive. This statement is always true.

(B) $P(E) + P(E') = 1$: This is the Law of Complement. $E'$ represents the complement of event E (the event that E does not occur). Since E and E' are mutually exclusive and exhaustive, their probabilities sum to 1. This statement is always true.

(C) $P(\emptyset) = 0$: The empty set ($\emptyset$) represents an impossible event, meaning it has no outcomes. By the axioms of probability, the probability of an impossible event is always 0. This statement is always true.

(D) $P(S) = 1$: The sample space (S) represents the set of all possible outcomes. Since it is certain that one of the outcomes in the sample space will occur, its probability is 1. This is also an axiom of probability. This statement is always true.

All four statements are fundamental properties and axioms of probability theory that hold true for any event E.

The correct answer is (A), (B), (C), and (D).

We are given the condition $P(A \cup B) = P(A) + P(B)$. We need to determine what this implies about the relationship between events A and B.

Recall the Addition Law of Probability:

For any two events A and B, the probability of their union is given by:

Comparing the Given Condition with the Addition Law:

We are given: $P(A \cup B) = P(A) + P(B)$.

For this to be true, the term $-P(A \cap B)$ in the Addition Law must be zero.

So, $P(A) + P(B) - P(A \cap B) = P(A) + P(B)$.

This equality holds if and only if:

Which means:

Interpreting $P(A \cap B) = 0$:

The condition $P(A \cap B) = 0$ means that the probability of both events A and B occurring together is zero. This is the definition of mutually exclusive events.

Evaluating the Options:

(A) A and B are independent: Independence means $P(A \cap B) = P(A)P(B)$. If $P(A)=0.5$ and $P(B)=0.5$, then $P(A)P(B)=0.25$. If $P(A \cap B)=0$, they are not independent unless one of the probabilities is 0.

(B) A and B are mutually exclusive: This is true because $P(A \cap B) = 0$.

(C) $P(A \cap B) \neq 0$: This contradicts our finding that $P(A \cap B) = 0$.

(D) $A=B$: This would imply $P(A \cup B) = P(A)$ (or $P(B)$), which is not generally true from the given information.

The condition $P(A \cup B) = P(A) + P(B)$ directly implies that $P(A \cap B) = 0$, which is the definition of mutually exclusive events.

The correct answer is (B) A and B are mutually exclusive.

When rolling dice, the outcome of each die is independent of the others. To find the total number of elements in the sample space when rolling multiple dice, we multiply the number of outcomes for each die.

Understanding the Outcomes for One Die:

A single fair die has 6 faces, numbered 1 through 6. So, there are 6 possible outcomes when rolling one die.

Calculating for Three Dice:

When rolling three dice simultaneously, each die has 6 possible outcomes. The total number of possible outcomes (elements in the sample space) is the product of the number of outcomes for each die:

Number of outcomes for Die 1 = 6

Number of outcomes for Die 2 = 6

Number of outcomes for Die 3 = 6

Total number of elements in the sample space = (Outcomes for Die 1) $\times$ (Outcomes for Die 2) $\times$ (Outcomes for Die 3)

This can be written as $6^3$:

Evaluating the Options:

(A) $6 \times 3 = 18$: This incorrectly multiplies the number of outcomes for one die by the number of dice, implying that each die only has 6 outcomes in total, or that we are summing the outcomes in a peculiar way.

(B) $6 + 6 + 6 = 18$: This incorrectly adds the number of outcomes for each die, which does not represent the total combinations.

(C) $6^3 = 216$: This correctly calculates the total number of outcomes by multiplying the possibilities for each independent event.

(D) $3^6$: This would be the number of ways to choose 6 outcomes from 3 categories (e.g., assigning one of 3 colors to 6 items), which is not relevant here.

The correct way to calculate the size of the sample space when rolling multiple dice is to raise the number of outcomes for a single die to the power of the number of dice.

The correct answer is (C) $6^3 = 216$.

We need to find the probability that both balls drawn are red when drawing two balls at random without replacement from a bag containing 4 red and 6 black balls.

Understanding the Scenario:

Total number of balls in the bag = 4 red + 6 black = 10 balls.

We are drawing two balls without replacement, meaning that after the first ball is drawn, it is not put back into the bag before the second ball is drawn. This affects the probabilities for the second draw.

Method 1: Using Conditional Probability (Sequential Draws)

The probability that the first ball drawn is red:

After drawing one red ball, there are now 9 balls left in the bag, and 3 of them are red.

The probability that the second ball drawn is red, given that the first ball drawn was red:

The probability that both balls are red is the product of these probabilities:

Let's calculate this value: $\frac{4}{10} \times \frac{3}{9} = \frac{12}{90} = \frac{2}{15}$.

Method 2: Using Combinations

The total number of ways to choose 2 balls from 10 balls (without regard to order) is given by the combination formula $C(n, k) = \frac{n!}{k!(n-k)!}$.

Total number of ways to choose 2 balls from 10 = $C(10, 2)$.

The number of ways to choose 2 red balls from the 4 red balls available:

Number of ways to choose 2 red balls = $C(4, 2)$.

The probability of drawing two red balls is the ratio of the number of ways to choose 2 red balls to the total number of ways to choose 2 balls:

Simplifying this fraction: $\frac{6}{45} = \frac{2}{15}$.

Evaluating the Options:

(A) $\frac{4}{10} \times \frac{4}{10}$: This would be the probability if the balls were drawn *with* replacement.

(B) $\frac{4}{10} \times \frac{3}{9}$: This correctly represents the probability of drawing two red balls without replacement using sequential probabilities.

(C) $\frac{C(4, 2)}{C(10, 2)} = \frac{6}{45}$: This correctly represents the probability of drawing two red balls without replacement using combinations.

(D) Both (B) and (C): Since both methods (sequential probability and combinations) yield the same correct probability of drawing two red balls without replacement, and both (B) and (C) represent these methods accurately, this option is correct.

Both methods are valid and yield the same result ($\frac{12}{90} = \frac{6}{45} = \frac{2}{15}$). Option (B) shows the sequential probability calculation, and option (C) shows the combination calculation.

The correct answer is (D) Both (B) and (C).

The probability of an event is a fundamental concept in probability theory. It quantifies the likelihood of a specific event occurring.

Definition of Probability:

For an event E, where all outcomes in the sample space are equally likely, the probability of event E is defined as:

Analyzing the Options:

(A) number of unfavorable outcomes: This would be the denominator if we were calculating the odds against the event, not the probability.

(B) total number of outcomes in the sample space: This correctly completes the definition of probability when outcomes are equally likely.

(C) probability of the complement event: This relates to the law of complements ($P(E) = 1 - P(E')$), but it's not the ratio that defines $P(E)$.

(D) number of times the experiment is repeated: This is related to empirical probability (frequency interpretation), but the question implies theoretical probability based on equally likely outcomes.

Therefore, the probability of an event E is the ratio of the number of favorable outcomes to the total number of outcomes in the sample space.

The correct answer is (B) total number of outcomes in the sample space.

We are given the probabilities of two events, A and B, and we are told that they are mutually exclusive. We need to find the probability of "A and B" occurring, which refers to the intersection of events A and B, denoted as $P(A \cap B)$.

Understanding Mutually Exclusive Events:

Mutually exclusive events are events that cannot happen at the same time. This means that if one event occurs, the other cannot. Mathematically, this implies that their intersection is the empty set: $A \cap B = \emptyset$.

Probability of the Intersection:

The probability of an empty set is always 0. Therefore, for mutually exclusive events, the probability of their intersection is:

The values of $P(A) = 0.3$ and $P(B) = 0.6$ are provided but are not needed to determine $P(A \cap B)$ if the events are stated to be mutually exclusive.

Evaluating the Options:

(A) 0.18: This would be the result of $P(A) \times P(B)$, which is for independent events.

(B) 0.9: This would be the result of $P(A) + P(B)$ if they were mutually exclusive, representing $P(A \cup B)$.

(C) 0: This is the correct probability for the intersection of mutually exclusive events.

(D) 0.3: This is the probability of event A.

Since A and B are mutually exclusive, they cannot occur at the same time, meaning the probability of both occurring ($P(A \text{ and } B)$ or $P(A \cap B)$) is 0.

The correct answer is (C) 0.

To find the probability of getting a number divisible by 3 when rolling a fair die, we need to identify the total possible outcomes and the outcomes that are divisible by 3.

1. Sample Space:

When a fair die is rolled, the sample space (set of all possible outcomes) is:

The total number of possible outcomes is 6.

2. Favorable Outcomes (Numbers Divisible by 3):

We need to identify the numbers in the sample space that are divisible by 3. A number is divisible by 3 if it can be divided by 3 with no remainder.

Let E be the event of getting a number divisible by 3. The numbers in S that are divisible by 3 are:

• 3 (since $3 \div 3 = 1$)
• 6 (since $6 \div 3 = 2$)

The favorable outcomes for event E are $\{3, 6\}$.

The number of favorable outcomes is 2.

3. Calculating the Probability:

The probability of event E is the ratio of the number of favorable outcomes to the total number of possible outcomes:

Simplifying the fraction:

Matching with Options:

(A) 1/6: Probability of getting a specific number, e.g., 1.

(B) 2/6 = 1/3: This matches our calculated probability.

(C) 3/6 = 1/2: Probability of getting an odd number or an even number.

(D) 1: Probability of a sure event.

The correct answer is (B) 2/6 = 1/3.

To find the probability of choosing a vowel from the letters of the word 'MATHEMATICS', we need to count the total number of letters and the number of vowels.

1. Identify the Letters and Count Total Letters:

The word is 'MATHEMATICS'.

Let's list all the letters: M, A, T, H, E, M, A, T, I, C, S.

Count the total number of letters: There are 11 letters in the word.

Total number of outcomes = 11.

2. Identify the Vowels and Count Vowels:

The vowels in the English alphabet are A, E, I, O, U.

Let's find the vowels in the word 'MATHEMATICS':

• M - Not a vowel
• A - Vowel
• T - Not a vowel
• H - Not a vowel
• E - Vowel
• M - Not a vowel
• A - Vowel
• T - Not a vowel
• I - Vowel
• C - Not a vowel
• S - Not a vowel

The vowels in the word are: A, E, A, I.

Count the number of vowels: There are 4 vowels.

Number of favorable outcomes = 4.

3. Calculate the Probability:

The probability of choosing a vowel is the ratio of the number of vowels to the total number of letters:

Evaluating the Options:

(A) There are 11 letters: M, A, T, H, E, M, A, T, I, C, S. Vowels are A, E, A, I. Total vowels = 4. Total letters = 11. Probability = 4/11: This option correctly identifies the letters, counts the total letters, identifies and counts the vowels, and states the correct probability.

(B) 4/11: This is the correct numerical probability.

(C) 5/11: This might be incorrect if the count of vowels or letters is wrong.

(D) 6/11: This might be the probability of choosing a consonant.

Option (A) provides the complete reasoning and the final answer, while option (B) provides only the numerical answer. Since option (A) gives a full explanation, it is the most descriptive correct answer.

The correct answer is (A) There are 11 letters: M, A, T, H, E, M, A, T, I, C, S. Vowels are A, E, A, I. Total vowels = 4. Total letters = 11. Probability = 4/11.

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

Assertion (A): The sum of probabilities of all simple events in a sample space is always 1.

A sample space $S$ is the set of all possible outcomes of a random experiment. Each outcome in the sample space is considered a simple event (or elementary event). By the axioms of probability, the probability of the entire sample space is 1. Since the sample space is the union of all disjoint simple events, the sum of the probabilities of all simple events that make up the sample space must equal the probability of the sample space itself, which is 1.

Thus, Assertion (A) is true.

Reason (R): This is one of the basic axioms of probability.

The statement that the sum of the probabilities of all simple events in a sample space is always 1 is a direct consequence of the axioms of probability, specifically the axiom that states $P(S) = 1$, where $S$ is the sample space. The set of all simple events constitutes the sample space, and their probabilities must sum to the probability of the sample space.

Thus, Reason (R) is true.

Relationship between Assertion (A) and Reason (R):

Assertion (A) states a fundamental property of probability related to simple events. Reason (R) correctly identifies that this property stems directly from the basic axioms of probability. The reason explains why the assertion is true by referencing the foundational rules of probability theory.

Therefore, Reason (R) is the correct explanation for Assertion (A).

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

We are given that A and B are mutually exclusive and exhaustive events, and there is a relationship between their probabilities. We need to find the probability of event A.

Understanding the Conditions:

1. Mutually Exclusive Events: This means that events A and B cannot occur at the same time. Therefore, their intersection is empty, $A \cap B = \emptyset$, and $P(A \cap B) = 0$.

2. Exhaustive Events: This means that the union of events A and B covers the entire sample space. Therefore, $A \cup B = S$, and $P(A \cup B) = P(S) = 1$.

3. Relationship between probabilities: $P(A) = 2P(B)$.

Applying the Addition Law of Probability:

The Addition Law of Probability states: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.

Since A and B are mutually exclusive and exhaustive:

$1 = P(A) + P(B) - 0$

Solving for P(A):

We have two equations:

1. $P(A) + P(B) = 1$
2. $P(A) = 2P(B)$

Substitute the second equation into the first equation:

$(2P(B)) + P(B) = 1$

Combine the terms:

Solve for $P(B)$:

Now, use the relationship $P(A) = 2P(B)$ to find $P(A)$:

Evaluating the Options:

(A) $P(A) + P(B) = 1 \implies 2P(B) + P(B) = 1 \implies 3P(B) = 1 \implies P(B) = 1/3$. $P(A) = 2/3$.: This option correctly shows the entire step-by-step derivation and arrives at the correct value for $P(A)$.

(B) 1/3: This is the value of $P(B)$, not $P(A)$.

(C) 1/2: This is an incorrect value.

(D) 2/3: This is the correct numerical value for $P(A)$.

Option (A) provides the complete derivation and the answer, making it the most comprehensive correct choice.

The correct answer is (A) $P(A) + P(B) = 1 \implies 2P(B) + P(B) = 1 \implies 3P(B) = 1 \implies P(B) = 1/3$. $P(A) = 2/3$.

To find the probability of getting a number less than 1 when rolling a fair die, we first need to identify the sample space and then the outcomes that satisfy the condition.

1. Sample Space:

When a fair die is rolled, the sample space (set of all possible outcomes) is:

The total number of possible outcomes is 6.

2. Favorable Outcomes (Numbers less than 1):

We need to identify the numbers in the sample space that are strictly less than 1.

Looking at the sample space $\{1, 2, 3, 4, 5, 6\}$, there are no numbers that are less than 1.

The set of favorable outcomes is the empty set, $\emptyset$.

The number of favorable outcomes is 0.

3. Calculating the Probability:

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

Calculating the value:

Interpreting the Result:

An event with a probability of 0 is called an impossible event, which makes sense because it's impossible to roll a number less than 1 on a standard die.

Matching with Options:

(A) 0: This matches our calculated probability for an impossible event.

(B) 1/6: This is the probability of getting any single specific outcome (e.g., rolling a 1).

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

(D) Undefined: Probabilities are always defined values between 0 and 1.

The correct answer is (A) 0.

We need to find the probability of drawing a card that is either red or black from a standard deck of 52 cards.

Understanding the Deck of Cards:

A standard deck of 52 cards consists of two colors: red and black.

Number of red cards = 26 (Hearts and Diamonds).

Number of black cards = 26 (Clubs and Spades).

Total number of cards = 52.

Defining the Events:

Let R be the event of drawing a red card.

Let B be the event of drawing a black card.

We want to find the probability of R or B, which is $P(R \cup B)$.

Analyzing the Events:

The events R (drawing a red card) and B (drawing a black card) are:

• Mutually Exclusive: A card cannot be both red and black at the same time. So, $R \cap B = \emptyset$, which means $P(R \cap B) = 0$.
• Exhaustive: Every card in the deck is either red or black. So, $R \cup B = S$ (the entire sample space), which means $P(R \cup B) = P(S) = 1$.

Calculating the Probability:

Using the Addition Law of Probability:

We know:

• $P(R) = \frac{\text{Number of red cards}}{\text{Total cards}} = \frac{26}{52}$
• $P(B) = \frac{\text{Number of black cards}}{\text{Total cards}} = \frac{26}{52}$
• $P(R \cap B) = 0$ (since they are mutually exclusive)

Substituting these values:

Interpreting the Result:

The probability of drawing a red card or a black card is 1. This means it is a sure event, which makes sense because every card in the deck is either red or black.

Evaluating the Options:

(A) $26/52 = 1/2$: This is the probability of drawing a red card OR the probability of drawing a black card, but not the probability of drawing a red card OR a black card.

(B) $26/52 + 26/52 = 1$: This option correctly shows the addition of probabilities for mutually exclusive events and the resulting probability.

(C) 1: This is the correct numerical probability.

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

Option (B) provides the calculation showing why the probability is 1, based on the sum of the probabilities of the two mutually exclusive and exhaustive events. Option (C) simply states the final answer.

The correct answer is (B) $26/52 + 26/52 = 1$.

We are given the probabilities of two events, A and B, and we are told that these events are mutually exclusive. We need to find the probability of their intersection, $P(A \cap B)$.

Understanding Mutually Exclusive Events:

Mutually exclusive events are events that cannot occur at the same time. This means that if one event happens, the other cannot. In terms of set theory, their intersection is empty: $A \cap B = \emptyset$.

Probability of the Intersection for Mutually Exclusive Events:

The probability of an empty set (an event that cannot occur) is always 0.

Therefore, for any two mutually exclusive events A and B:

The specific values of $P(A) = 0.4$ and $P(B) = 0.5$ are not needed to determine $P(A \cap B)$ once we know the events are mutually exclusive.

Evaluating the Options:

(A) 0.2: This might be $P(A) \times P(B)$, which is for independent events.

(B) 0.9: This might be $P(A) + P(B)$ for mutually exclusive events, representing $P(A \cup B)$.

(C) 0: This is the correct probability for the intersection of mutually exclusive events.

(D) Cannot be determined: This would be true if we didn't know if the events were mutually exclusive or independent, but we are given that they are mutually exclusive.

Since A and B are mutually exclusive, they cannot happen at the same time. Therefore, the probability of both A and B occurring ($P(A \cap B)$) is 0.

The correct answer is (C) 0.

We need to find the probability that a randomly selected student passed in Mathematics or Physics. This involves the union of two events.

Let M be the event that a student passed in Mathematics.

Let P be the event that a student passed in Physics.

We are given:

• $P(M) = 30\% = 0.3$ (Probability of passing Mathematics)
• $P(P) = 20\% = 0.2$ (Probability of passing Physics)
• $P(M \cap P) = 10\% = 0.1$ (Probability of passing both Mathematics and Physics)

We want to find $P(M \cup P)$, the probability that a student passed in Mathematics OR Physics.

Using the Addition Law of Probability:

The Addition Law of Probability states:

Substitute the given values:

Perform the addition and subtraction:

Evaluating the Options:

(A) $0.3 + 0.2 - 0.1 = 0.4$: This option correctly shows the formula, substitution, calculation, and the final answer.

(B) 0.4: This is the correct numerical answer.

(C) 0.5: This would be $P(M) + P(P)$ without subtracting the intersection.

(D) 0.6: This is $P(M) + P(P)$ if they were mutually exclusive and the probabilities added up to less than 1, or potentially an incorrect calculation.

Option (A) provides the derivation and the result, making it the most complete correct answer.

The correct answer is (A) $0.3 + 0.2 - 0.1 = 0.4$.

The experiment consists of tossing a coin repeatedly until a head appears. This means the experiment stops as soon as a head is observed.

Understanding the Experiment:

The possible outcomes are sequences of coin tosses, where the sequence ends with the first head.

• The shortest possible sequence is getting a head on the first toss: H.
• If the first toss is a tail (T), then we toss again. If the second toss is a head (H), the sequence is TH.
• If the first two tosses are tails (TT), then we toss again. If the third toss is a head (H), the sequence is TTH.
• This pattern continues: if the first k tosses are tails, and the (k+1)th toss is a head, the sequence is $T^kH$.

The sample space is the set of all such possible sequences.

Constructing the Sample Space:

The possible outcomes are:

• H (Head on the 1st toss)
• TH (Tail on 1st, Head on 2nd)
• TTH (Tail on 1st, Tail on 2nd, Head on 3rd)
• TTTH (Tail on 1st, Tail on 2nd, Tail on 3rd, Head on 4th)
• And so on, indefinitely...

This can be represented as $\{H, TH, TTH, TTTH, \dots\}$, where the ellipsis indicates that the pattern continues infinitely.

Evaluating the Options:

(A) $\{H, TH, TTH, TTTH, ...\}$: This correctly lists the possible sequences ending with the first head and indicates the continuation of the pattern.

(B) $\{H, T\}$: This is the sample space for a single coin toss.

(C) $\{H, T\}$: This is the same as option (B).

(D) $\{H, TH, TTH, ...\}$: This option is very similar to (A) but lacks the explicit "TTTH" term which helps illustrate the pattern. However, the ellipsis implies continuation, making it a valid representation of the infinite sample space.

Both (A) and (D) represent the correct sample space. Option (A) is slightly more explicit in showing the pattern of tails before the head. However, given standard mathematical notation, both are acceptable representations of an infinite sample space where each outcome is a sequence of tails followed by a head.

Typically, when a pattern is clear, the ellipsis is sufficient. Let's look closely at the options. Option (A) lists the first few terms and then the ellipsis, clearly defining the structure. Option (D) omits one of the intermediate terms, but the ellipsis still implies the continuation. In many contexts, (D) would be considered correct. However, (A) is a more complete depiction of the initial terms of the sequence.

Assuming the question intends the most descriptive representation of the sample space:

The correct answer is (A) $\{H, TH, TTH, TTTH, ...\}$.

The event 'not A' refers to the complement of event A. The complement of an event A, denoted in various ways, includes all outcomes in the sample space that are not in event A.

Understanding Notations for Complement:

• $A'$ or $\bar{A}$: These are the most common and standard notations used to represent the complement of an event A.
• $S \setminus A$: This notation represents the set difference between the sample space S and the event A. It means all elements in S that are not in A, which is precisely the definition of the complement of A.

Analyzing the Options:

(A) $A \cup A$: The union of an event with itself is simply the event itself ($A \cup A = A$). This is not the complement.

(B) $A \cap A$: The intersection of an event with itself is also the event itself ($A \cap A = A$). This is not the complement.

(C) $A'$ or $\bar{A}$: These are standard notations for the complement of event A.

(D) $S \setminus A$: This is a valid notation for the complement of event A, representing the set difference.

The question asks how the event 'not A' is denoted. Options (C) and (D) both provide valid and commonly used notations for the complement of an event.

Since the question asks "which of the following denotes", and both (C) and (D) are correct, we should consider if there's a preference or if the question implies multiple correct answers.

However, typically in multiple-choice questions, if there are multiple technically correct answers, one might be considered more direct or common. $A'$ and $\bar{A}$ are the most universally recognized symbols for the complement.

Let's re-read the question. It says "the event 'not A' is denoted by:". This implies one or more correct notations.

Both (C) and (D) are correct. If this were a "select all that apply" question, both would be chosen. In a single-choice format, there might be an intended answer. However, based on standard mathematical notation, both are valid. If we have to choose the most direct symbolic representation, $A'$ or $\bar{A}$ are the primary symbols.

Given the options, and the commonality of notations:

The correct answer is (C) $A'$ or $\bar{A}$, as these are the most direct symbolic representations of the complement. While (D) is also correct in meaning, (C) uses dedicated symbols.

The statement describes a fundamental rule used in probability theory for calculating the probability of the union of multiple mutually exclusive events.

Axioms of Probability:

Kolmogorov's axioms of probability are:

1. Axiom 1: Non-negativity. For any event E, $P(E) \geq 0$.
2. Axiom 2: Normalization. The probability of the sample space S is 1, $P(S) = 1$.
3. Axiom 3: Additivity. If $E_1, E_2, \dots$ are mutually exclusive events, then $P(E_1 \cup E_2 \cup \dots) = P(E_1) + P(E_2) + \dots$.

The given statement, $P(E_1 \cup E_2 \cup ... \cup E_n) = \sum\limits_{i=1}^n P(E_i)$ for mutually exclusive events $E_1, E_2, ..., E_n$, is precisely the statement of the third axiom of probability, often referred to as the axiom of additivity or the addition rule for mutually exclusive events.

Evaluating the Options:

(A) first: Refers to the non-negativity axiom ($P(E) \geq 0$).

(B) second: Refers to the normalization axiom ($P(S) = 1$).

(C) third: Refers to the additivity axiom for mutually exclusive events.

(D) addition: While the axiom is about addition, the question asks for the *axiom number* or its common name. "Addition axiom" is a descriptive name, but the numbering refers to the sequence of axioms.

The statement provided is the third axiom of probability, specifically concerning the addition of probabilities for mutually exclusive events.

The correct answer is (C) third.

To find the probability of drawing a King of Hearts from a deck of 52 cards, we need to identify the total number of possible outcomes and the number of favorable outcomes (drawing the King of Hearts).

1. Total Number of Possible Outcomes:

A standard deck of playing cards has 52 cards. Therefore, the total number of possible outcomes when drawing one card is 52.

2. Number of Favorable Outcomes:

The favorable outcome is drawing the specific card "King of Hearts". In a standard deck, there is only one King of Hearts.

Number of favorable outcomes = 1.

3. Calculating the Probability:

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

Evaluating the Options:

(A) 1/52: This matches our calculated probability.

(B) 4/52: This is the probability of drawing any King (King of Hearts, King of Diamonds, King of Clubs, King of Spades).

(C) 13/52: This is the probability of drawing a card from a specific suit (e.g., Hearts or Spades).

(D) 26/52: This is the probability of drawing a red card.

The correct answer is (A) 1/52.

We need to find the probability that neither A nor B occurs. This is the complement of the event that A or B occurs (or both).

Given Information:

• $P(A) = 0.6$
• $P(B) = 0.3$
• A and B are mutually exclusive events.

We need to find the probability that neither A nor B occurs, which is $P(\text{neither A nor B})$.

Understanding "Neither A nor B":

The event "neither A nor B" means that A does not occur AND B does not occur. This can be written using set notation as $A' \cap B'$.

By De Morgan's Laws, $(A \cup B)' = A' \cap B'$. Therefore, the event "neither A nor B" is the complement of the event "A or B" ($A \cup B$).

So, $P(\text{neither A nor B}) = P((A \cup B)')$.

Calculating $P(A \cup B)$ for Mutually Exclusive Events:

Since A and B are mutually exclusive, $P(A \cap B) = 0$.

Using the Addition Law of Probability:

Substituting the values:

Calculating $P((A \cup B)')$:

Using the Law of Complement:

Substituting the value of $P(A \cup B)$:

Evaluating the Options:

(A) $P((A \cup B)') = 1 - P(A \cup B) = 1 - (P(A) + P(B))$ since mutually exclusive: This option correctly sets up the problem using the complement of the union and correctly substitutes the formula for mutually exclusive events.

(B) $1 - (0.6 + 0.3) = 1 - 0.9 = 0.1$: This option shows the calculation using the values provided.

(C) 0.1: This is the correct numerical answer.

(D) 0.9: This is the probability of $P(A \cup B)$, not its complement.

Option (A) explains the approach, and Option (B) executes the calculation. Option (C) provides the final numerical answer. Option (A) is the most comprehensive as it shows the reasoning and the specific calculation setup.

The correct answer is (A) $P((A \cup B)') = 1 - P(A \cup B) = 1 - (P(A) + P(B))$ since mutually exclusive.

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

Assertion (A): If A is an event, then $P(A') = 1 - P(A)$.

The notation $A'$ represents the complement of event A, meaning all outcomes in the sample space that are not in A. The relationship $P(A') = 1 - P(A)$ is a fundamental rule in probability theory, known as the Law of Complement. It arises directly from the axioms of probability.

Thus, Assertion (A) is true.

Reason (R): A and A' are mutually exclusive and exhaustive events, so $P(A \cup A') = P(A) + P(A') = P(S) = 1$.

Let's break down the components of Reason (R):

• "A and A' are mutually exclusive": This is true, as an event and its complement cannot occur at the same time. Their intersection is the empty set: $A \cap A' = \emptyset$.
• "A and A' are exhaustive events": This is also true, as their union covers the entire sample space: $A \cup A' = S$.
• "so $P(A \cup A') = P(A) + P(A')$": This uses the Addition Law for mutually exclusive events, which is correct.
• "$P(A \cup A') = P(S) = 1$": This correctly states that the union of an event and its complement is the sample space, and the probability of the sample space is 1.

Combining these points, the reason correctly demonstrates how the relationship $P(A) + P(A') = 1$ is derived from the axioms of probability, which then leads to $P(A') = 1 - P(A)$.

Thus, Reason (R) is true.

Relationship between Assertion (A) and Reason (R):

Assertion (A) states the Law of Complement: $P(A') = 1 - P(A)$. Reason (R) provides the justification for this law by explaining that A and A' are mutually exclusive and exhaustive, and their probabilities sum to 1, which is a direct consequence of the axioms of probability.

Therefore, Reason (R) correctly explains Assertion (A).

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

When two independent experiments are performed simultaneously, the total number of elements in the sample space is the product of the number of outcomes for each individual experiment.

Understanding the Individual Experiments:

1. Tossing a Coin: A coin toss has two possible outcomes: Heads (H) or Tails (T). So, there are 2 outcomes.

2. Rolling a Die: A fair die has six faces, numbered 1 through 6. So, there are 6 possible outcomes.

Calculating the Total Number of Outcomes:

Since the coin toss and the die roll are independent events, the total number of elements in the sample space is the product of the number of outcomes for each event.

Total number of outcomes = (Number of outcomes for coin toss) $\times$ (Number of outcomes for die roll)

For example, the sample space would be {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)}.

Evaluating the Options:

(A) $2+6=8$: This incorrectly adds the number of outcomes.

(B) $2 \times 6 = 12$: This correctly multiplies the number of outcomes and shows the calculation.

(C) 12: This is the correct numerical answer.

(D) 8: This is the result of adding the outcomes.

Option (B) correctly shows the calculation for finding the size of the sample space by multiplying the number of outcomes from each independent event. Option (C) provides the result of that calculation.

The correct answer is (B) $2 \times 6 = 12$.

We are asked to find an expression for $P(A \cap B)$, the probability of the intersection of two events A and B.

Recall the Addition Law of Probability:

The Addition Law of Probability states that for any two events A and B:

Rearranging the Formula:

We can rearrange this formula to solve for $P(A \cap B)$:

Add $P(A \cap B)$ to both sides:

Subtract $P(A \cup B)$ from both sides:

Evaluating the Options:

(A) $P(A) + P(B) - P(A \cup B)$: This expression is exactly what we derived from the Addition Law of Probability.

(B) $P(A \cup B) - P(A) - P(B)$: This is equivalent to $-(P(A) + P(B) - P(A \cup B))$, which is $-P(A \cap B)$.

(C) $P(A)P(B)$: This is the formula for the intersection of independent events.

(D) $P(A) + P(B)$: This is the formula for the union of mutually exclusive events.

The expression for $P(A \cap B)$ derived from the Addition Law of Probability is $P(A) + P(B) - P(A \cup B)$.

The correct answer is (A) $P(A) + P(B) - P(A \cup B)$.

We need to find the probability of getting an even number OR a prime number when rolling a fair die. This requires using the Addition Law of Probability.

1. Sample Space:

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

2. Defining Events:

Let E be the event of getting an even number.

The even numbers in S are $\{2, 4, 6\}$. So, $E = \{2, 4, 6\}$.

Let P be the event of getting a prime number.

The prime numbers in S are $\{2, 3, 5\}$. (Note: 1 is not considered a prime number). So, $P = \{2, 3, 5\}$.

3. Probabilities of Individual Events:

Probability of getting an even number:

Probability of getting a prime number:

4. Finding the Intersection:

The intersection of E and P ($E \cap P$) is the set of outcomes that are both even AND prime. The only number in the sample space that is both even and prime is 2.

So, $E \cap P = \{2\}$.

Probability of getting a number that is both even and prime:

5. Calculating the Probability of the Union:

We use the Addition Law of Probability: $P(E \cup P) = P(E) + P(P) - P(E \cap P)$.

Substituting the probabilities:

Perform the calculation:

Evaluating the Options:

(A) Event E (even) = $\{2, 4, 6\}$, Event P (prime) = $\{2, 3, 5\}$. $E \cap P = \{2\}$.: This option correctly identifies the events and their intersection.

(B) $P(E) = 3/6, P(P) = 3/6, P(E \cap P) = 1/6$. $P(E \cup P) = P(E) + P(P) - P(E \cap P) = 3/6 + 3/6 - 1/6 = 5/6$.: This option correctly states the probabilities and the calculation using the Addition Law, arriving at the correct answer.

(C) 5/6: This is the correct numerical probability.

(D) 6/6 = 1: This would be the probability of a sure event.

Option (B) provides the complete reasoning and calculation leading to the correct probability, while option (C) provides just the numerical answer.

The correct answer is (B) $P(E) = 3/6, P(P) = 3/6, P(E \cap P) = 1/6$. $P(E \cup P) = P(E) + P(P) - P(E \cap P) = 3/6 + 3/6 - 1/6 = 5/6$.

We need to find the probability that when two balls are drawn at random without replacement from a bag containing 2 red and 3 blue balls, one ball is red and the other is blue.

Understanding the Scenario:

Total number of balls in the bag = 2 red + 3 blue = 5 balls.

We are drawing two balls without replacement.

Method 1: Considering Order (Sequential Draws)

There are two possible scenarios for drawing one red and one blue ball:

1. Scenario 1: First ball is Red, Second ball is Blue.

   Probability of drawing a red ball first: $P(\text{1st Red}) = \frac{\text{Number of red balls}}{\text{Total balls}} = \frac{2}{5}$

   After drawing one red ball, there are 4 balls left: 1 red and 3 blue.

   Probability of drawing a blue ball second, given the first was red: $P(\text{2nd Blue} | \text{1st Red}) = \frac{\text{Number of blue balls}}{\text{Remaining balls}} = \frac{3}{4}$

   Probability of Scenario 1 = $P(\text{1st Red}) \times P(\text{2nd Blue} | \text{1st Red}) = \frac{2}{5} \times \frac{3}{4} = \frac{6}{20}$

2. Scenario 2: First ball is Blue, Second ball is Red.

   Probability of drawing a blue ball first: $P(\text{1st Blue}) = \frac{\text{Number of blue balls}}{\text{Total balls}} = \frac{3}{5}$

   After drawing one blue ball, there are 4 balls left: 2 red and 2 blue.

   Probability of drawing a red ball second, given the first was blue: $P(\text{2nd Red} | \text{1st Blue}) = \frac{\text{Number of red balls}}{\text{Remaining balls}} = \frac{2}{4}$

   Probability of Scenario 2 = $P(\text{1st Blue}) \times P(\text{2nd Red} | \text{1st Blue}) = \frac{3}{5} \times \frac{2}{4} = \frac{6}{20}$

Since these two scenarios are mutually exclusive, the total probability of getting one red and one blue is the sum of their probabilities:

Total Probability = Probability of Scenario 1 + Probability of Scenario 2

Simplifying the fraction:

Method 2: Using Combinations

The total number of ways to choose 2 balls from the 5 balls in the bag is $C(5, 2)$.

The number of ways to choose 1 red ball from the 2 red balls is $C(2, 1)$.

The number of ways to choose 1 blue ball from the 3 blue balls is $C(3, 1)$.

The number of ways to choose 1 red ball AND 1 blue ball is the product of these two combinations:

Number of ways to get one red and one blue = $C(2, 1) \times C(3, 1) = 2 \times 3 = 6$.

The probability is the ratio of favorable outcomes to the total number of outcomes:

Simplifying the fraction:

Evaluating the Options:

(A) $\frac{2}{5} \times \frac{3}{4}$ (Order matters, one specific order) + $\frac{3}{5} \times \frac{2}{4}$ (Other order) = $6/20 + 6/20 = 12/20 = 3/5$: This option correctly applies the sequential probability method, considering both orders and summing them up.

(B) $\frac{C(2, 1) \times C(3, 1)}{C(5, 2)} = \frac{2 \times 3}{10} = 6/10 = 3/5$: This option correctly applies the combinations method and shows the calculation.

(C) 3/5: This is the correct numerical answer.

(D) 6/25: This is incorrect. It might be $\frac{2}{5} \times \frac{3}{5}$ which implies replacement.

Both options (A) and (B) provide a complete and correct derivation of the answer, using different valid methods. Option (C) is the numerical answer. Since the question asks "What is the probability", and options (A) and (B) show the methods to arrive at that probability, they are more comprehensive.

The correct answer is (A) $\frac{2}{5} \times \frac{3}{4}$ (Order matters, one specific order) + $\frac{3}{5} \times \frac{2}{4}$ (Other order) = $6/20 + 6/20 = 12/20 = 3/5$.

The sample space of a random experiment is the set of all possible outcomes. In this case, the experiment is drawing a card from a deck of 52 cards.

Analyzing the Options:

(A) $\{ \text{Hearts, Diamonds, Clubs, Spades} \}$: This set represents the four suits of the cards, not the individual cards themselves. It's a description of categories, not the sample space of drawing a single card.

(B) $\{ \text{King, Queen, Jack, ...} \}$: This set lists some of the ranks of the cards, but it's incomplete and doesn't represent the entire set of all possible outcomes of drawing a single card.

(C) The set of all 52 cards: This option describes the complete collection of every possible card that can be drawn from the deck. Each individual card (e.g., Ace of Spades, 7 of Hearts, King of Diamonds) is a distinct outcome.

(D) $\{ \text{Red, Black} \}$: This set represents the two colors of cards, not the individual cards. It describes categories of outcomes, not the sample space of drawing a single card.

The sample space for drawing a card from a deck of 52 cards is the collection of all 52 unique cards, as each card represents a distinct possible outcome of the experiment.

The correct answer is (C) The set of all 52 cards.

The formula $P(A \cup B) = P(A) + P(B)$ is used when events A and B are mutually exclusive.

Understanding the Laws of Probability:

1. General Addition Law: For any two events A and B, $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. This law accounts for the possibility of an overlap (intersection) between the events.

2. Special Addition Law: This is a specific case of the General Addition Law that applies when events are mutually exclusive. If A and B are mutually exclusive, then $P(A \cap B) = 0$. Substituting this into the General Addition Law gives $P(A \cup B) = P(A) + P(B) - 0$, which simplifies to $P(A \cup B) = P(A) + P(B)$.

3. Multiplication Law: This law is used for finding the probability of the intersection of events. For independent events, $P(A \cap B) = P(A)P(B)$. For dependent events, $P(A \cap B) = P(A)P(B|A)$ or $P(B)P(A|B)$.

4. Complement Law: This law relates the probability of an event to the probability of its complement: $P(A') = 1 - P(A)$.

The formula $P(A \cup B) = P(A) + P(B)$ is specifically used when events are mutually exclusive, differentiating it from the general addition law which includes the subtraction of the intersection.

Therefore, the law that states $P(A \cup B) = P(A) + P(B)$ for mutually exclusive events is known as the Special Addition Law.

The correct answer is (B) Special Addition Law.

In probability theory, an impossible event is an event that cannot occur. This means that there are no outcomes in the sample space that correspond to this event.

Axioms of Probability:

One of the fundamental axioms of probability is that the probability of any event E must be between 0 and 1, inclusive ($0 \leq P(E) \leq 1$). Specifically, the probability of an impossible event (represented by the empty set, $\emptyset$) is always 0.

Interpreting the Options:

(A) 1: This is the probability of a sure event (an event that is certain to occur).

(B) 0.5: This represents an event that has a 50% chance of occurring, like getting a head when flipping a fair coin.

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

(D) Undefined: Probabilities are always defined values within the range [0, 1].

Therefore, the probability of an impossible event is 0.

The correct answer is (C) 0.

We need to find the number of students who play neither Cricket nor Football. This involves using the Principle of Inclusion-Exclusion for sets.

Given Information:

• Total number of students in the class = 50.
• Number of students who play Cricket, $n(C) = 30$.
• Number of students who play Football, $n(F) = 25$.
• Number of students who play both Cricket and Football, $n(C \cap F) = 10$.

Finding the Number of Students Who Play at Least One Sport:

The number of students who play Cricket OR Football (or both) is given by the formula:

Substitute the given values:

Calculate the result:

So, 45 students play at least one of the two sports.

Finding the Number of Students Who Play Neither Sport:

The number of students who play neither Cricket nor Football is the total number of students minus the number of students who play at least one sport.

Number of students who play neither = Total students - $n(C \cup F)$

Matching with Options:

(A) 5: This matches our calculated number of students who play neither sport.

(B) 10: This is the number of students who play both sports.

(C) 15: This is the number of students who play only Football ($n(F) - n(C \cap F) = 25 - 10 = 15$).

(D) 20: This is the number of students who play only Cricket ($n(C) - n(C \cap F) = 30 - 10 = 20$).

The correct answer is (A) 5.

The experiment involves two independent actions: rolling a die and tossing a coin. The sample space is the set of all possible combined outcomes.

Understanding the Outcomes:

When rolling a die, the possible outcomes are the numbers {1, 2, 3, 4, 5, 6}.

When tossing a coin, the possible outcomes are {H, T}.

Constructing the Sample Space:

The sample space consists of all possible pairs where the first element is the outcome of the die roll and the second element is the outcome of the coin toss.

We can list these pairs:

• If the die shows 1, the coin can be H or T: (1, H), (1, T)
• If the die shows 2, the coin can be H or T: (2, H), (2, T)
• If the die shows 3, the coin can be H or T: (3, H), (3, T)
• If the die shows 4, the coin can be H or T: (4, H), (4, T)
• If the die shows 5, the coin can be H or T: (5, H), (5, T)
• If the die shows 6, the coin can be H or T: (6, H), (6, T)

The complete sample space is the set of all these pairs:

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

The total number of outcomes is $6 \times 2 = 12$.

Evaluating the Options:

(A) $\{1H, 2H, 3H, 4H, 5H, 6H\}$: This set only includes outcomes where the coin toss is Heads. It's incomplete.

(B) $\{1T, 2T, 3T, 4T, 5T, 6T\}$: This set only includes outcomes where the coin toss is Tails. It's incomplete.

(C) $\{1H, 2H, 3H, 4H, 5H, 6H, 1T, 2T, 3T, 4T, 5T, 6T\}$: This set correctly lists all possible combinations of a die roll (1-6) and a coin toss (H or T). Each element represents one outcome.

(D) $\{1, 2, 3, 4, 5, 6, H, T\}$: This set lists the outcomes of the die roll and the coin toss separately, not as combined outcomes.

The sample space should represent all possible joint outcomes. Option (C) correctly lists these combinations.

The correct answer is (C) $\{1H, 2H, 3H, 4H, 5H, 6H, 1T, 2T, 3T, 4T, 5T, 6T\}$.

In probability theory, the probability of an event is a measure of the likelihood that the event will occur. Probabilities are assigned values on a scale from 0 to 1.

Interpreting Probability Values:

• Probability of 0: An event with a probability of 0 is considered an impossible event. This means that the event cannot occur under any circumstances within the given sample space.
• Probability of 1: An event with a probability of 1 is considered a sure event. This means that the event is certain to occur.
• Probability between 0 and 1 (exclusive): An event with a probability strictly between 0 and 1 is a possible event that is neither impossible nor sure.

A simple event is an event with only one outcome. A compound event is an event with two or more outcomes.

Applying the Interpretation to the Question:

The question states that the probability of event E occurring is 0 ($P(E) = 0$). Based on the interpretation of probability values:

• If $P(E) = 0$, then E is an impossible event.
• If E were a sure event, its probability would be 1.
• The probability value does not determine if an event is simple or compound; those classifications relate to the number of outcomes in the event.

Therefore, if the probability of an event is 0, it means the event is impossible.

The correct answer is (B) E is an impossible event.

To find the probability of drawing a black King from a deck of 52 cards, we need to identify the total number of cards and the number of black Kings.

1. Total Number of Cards:

A standard deck of playing cards has 52 cards. So, the total number of possible outcomes is 52.

2. Number of Black Kings:

In a standard deck, there are four Kings: King of Hearts, King of Diamonds, King of Clubs, and King of Spades.

The suits of Hearts and Diamonds are red.

The suits of Clubs and Spades are black.

Therefore, the black Kings are the King of Clubs and the King of Spades.

Number of black Kings = 2.

These are the favorable outcomes.

3. Calculating the Probability:

The probability of drawing a black King is the ratio of the number of black Kings to the total number of cards:

Simplifying the fraction:

Evaluating the Options:

(A) 1/52: This is the probability of drawing a specific card, e.g., the King of Hearts.

(B) 2/52 = 1/26: This matches our calculated probability.

(C) 4/52 = 1/13: This is the probability of drawing any King.

(D) 13/52 = 1/4: This is the probability of drawing a card from a specific suit.

The correct answer is (B) 2/52 = 1/26.

In probability theory, the term "or" when referring to events signifies the union of those events.

Understanding Probability Notation:

• $P(A \cap B)$: This notation represents the probability that both event A AND event B occur. The symbol $\cap$ means intersection.
• $P(A \cup B)$: This notation represents the probability that event A OR event B (or both) occur. The symbol $\cup$ means union.
• $P(A) + P(B)$: This is part of the formula for the probability of the union of events. It's only equal to $P(A \cup B)$ if A and B are mutually exclusive.
• $P(A)P(B)$: This is typically used for the probability of the intersection of independent events.

Relating "or" to Notation:

The phrase "$P(A \text{ or } B)$" means the probability that event A occurs, or event B occurs, or both occur. This is the definition of the union of events A and B.

Therefore, $P(A \text{ or } B)$ is written as $P(A \cup B)$.

Evaluating the Options:

(A) $P(A \cap B)$: This represents "A and B".

(B) $P(A \cup B)$: This represents "A or B".

(C) $P(A) + P(B)$: This is part of the calculation for $P(A \cup B)$ but is not the final notation itself and only applies directly if A and B are mutually exclusive.

(D) $P(A)P(B)$: This is related to the intersection of independent events.

The standard notation for the probability of "A or B" is $P(A \cup B)$.

We need to identify the statement that is NOT always true for mutually exclusive events A and B.

Understanding Mutually Exclusive Events:

Mutually exclusive events are events that cannot occur at the same time. This means that if one event occurs, the other cannot.

Analyzing the Options:

(A) $A \cap B = \emptyset$: This is the definition of mutually exclusive events in set notation. It means the intersection of A and B is empty, implying they have no common outcomes. This statement IS true for mutually exclusive events.

(B) $P(A \cap B) = 0$: This is a consequence of $A \cap B = \emptyset$. The probability of an impossible event (the empty set) is 0. This statement IS true for mutually exclusive events.

(C) $P(A \cup B) = P(A) + P(B)$: This is the Special Addition Law of Probability, which applies when events are mutually exclusive. It is derived from the general addition law by setting $P(A \cap B) = 0$. This statement IS true for mutually exclusive events.

(D) $A \cup B = S$: This statement means that the union of events A and B covers the entire sample space. Events for which this is true are called exhaustive events. Mutually exclusive events are not necessarily exhaustive. For example, if we roll a die, the events "getting a 1" (A) and "getting a 2" (B) are mutually exclusive ($A \cap B = \emptyset$), but they are not exhaustive because outcomes like 3, 4, 5, or 6 are not included in $A \cup B$.

Since the condition $A \cup B = S$ is not always true for mutually exclusive events (only when they are also exhaustive), this statement is NOT always true.

The correct answer is (D) $A \cup B = S$.

To find the probability of getting a sum greater than 10 when rolling two fair dice, we need to determine the total number of possible outcomes and the number of outcomes that result in a sum greater than 10.

1. Total Number of Possible Outcomes:

When rolling two fair dice, there are $6 \times 6 = 36$ possible outcomes. These are pairs $(d_1, d_2)$, where $d_1$ is the result of the first die and $d_2$ is the result of the second die.

2. Favorable Outcomes (Sum Greater Than 10):

We need to find the pairs of outcomes $(d_1, d_2)$ such that their sum $d_1 + d_2 > 10$. The possible sums greater than 10 are 11 and 12.

Let's list the pairs that give these sums:

• Sum of 11:
  • (5, 6)
  • (6, 5)
• Sum of 12:
  • (6, 6)

The favorable outcomes are {(5, 6), (6, 5), (6, 6)}.

The total number of favorable outcomes is 3.

3. Calculating the Probability:

The probability of getting a sum greater than 10 is the ratio of the number of favorable outcomes to the total number of possible outcomes:

Simplifying the fraction:

Evaluating the Options:

(A) Sums > 10 are 11, 12. Outcomes: (5,6), (6,5), (6,6). Total 3 outcomes.: This option correctly identifies the possible sums, lists the outcomes, and counts them.

(B) 3/36 = 1/12: This option correctly states the probability based on the counts identified in (A).

(C) 1/36: This is the probability of a single specific outcome, like (5,6).

(D) 6/36 = 1/6: This is the probability of getting a sum of 7.

Option (A) correctly analyzes the problem by listing the favorable outcomes. Option (B) correctly calculates the probability based on that analysis.

The correct answer is (B) 3/36 = 1/12.

We need to find the probability of A or B occurring, given their individual probabilities and the fact that they are mutually exclusive events.

Given Information:

• $P(A) = 0.5$
• $P(B) = 0.4$
• A and B are mutually exclusive events.

We need to find $P(A \text{ or } B)$, which is equivalent to $P(A \cup B)$.

Understanding Mutually Exclusive Events:

Mutually exclusive events cannot occur at the same time. This means their intersection is empty ($A \cap B = \emptyset$), and thus, $P(A \cap B) = 0$.

Applying the Special Addition Law:

For mutually exclusive events, the Addition Law simplifies:

Substitute the given probabilities:

Calculate the sum:

Evaluating the Options:

(A) $0.5 + 0.4 = 0.9$: This option correctly shows the application of the Special Addition Law and the resulting calculation.

(B) 0.9: This is the correct numerical answer.

(C) 0.1: This would be $1 - P(A \cup B)$ if $P(A \cup B)$ was 0.9, which is the probability of the complement of $A \cup B$.

(D) 0: This is the probability of the intersection of mutually exclusive events ($P(A \cap B)$).

Option (A) provides the derivation and calculation, while Option (B) provides the final answer.

The correct answer is (A) $0.5 + 0.4 = 0.9$.

We need to find the probability that the first ball drawn is red AND the second ball drawn is blue, without replacement.

Understanding the Scenario:

Total number of balls in the bag = 6 red + 4 blue = 10 balls.

We are drawing two balls without replacement.

Method 1: Using Conditional Probability (Sequential Draws)

We need to find the probability of two events happening in sequence:

Event 1: The first ball drawn is red.

The probability of drawing a red ball first is:

Event 2: The second ball drawn is blue, GIVEN that the first ball drawn was red.

After drawing one red ball, there are now 9 balls left in the bag. The number of red balls has decreased by one, but the number of blue balls remains the same.

• Remaining red balls = $6 - 1 = 5$.
• Remaining blue balls = 4.
• Remaining total balls = $10 - 1 = 9$.

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

The probability of both events happening in this specific order (First Red AND Second Blue) is the product of their probabilities:

Method 2: Using Combinations (Less direct for ordered events)

While combinations are typically used for unordered selections, it's possible to frame this using permutations if order matters, or to be careful with the interpretation of combinations for this specific ordered outcome.

The total number of ordered ways to draw 2 balls from 10 is $P(10, 2) = 10 \times 9 = 90$.

The number of ways to draw a red ball first and a blue ball second is $6 \times 4 = 24$.

The probability is $\frac{24}{90}$.

Evaluating the Options:

(A) $\frac{6}{10} \times \frac{4}{10}$: This implies drawing *with* replacement, as the denominator remains 10 for the second draw.

(B) $\frac{6}{10} \times \frac{4}{9}$: This correctly represents the probability of drawing a red ball first, then a blue ball second, without replacement.

(C) $\frac{C(6, 1) \times C(4, 1)}{C(10, 2)}$: This calculates the probability of drawing one red AND one blue ball *in any order* (as combinations are unordered). It is the probability for getting one of each color, not necessarily in the specified order (first red, then blue).

(D) $\frac{24}{90}$: This is the simplified probability of drawing a red ball first and a blue ball second, calculated as $\frac{6 \times 4}{10 \times 9}$. It's numerically equivalent to option (B).

The question specifically asks for the probability that the *first* ball is red AND the *second* ball is blue. Option (B) directly models this sequence of events. Option (D) is the simplified numerical result of option (B).

If the question asks for the *probability*, both (B) and (D) represent the correct probability value, with (B) showing the calculation process and (D) showing the simplified result.

However, option (B) is a more explicit representation of the calculation for the specific ordered event.

The correct answer is (B) $\frac{6}{10} \times \frac{4}{9}$.

The experiment is to draw a card from a deck of 52 cards and note its suit. This means the outcome we are interested in is the suit of the drawn card, not the specific card itself.

Understanding the Experiment:

When a card is drawn from a standard deck, the possible suits are:

• Hearts (♥)
• Diamonds (♦)
• Clubs (♣)
• Spades (♠)

The experiment asks us to note the suit of the card, so these four suits are the possible outcomes.

Analyzing the Options:

(A) $\{ \text{Hearts, Diamonds, Clubs, Spades} \}$: This set represents the four possible suits of a card. Since the experiment specifies noting the suit, this is the correct sample space.

(B) $\{ \text{Red, Black} \}$: This set represents the colors of the cards, which is a subset of the suit information but not the complete suit information itself. It's a coarser classification.

(C) $\{1, 2, ..., 13\}$: This set represents the ranks of the cards within a suit (Ace through King), not the suits themselves.

(D) The set of all 52 cards: This would be the sample space if the experiment were to draw a card and note its specific identity (e.g., "King of Hearts"), not just its suit.

Since the experiment is about noting the suit, the sample space is the set of all possible suits.

The correct answer is (A) $\{ \text{Hearts, Diamonds, Clubs, Spades} \}$.

We are asked to determine if the statement $P(A \cap B') = P(A)$ is true when A and B are mutually exclusive events.

Understanding Mutually Exclusive Events:

Mutually exclusive events A and B means that they cannot occur at the same time. In terms of sets, their intersection is empty: $A \cap B = \emptyset$.

Understanding $B'$:

$B'$ denotes the complement of event B. It includes all outcomes in the sample space that are not in B.

Analyzing the Intersection $A \cap B'$:

We are considering the intersection of event A with the complement of event B ($A \cap B'$). This represents the outcomes that are in A AND are NOT in B.

Since A and B are mutually exclusive, any outcome in A is by definition NOT in B. This means that all outcomes in A are also in $B'$ (the complement of B).

Therefore, if A and B are mutually exclusive, the set of outcomes in A is a subset of the outcomes not in B. That is, $A \subseteq B'$.

When A is a subset of B', the intersection of A and B' is simply A itself: $A \cap B' = A$.

Relating to Probability:

If $A \cap B' = A$, then the probability of this intersection is:

This statement is true for any mutually exclusive events A and B.

Evaluating the Options:

(A) Yes (If A and B are mutually exclusive, A is completely outside B, so $A \cap B' = A$): This option correctly states that the statement is true and provides the correct reasoning ($A \cap B' = A$ because A is outside B). This is the correct explanation.

(B) No: This is incorrect as the statement is true.

(C) Only if B is the impossible event: While if B is impossible ($B = \emptyset$), then $B' = S$, and $A \cap S = A$, making the statement true. However, the statement is true for ANY mutually exclusive events, not just when B is impossible.

(D) Only if A is the sample space: If A is the sample space ($A = S$), then $P(A)=1$. Since A and B are mutually exclusive, B must be the empty set ($B = \emptyset$) and $P(B)=0$. In this case, $P(A \cap B') = P(S \cap S) = P(S) = 1$, and $P(A) = P(S) = 1$. So it holds. But again, the statement is true in more general cases.

The statement $P(A \cap B') = P(A)$ is indeed true for any mutually exclusive events A and B, and option (A) provides the correct reasoning.

The correct answer is (A) Yes (If A and B are mutually exclusive, A is completely outside B, so $A \cap B' = A$).

Random Experiment: A random experiment is an experiment whose outcome cannot be predicted with certainty, even though all possible outcomes are known.

Sample Space: The sample space of a random experiment is the set of all possible outcomes of the experiment.

Example of a Random Experiment:

Experiment: Tossing a fair coin once.

Possible Outcomes: When a fair coin is tossed once, there are two possible outcomes: Head (H) or Tail (T).

Sample Space: The sample space for this experiment is the set containing these two outcomes.

Let S be the sample space. Then, $S = \{H, T\}$.

When a coin is tossed repeatedly until a head appears for the first time, the possible outcomes are:

The first toss is a head (H).

The first toss is a tail, and the second toss is a head (TH).

The first two tosses are tails, and the third toss is a head (TTH).

And so on...

The sample space (S) can be represented as:

$S = \{H, TH, TTH, TTTH, TTTTH, \ldots \}$

This is an infinite sample space, where each outcome consists of a sequence of tails followed by a single head.

When a family has 3 children, we can represent the gender of each child by B for Boy and G for Girl. The order matters, indicating the birth order.

Sample Space (S): The sample space for a family with 3 children is the set of all possible combinations of genders:

$S = \{BBB, BBG, BGB, GBB, BGG, GBG, GGB, GGG\}$

There are $2^3 = 8$ possible outcomes.

Event A: At most two children.

The event "at most two children" means the family has either 0, 1, or 2 children. However, the problem states we are considering a family that *has* 3 children. In the context of the given sample space of 3 children, "at most two children" implies that we are looking for outcomes within this sample space that satisfy this condition. This phrasing might seem contradictory, but it usually refers to a characteristic of the outcomes themselves if we were to select from a larger population.

If we strictly interpret "at most two children" within the sample space of 3 children, it means we are interested in the outcomes where the number of children who are, for example, boys (or girls, depending on how "children" is implicitly defined in the event) is at most two. However, a more common interpretation in such problems is that the event A refers to the subset of the sample space where the characteristic being counted (e.g., number of boys, or number of girls) is at most two. Without further clarification on what "children" refers to in the context of the event A, let's assume it refers to the number of boys.

Describing Event A (Number of Boys is at most 2) in terms of the sample space:

Event A consists of all outcomes in S where there are 0, 1, or 2 boys.

Outcomes with 0 boys: {GGG}

Outcomes with 1 boy: {BGG, GBG, GGB}

Outcomes with 2 boys: {BBG, BGB, GBB}

Therefore, Event A is:

$A = \{GGG, BGG, GBG, GGB, BBG, BGB, GBB\}$

Note that the outcome BBB (3 boys) is excluded from event A.

Elementary Event: An elementary event is an event that consists of a single outcome from the sample space of a random experiment.

Compound Event: A compound event is an event that consists of two or more elementary events, or outcomes, from the sample space of a random experiment.

Example when a single die is rolled:

When a single die is rolled, the sample space is $S = \{1, 2, 3, 4, 5, 6\}$.

Example of an Elementary Event:

Let E be the event of rolling a '4'.

Then, $E = \{4\}$. This event consists of only one outcome.

Example of a Compound Event:

Let F be the event of rolling an even number.

Then, $F = \{2, 4, 6\}$. This event consists of three elementary events (outcomes 2, 4, and 6).

We are given the following probabilities:

$P(E) = 0.5$

$P(F) = 0.4$

$P(E \cap F) = 0.2$

We need to find $P(E \cup F)$.

We use the addition rule for probabilities:

Substitute the given values into the formula:

$P(E \cup F) = 0.5 + 0.4 - 0.2$

$P(E \cup F) = 0.9 - 0.2$

$P(E \cup F) = 0.7$

Thus, the probability of the union of events E and F is 0.7.

When a single die is rolled, the sample space is $S = \{1, 2, 3, 4, 5, 6\}$.

Event E: Getting an even number.

The even numbers in the sample space are 2, 4, and 6.

So, $E = \{2, 4, 6\}$.

Event F: Getting a number less than 4.

The numbers less than 4 in the sample space are 1, 2, and 3.

So, $F = \{1, 2, 3\}$.

Event $E \cup F$ (E union F):

This event represents getting an even number OR getting a number less than 4 (or both).

To find $E \cup F$, we combine all the distinct outcomes from both E and F.

$E \cup F = E \cup F = \{2, 4, 6\} \cup \{1, 2, 3\} = \{1, 2, 3, 4, 6\}$

Therefore, $E \cup F$ is the event of "getting a number that is either even or less than 4 (or both)".

Event $E \cap F$ (E intersection F):

This event represents getting a number that is BOTH an even number AND less than 4.

To find $E \cap F$, we find the common outcomes present in both E and F.

$E \cap F = E \cap F = \{2, 4, 6\} \cap \{1, 2, 3\} = \{2\}$

Therefore, $E \cap F$ is the event of "getting the number 2".

In a well-shuffled deck of 52 cards:

Total number of cards = 52.

Let R be the event that the card drawn is a red card.

There are 26 red cards in a deck (13 hearts and 13 diamonds).

So, the number of outcomes favorable to R is 26.

The probability of drawing a red card is $P(R) = \frac{\text{Number of red cards}}{\text{Total number of cards}} = \frac{26}{52} = \frac{1}{2}$.

Let K be the event that the card drawn is a king.

There are 4 kings in a deck (King of Hearts, King of Diamonds, King of Clubs, King of Spades).

So, the number of outcomes favorable to K is 4.

The probability of drawing a king is $P(K) = \frac{\text{Number of kings}}{\text{Total number of cards}} = \frac{4}{52} = \frac{1}{13}$.

We want to find the probability that the card is a red card OR a king, which is $P(R \cup K)$.

We use the addition rule for probabilities: $P(R \cup K) = P(R) + P(K) - P(R \cap K)$.

Now we need to find $P(R \cap K)$, the probability that the card is both red AND a king.

The cards that are both red and a king are the King of Hearts and the King of Diamonds.

So, there are 2 cards that are both red and a king.

The probability of drawing a card that is both red and a king is $P(R \cap K) = \frac{\text{Number of red kings}}{\text{Total number of cards}} = \frac{2}{52} = \frac{1}{26}$.

Now, substitute these probabilities into the addition rule:

$P(R \cup K) = \frac{1}{2} + \frac{1}{13} - \frac{1}{26}$

To add these fractions, we find a common denominator, which is 26.

$P(R \cup K) = \frac{1 \times 13}{2 \times 13} + \frac{1 \times 2}{13 \times 2} - \frac{1}{26}$

$P(R \cup K) = \frac{13}{26} + \frac{2}{26} - \frac{1}{26}$

$P(R \cup K) = \frac{13 + 2 - 1}{26}$

$P(R \cup K) = \frac{14}{26}$

$P(R \cup K) = \frac{7}{13}$

Therefore, the probability that the card is a red card or a king is $\frac{7}{13}$.

When two dice are thrown, each die has 6 possible outcomes. The total number of possible outcomes is $6 \times 6 = 36$.

We are interested in the event where the sum of the numbers on the two dice is 9.

Let's list the pairs of outcomes that result in a sum of 9:

• (3, 6)
• (4, 5)
• (5, 4)
• (6, 3)

There are 4 favorable outcomes where the sum is 9.

The probability of an event is calculated as:

In this case:

Number of favorable outcomes (sum is 9) = 4

Total number of possible outcomes = 36

Therefore, the probability of getting a sum of 9 is:

$P(\text{Sum is 9}) = \frac{4}{36}$

Simplifying the fraction:

$P(\text{Sum is 9}) = \frac{1}{9}$

So, the probability that the numbers appearing on the two dice have a sum of 9 is $\frac{1}{9}$.

For a finite sample space S, the axiomatic approach to probability, as formulated by Andrey Kolmogorov, defines probability based on three fundamental axioms:

Axiom 1: Non-negativity

The probability of any event A is greater than or equal to zero.

This axiom states that probability cannot be negative. It implies that any event must have a probability of zero or a positive value.

Axiom 2: Probability of the Sample Space

The probability of the entire sample space S is equal to 1.

This axiom signifies that the occurrence of at least one outcome in the sample space is certain.

Axiom 3: Additivity for Mutually Exclusive Events

If $A_1, A_2, A_3, \ldots$ are any sequence of mutually exclusive events (i.e., no two events have any outcome in common, $A_i \cap A_j = \emptyset$ for $i \neq j$), then the probability of their union is the sum of their individual probabilities.

For a finite sample space, this often reduces to the case of two mutually exclusive events:

This axiom states that for events that cannot happen at the same time, the probability that at least one of them occurs is simply the sum of their individual probabilities.

We are given the probability of an event E:

$P(E) = 0.6$

We need to find the probability of the complement of event E, denoted as $P(\text{not } E)$ or $P(E^c)$.

The event "not E" is the event that E does not occur.

The sample space S can be divided into two mutually exclusive events: E (the event E occurs) and not E (the event E does not occur). This means that $E \cup (\text{not } E) = S$ and $E \cap (\text{not } E) = \emptyset$.

Using Axiom 3 (Additivity for mutually exclusive events) for these two events:

Since $E \cup (\text{not } E) = S$, we can write:

From Axiom 2, we know that $P(S) = 1$. Therefore:

This equation expresses the relationship between $P(E)$ and $P(\text{not } E)$ in terms of the axioms.

Now, we can find $P(\text{not } E)$ by rearranging the formula:

$P(\text{not } E) = 1 - P(E)$

Substitute the given value of $P(E) = 0.6$:

$P(\text{not } E) = 1 - 0.6$

$P(\text{not } E) = 0.4$

So, the probability of "not E" is 0.4.

First, let's find the total number of balls in the bag:

Number of red balls = 4

Number of black balls = 5

Number of white balls = 3

Total number of balls = $4 + 5 + 3 = 12$.

The probability of an event is given by:

(i) Probability that the ball drawn is red:

Number of red balls (favorable outcomes) = 4

Total number of balls (total possible outcomes) = 12

Probability of drawing a red ball = $P(\text{Red}) = \frac{4}{12} = \frac{1}{3}$.

(ii) Probability that the ball drawn is not white:

The event "not white" means the ball drawn is either red or black.

Number of balls that are not white = Number of red balls + Number of black balls = $4 + 5 = 9$.

Number of balls that are not white (favorable outcomes) = 9

Total number of balls (total possible outcomes) = 12

Probability of drawing a ball that is not white = $P(\text{Not White}) = \frac{9}{12} = \frac{3}{4}$.

Alternatively, we can first find the probability of drawing a white ball and then use the complement rule.

Probability of drawing a white ball = $P(\text{White}) = \frac{\text{Number of white balls}}{\text{Total number of balls}} = \frac{3}{12} = \frac{1}{4}$.

The probability of the ball not being white is $P(\text{Not White}) = 1 - P(\text{White})$.

$P(\text{Not White}) = 1 - \frac{1}{4} = \frac{4}{4} - \frac{1}{4} = \frac{3}{4}$.

So, the probability that the ball drawn is red is $\frac{1}{3}$, and the probability that it is not white is $\frac{3}{4}$.

Let H represent heads and T represent tails.

The experiment stops when a tail appears or after 4 tosses if no tail appears.

Let's list all the possible outcomes:

Case 1: Tail appears on the first toss.

Outcome: T

Case 2: Tail appears on the second toss (first toss was heads).

Outcome: HT

Case 3: Tail appears on the third toss (first two tosses were heads).

Outcome: HHT

Case 4: Tail appears on the fourth toss (first three tosses were heads).

Outcome: HHHT

Case 5: No tail appears in 4 tosses (experiment stops after 4 tosses).

Outcome: HHHH

The sample space (S) is the collection of all these possible outcomes:

$S = \{T, HT, HHT, HHHT, HHHH\}$

Classical Definition of Probability:

The classical definition of probability states that if an experiment has $n$ equally likely, mutually exclusive outcomes, and an event A is associated with $m$ of these outcomes, then the probability of event A, denoted by $P(A)$, is given by:

Where:

• $n$ is the total number of possible outcomes (the size of the sample space).
• $m$ is the number of outcomes favorable to event A.

This definition relies on the assumption that all outcomes are equally likely, meaning each outcome has the same chance of occurring.

Limitations of the Classical Definition:

The classical definition of probability is useful in many situations, but it has several significant limitations:

1. Equally Likely Outcomes: The most crucial limitation is that it requires all possible outcomes to be equally likely. This assumption is not always valid in real-world scenarios. For example, if we consider the probability of a specific sports team winning a match, the outcomes (win, lose, draw) are generally not equally likely.
2. Finite Sample Space: This definition is typically applicable only to experiments with a finite number of outcomes. It cannot be directly applied to experiments with an infinite number of possible outcomes, such as tossing a coin until the first head appears (as seen in a previous example).
3. Definition of "Equally Likely": The definition itself is somewhat circular because it uses the concept of "equally likely" outcomes to define probability, and yet "equally likely" is often understood in terms of probability. If we don't already know that outcomes are equally likely, this definition doesn't help us determine their probabilities.
4. Applicability to Theoretical Problems: It is best suited for theoretical problems like tossing coins, rolling dice, or drawing cards from a deck, where symmetry ensures equal likelihood. It is less practical for empirical situations where outcomes might be influenced by various factors.

We need to prove that for any event E, $0 \leq P(E) \leq 1$, using the axioms of probability.

Part 1: Proving $P(E) \geq 0$

From Axiom 1 of probability, we know that the probability of any event is non-negative.

Since E is an event, by Axiom 1:

This establishes the first part of the inequality.

Part 2: Proving $P(E) \leq 1$

Let S be the sample space for a random experiment. The event E is a subset of the sample space S, meaning all outcomes in E are also in S.

We know that for any event E, $E \subseteq S$.

Also, the complement of event E, denoted by $E^c$ or "not E", is the set of all outcomes in S that are not in E. The event E and its complement $E^c$ are mutually exclusive, meaning $E \cap E^c = \emptyset$.

The union of an event and its complement is the entire sample space: $E \cup E^c = S$.

Using Axiom 3 (Additivity for mutually exclusive events), since E and $E^c$ are mutually exclusive:

Since $E \cup E^c = S$, we can substitute S:

From Axiom 2, we know that $P(S) = 1$. So:

Now, consider $P(E^c)$. From Axiom 1, we know that $P(E^c) \geq 0$.

Substituting this into the equation above:

$1 = P(E) + (\text{a non-negative number})$

This implies that $P(E)$ must be less than or equal to 1 for the sum to be 1.

Rearranging the equation $1 = P(E) + P(E^c)$ to solve for $P(E)$ gives:

$P(E) = 1 - P(E^c)$

Since $P(E^c) \geq 0$, subtracting a non-negative number from 1 will result in a value less than or equal to 1.

Therefore, $P(E) \leq 1$.

Combining both parts, we have shown that $0 \leq P(E)$ and $P(E) \leq 1$. Thus, for any event E, $0 \leq P(E) \leq 1$.

A standard deck of 52 cards has 4 suits: hearts, diamonds, clubs, and spades. Each suit has 13 cards: Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King.

The face cards are Jack (J), Queen (Q), and King (K).

Total number of outcomes = 52 (total cards in the deck).

The probability of an event is given by:

(i) Probability that the card is a face card:

There are 3 face cards (Jack, Queen, King) in each of the 4 suits.

Number of face cards = 3 face cards/suit $\times$ 4 suits = 12 face cards.

Number of favorable outcomes (face cards) = 12.

Probability of drawing a face card = $P(\text{Face card}) = \frac{12}{52}$.

Simplifying the fraction:

$P(\text{Face card}) = \frac{12}{52} = \frac{3 \times 4}{13 \times 4} = \frac{3}{13}$.

(ii) Probability that the card is a card of diamond:

There are 13 cards in each suit, and diamond is one of the suits.

Number of favorable outcomes (diamond cards) = 13.

Probability of drawing a diamond card = $P(\text{Diamond}) = \frac{13}{52}$.

Simplifying the fraction:

$P(\text{Diamond}) = \frac{13}{52} = \frac{1 \times 13}{4 \times 13} = \frac{1}{4}$.

So, the probability that the card is a face card is $\frac{3}{13}$, and the probability that it is a card of diamond is $\frac{1}{4}$.

When three coins are tossed simultaneously, each coin can land as either Heads (H) or Tails (T). The total number of possible outcomes is $2 \times 2 \times 2 = 2^3 = 8$.

Let's list all the possible outcomes:

• HHH
• HHT
• HTH
• THH
• HTT
• THT
• TTH
• TTT

The sample space is $S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$.

We are interested in the event of getting exactly two heads.

Let's identify the outcomes with exactly two heads from the list above:

• HHT
• HTH
• THH

There are 3 outcomes where exactly two heads appear.

The probability of an event is calculated as:

In this case:

Number of favorable outcomes (exactly two heads) = 3

Total number of possible outcomes = 8

Therefore, the probability of getting exactly two heads is:

$P(\text{Exactly two heads}) = \frac{3}{8}$

So, the probability of getting exactly two heads when three coins are tossed simultaneously is $\frac{3}{8}$.

Definition of Mutually Exclusive Events:

Two events are said to be mutually exclusive if they cannot occur at the same time. In other words, if one event occurs, the other event cannot occur. Mathematically, if E and F are mutually exclusive events, their intersection is an empty set: $E \cap F = \emptyset$.

Formula for $P(E \cup F)$ when E and F are mutually exclusive:

The general formula for the probability of the union of two events is:

However, if E and F are mutually exclusive events, it means they have no common outcomes, so their intersection is the empty set ($E \cap F = \emptyset$).

According to the axiomatic approach (specifically, related to Axiom 1 and the interpretation of probabilities of impossible events), the probability of an impossible event (like the empty set) is 0.

So, for mutually exclusive events, $P(E \cap F) = P(\emptyset) = 0$.

Substituting $P(E \cap F) = 0$ into the general formula:

This simplifies to:

This formula shows that for mutually exclusive events, the probability that either one or the other occurs is simply the sum of their individual probabilities.

Explanation for $P(E \cap F) = 0$:

As stated in the definition, mutually exclusive events cannot occur simultaneously. The intersection of two events, $E \cap F$, represents the event where both E and F occur. If E and F are mutually exclusive, it is impossible for both to occur at the same time. An impossible event has a probability of 0. Therefore, $P(E \cap F) = 0$ for mutually exclusive events.

The set of integers from 1 to 20 is $S = \{1, 2, 3, \ldots, 20\}$.

The total number of possible outcomes is 20.

We are interested in the event that the chosen number is divisible by 4.

Let A be the event that the chosen number is divisible by 4.

The numbers in the set S that are divisible by 4 are:

• 4
• 8
• 12
• 16
• 20

So, the favorable outcomes for event A are $\{4, 8, 12, 16, 20\}$.

The number of favorable outcomes is 5.

The probability of an event is calculated as:

In this case:

Number of favorable outcomes (divisible by 4) = 5

Total number of possible outcomes = 20

Therefore, the probability that the chosen number is divisible by 4 is:

$P(A) = \frac{5}{20}$

Simplifying the fraction:

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

So, the probability that a randomly chosen number from 1 to 20 is divisible by 4 is $\frac{1}{4}$.

Impossible Event:

An impossible event is an event that cannot happen. Its occurrence is not possible under any circumstances.

Example when rolling a single die:

Getting a number greater than 6 on a single roll of a die.

Since there are no outcomes greater than 6 when rolling a die, the number of favourable outcomes is 0.

The probability of an impossible event is 0.

Sure Event:

A sure event is an event that is certain to happen. Its occurrence is guaranteed.

Example when rolling a single die:

Getting a number less than 7 on a single roll of a die.

When rolling a die, the possible outcomes are {1, 2, 3, 4, 5, 6}. All these outcomes are less than 7. Therefore, the number of favourable outcomes is 6.

The probability of a sure event is 1.

We are given the following probabilities:

$P(E) = 0.6$

$P(F) = 0.3$

$P(E \cup F) = 0.7$

We need to find $P(E \cap F)$.

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

Substitute the given values into the formula:

$0.7 = 0.6 + 0.3 - P(E \cap F)$

$0.7 = 0.9 - P(E \cap F)$

Now, we solve for $P(E \cap F)$:

So, the probability of the intersection of events E and F is $0.2$.

Are E and F mutually exclusive events?

Mutually exclusive events are events that cannot occur at the same time. This means that the probability of their intersection is 0, i.e., $P(E \cap F) = 0$.

In this case, we found that $P(E \cap F) = 0.2$.

Since the probability of the intersection of E and F is not zero, events E and F are not mutually exclusive events.

Let R denote drawing a red ball and B denote drawing a black ball.

The box initially contains 5 red balls and 7 black balls, making a total of 12 balls.

We need to consider two cases for the first draw:

Case 1: The first ball drawn is red (R).

If the first ball drawn is red, it is replaced. This means the composition of the box remains the same for the second draw (5 red, 7 black, total 12 balls).

The possible outcomes for the first two draws in this case are:

• First draw Red, Second draw Red (RR)
• First draw Red, Second draw Black (RB)

Case 2: The first ball drawn is black (B).

If the first ball drawn is black, it is not replaced. This means for the second draw, there will be one less black ball, and the total number of balls will also decrease by one.

After drawing a black ball (and not replacing it):

• Number of red balls = 5
• Number of black balls = $7 - 1 = 6$
• Total number of balls = $12 - 1 = 11$

The possible outcomes for the first two draws in this case are:

• First draw Black, Second draw Red (BR)
• First draw Black, Second draw Black (BB)

Therefore, a partial sample space for the first two draws is:

{RR, RB, BR, BB}

Let M be the event that a student passes the mathematics test.

Let P be the event that a student passes the physics test.

We are given the following probabilities:

$P(M) = 0.7$

$P(P) = 0.6$

$P(M \cap P) = 0.4$ (The probability of passing both tests)

We need to find the probability of passing at least one test, which means we need to find $P(M \cup P)$.

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

Substitute the given values into the formula:

$P(M \cup P) = 0.7 + 0.6 - 0.4$

$P(M \cup P) = 1.3 - 0.4$

Therefore, the probability of passing at least one test is $0.9$.

When two dice are thrown, the total number of possible outcomes is $6 \times 6 = 36$. We can represent the sample space S as follows:

S = {(1,1), (1,2), ..., (1,6), (2,1), ..., (6,6)}

Event A: Sum of numbers is 7

The outcomes for event A are:

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

The number of outcomes in A is $n(A) = 6$.

The probability of event A is:

Event B: Sum of numbers is 11

The outcomes for event B are:

B = {(5,6), (6,5)}

The number of outcomes in B is $n(B) = 2$.

The probability of event B is:

Event C: The number on the first die is 3

The outcomes for event C are:

C = {(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)}

The number of outcomes in C is $n(C) = 6$.

The probability of event C is:

Event $A \cap B$: Sum of numbers is 7 AND sum of numbers is 11

There are no outcomes where the sum of numbers on two dice is both 7 and 11 simultaneously. Therefore, $A \cap B$ is an empty set.

A $\cap$ B = {}

The number of outcomes in $A \cap B$ is $n(A \cap B) = 0$.

The probability of event $A \cap B$ is:

Event $B \cup C$: Sum of numbers is 11 OR the number on the first die is 3

The outcomes for event B are {(5,6), (6,5)}.

The outcomes for event C are {(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)}.

The union of B and C includes all outcomes that are in B or in C (or both). Since there are no common outcomes between B and C, the union is simply the set of all outcomes from B and C.

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

The number of outcomes in $B \cup C$ is $n(B \cup C) = n(B) + n(C) = 2 + 6 = 8$.

The probability of event $B \cup C$ is:

Summary of Probabilities:

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

$P(B) = \frac{1}{18}$

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

$P(A \cap B) = 0$

$P(B \cup C) = \frac{2}{9}$

Given:

Number of boys = 8

Number of girls = 5

Total number of people = $8 + 5 = 13$

A committee of 6 is to be formed.

Total number of ways to form the committee:

The total number of ways to choose 6 members from 13 people is given by the combination formula $C(n, k) = \frac{n!}{k!(n-k)!}$.

(i) Probability that the committee contains exactly 3 girls:

To have exactly 3 girls in the committee, we need to choose 3 girls from 5 girls and 3 boys from 8 boys.

Number of ways to choose 3 girls from 5 = $C(5, 3) = \frac{5!}{3!(5-3)!} = \frac{5!}{3!2!} = \frac{5 \times 4}{2 \times 1} = 10$

Number of ways to choose 3 boys from 8 = $C(8, 3) = \frac{8!}{3!(8-3)!} = \frac{8!}{3!5!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$

Number of ways to form the committee with exactly 3 girls = $C(5, 3) \times C(8, 3) = 10 \times 56 = 560$

Probability (exactly 3 girls) = $\frac{\text{Number of ways to form committee with 3 girls}}{\text{Total number of ways to form the committee}}$

Simplify the fraction:

(ii) Probability that the committee contains at least 2 girls:

"At least 2 girls" means 2 girls, 3 girls, 4 girls, or 5 girls.

Case 1: Exactly 2 girls and 4 boys

Number of ways = $C(5, 2) \times C(8, 4) = \frac{5!}{2!3!} \times \frac{8!}{4!4!} = 10 \times \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 10 \times 70 = 700$

Case 2: Exactly 3 girls and 3 boys (calculated above)

Number of ways = 560

Case 3: Exactly 4 girls and 2 boys

Number of ways = $C(5, 4) \times C(8, 2) = 5 \times \frac{8!}{2!6!} = 5 \times \frac{8 \times 7}{2 \times 1} = 5 \times 28 = 140$

Case 4: Exactly 5 girls and 1 boy

Number of ways = $C(5, 5) \times C(8, 1) = 1 \times 8 = 8$

Total number of ways to have at least 2 girls = $700 + 560 + 140 + 8 = 1408$

Probability (at least 2 girls) = $\frac{\text{Number of ways to form committee with at least 2 girls}}{\text{Total number of ways to form the committee}}$

Simplify the fraction:

(iii) Probability that the committee contains at most 1 boy:

"At most 1 boy" means either 0 boys or 1 boy.

Case 1: 0 boys and 6 girls

Number of ways = $C(8, 0) \times C(5, 6)$. Since we cannot choose 6 girls from 5, this is 0 ways.

Case 2: 1 boy and 5 girls

Number of ways = $C(8, 1) \times C(5, 5) = 8 \times 1 = 8$

Total number of ways to have at most 1 boy = $0 + 8 = 8$

Probability (at most 1 boy) = $\frac{\text{Number of ways to form committee with at most 1 boy}}{\text{Total number of ways to form the committee}}$

Simplify the fraction:

Summary of Probabilities:

(i) Probability of exactly 3 girls = $\frac{20}{61}$

(ii) Probability of at least 2 girls = $\frac{32}{39}$

(iii) Probability of at most 1 boy = $\frac{2}{429}$

Given:

Total number of students = 100

Number of students studying Mathematics, $n(M) = 40$

Number of students studying Physics, $n(P) = 35$

Number of students studying Chemistry, $n(C) = 25$

Number of students studying Mathematics and Physics, $n(M \cap P) = 10$

Number of students studying Physics and Chemistry, $n(P \cap C) = 8$

Number of students studying Mathematics and Chemistry, $n(M \cap C) = 12$

Number of students studying all three subjects, $n(M \cap P \cap C) = 5$

We will use the Principle of Inclusion-Exclusion for three sets:

(i) Probability that the student studies Mathematics or Physics:

We need to find $P(M \cup P)$. Using the formula for two sets:

The probability that the student studies Mathematics or Physics is:

(ii) Probability that the student studies Physics or Chemistry:

We need to find $P(P \cup C)$. Using the formula for two sets:

The probability that the student studies Physics or Chemistry is:

(iii) Probability that the student studies Mathematics only:

To find the number of students studying Mathematics only, we subtract the overlaps from $n(M)$:

We add $n(M \cap P \cap C)$ back because it was subtracted twice (once for $M \cap P$ and once for $M \cap C$).

The probability that the student studies Mathematics only is:

(iv) Probability that the student studies none of the three subjects:

First, let's find the number of students who study at least one of the three subjects using the Principle of Inclusion-Exclusion:

The number of students who study none of the three subjects is:

The probability that the student studies none of the three subjects is:

Summary of Probabilities:

(i) Probability of Mathematics or Physics = $\frac{13}{20}$

(ii) Probability of Physics or Chemistry = $\frac{13}{25}$

(iii) Probability of Mathematics only = $\frac{23}{100}$

(iv) Probability of none of the three subjects = $\frac{1}{4}$

Given:

A well-shuffled deck of 52 cards.

Four cards are drawn simultaneously.

Total number of ways to draw 4 cards from 52 cards:

This is given by the combination formula $C(n, k) = \frac{n!}{k!(n-k)!}$.

(i) Probability that all four are spades:

There are 13 spades in a deck of cards.

Number of ways to choose 4 spades from 13 = $C(13, 4)$

Probability (all four are spades) = $\frac{\text{Number of ways to choose 4 spades}}{\text{Total number of ways to choose 4 cards}}$

Simplify the fraction:

(ii) Probability that there is one card of each suit:

There are 4 suits: Spades, Hearts, Diamonds, Clubs. Each suit has 13 cards.

Number of ways to choose 1 spade from 13 = $C(13, 1) = 13$

Number of ways to choose 1 heart from 13 = $C(13, 1) = 13$

Number of ways to choose 1 diamond from 13 = $C(13, 1) = 13$

Number of ways to choose 1 club from 13 = $C(13, 1) = 13$

Number of ways to choose one card of each suit = $C(13, 1) \times C(13, 1) \times C(13, 1) \times C(13, 1) = 13^4$

Probability (one card of each suit) = $\frac{\text{Number of ways to choose one of each suit}}{\text{Total number of ways to choose 4 cards}}$

Simplify the fraction:

(iii) Probability that there are two red cards and two black cards:

There are 26 red cards (Hearts and Diamonds) and 26 black cards (Spades and Clubs) in a deck.

Number of ways to choose 2 red cards from 26 = $C(26, 2) = \frac{26!}{2!(26-2)!} = \frac{26 \times 25}{2 \times 1} = 13 \times 25 = 325$

Number of ways to choose 2 black cards from 26 = $C(26, 2) = \frac{26!}{2!(26-2)!} = \frac{26 \times 25}{2 \times 1} = 13 \times 25 = 325$

Number of ways to choose two red cards and two black cards = $C(26, 2) \times C(26, 2) = 325 \times 325 = 105625$

Probability (two red and two black cards) = $\frac{\text{Number of ways to choose 2 red and 2 black cards}}{\text{Total number of ways to choose 4 cards}}$

Simplify the fraction:

Summary of Probabilities:

(i) Probability that all four are spades = $\frac{11}{4165}$

(ii) Probability that there is one card of each suit = $\frac{2197}{20825}$

(iii) Probability that there are two red cards and two black cards = $\frac{325}{833}$

Given:

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

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

Probability of the intersection of E and F, $P(E \cap F) = 0.2$

(i) Find $P(E \cup F)$:

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

(ii) Find $P(\text{not } E)$:

The probability of the complement of an event E is given by:

(iii) Find $P(\text{not } F)$:

The probability of the complement of an event F is given by:

(iv) Find $P(E \cap (\text{not } F))$:

This represents the probability that event E occurs but event F does not occur. This can be calculated as:

(v) Find $P(F \cap (\text{not } E))$:

This represents the probability that event F occurs but event E does not occur. This can be calculated as:

Summary of Results:

$P(E \cup F) = 0.7$

$P(\text{not } E) = 0.4$

$P(\text{not } F) = 0.7$

$P(E \cap (\text{not } F)) = 0.4$

$P(F \cap (\text{not } E)) = 0.1$

Given:

Number of white balls = 6

Number of red balls = 8

Number of black balls = 10

Total number of balls in the bag = $6 + 8 + 10 = 24$

Three balls are drawn one by one without replacement.

(i) Probability that all three are red:

Probability of drawing the first red ball = $\frac{8}{24}$

After drawing one red ball, there are 7 red balls left and a total of 23 balls.

Probability of drawing the second red ball = $\frac{7}{23}$

After drawing two red balls, there are 6 red balls left and a total of 22 balls.

Probability of drawing the third red ball = $\frac{6}{22}$

The probability that all three are red is the product of these probabilities:

(ii) Probability that the first is white, the second is red, and the third is black:

Probability of drawing the first ball as white = $\frac{6}{24}$

After drawing one white ball, there are 8 red balls left and a total of 23 balls.

Probability of drawing the second ball as red = $\frac{8}{23}$

After drawing one white and one red ball, there are 10 black balls left and a total of 22 balls.

Probability of drawing the third ball as black = $\frac{10}{22}$

The probability of this specific sequence is:

(iii) Probability that one of each colour is drawn:

This means we can draw the balls in any order (White, Red, Black; White, Black, Red; Red, White, Black; Red, Black, White; Black, White, Red; Black, Red, White).

The probability of drawing one of each colour in a specific order (e.g., W, R, B) is $\frac{10}{253}$ (from part ii).

There are $3! = 3 \times 2 \times 1 = 6$ possible orders for drawing one of each colour.

Let's calculate the probability for each order:

P(W, R, B) = $\frac{6}{24} \times \frac{8}{23} \times \frac{10}{22} = \frac{480}{12144}$

P(W, B, R) = $\frac{6}{24} \times \frac{10}{23} \times \frac{8}{22} = \frac{480}{12144}$

P(R, W, B) = $\frac{8}{24} \times \frac{6}{23} \times \frac{10}{22} = \frac{480}{12144}$

P(R, B, W) = $\frac{8}{24} \times \frac{10}{23} \times \frac{6}{22} = \frac{480}{12144}$

P(B, W, R) = $\frac{10}{24} \times \frac{6}{23} \times \frac{8}{22} = \frac{480}{12144}$

P(B, R, W) = $\frac{10}{24} \times \frac{8}{23} \times \frac{6}{22} = \frac{480}{12144}$

The total probability of drawing one of each colour is the sum of the probabilities of these mutually exclusive events:

Alternatively, using combinations:

Total number of ways to draw 3 balls from 24 = $C(24, 3) = \frac{24 \times 23 \times 22}{3 \times 2 \times 1} = 4 \times 23 \times 22 = 2024$

Number of ways to draw one white ball = $C(6, 1) = 6$

Number of ways to draw one red ball = $C(8, 1) = 8$

Number of ways to draw one black ball = $C(10, 1) = 10$

Number of ways to draw one of each colour = $C(6, 1) \times C(8, 1) \times C(10, 1) = 6 \times 8 \times 10 = 480$

Probability (one of each colour) = $\frac{480}{2024}$

Summary of Probabilities:

(i) Probability that all three are red = $\frac{7}{253}$

(ii) Probability that the first is white, the second is red, and the third is black = $\frac{10}{253}$

(iii) Probability that one of each colour is drawn = $\frac{60}{253}$

When a die is thrown twice, the total number of possible outcomes is $6 \times 6 = 36$. We can represent the outcomes as ordered pairs $(x, y)$, where $x$ is the result of the first throw and $y$ is the result of the second throw.

(i) Probability of getting an even number on the first throw and a prime number on the second throw:

Even numbers on a die are {2, 4, 6}. There are 3 even numbers.

Prime numbers on a die are {2, 3, 5}. There are 3 prime numbers.

The outcomes where the first throw is even and the second throw is prime are:

(2, 2), (2, 3), (2, 5)

(4, 2), (4, 3), (4, 5)

(6, 2), (6, 3), (6, 5)

There are $3 \times 3 = 9$ favourable outcomes.

The probability is the number of favourable outcomes divided by the total number of outcomes.

(ii) Probability that the sum of the numbers is less than or equal to 5:

We list the pairs $(x, y)$ such that $x + y \leq 5$:

If $x=1$: (1, 1), (1, 2), (1, 3), (1, 4) (sum = 2, 3, 4, 5)

If $x=2$: (2, 1), (2, 2), (2, 3) (sum = 3, 4, 5)

If $x=3$: (3, 1), (3, 2) (sum = 4, 5)

If $x=4$: (4, 1) (sum = 5)

If $x=5$: No possible value for y ($5+1=6 > 5$)

If $x=6$: No possible value for y

The favourable outcomes are: (1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1).

There are 10 favourable outcomes.

The probability is the number of favourable outcomes divided by the total number of outcomes.

(iii) Probability that at least one 6 appears:

It's easier to calculate the probability that no 6 appears and subtract it from 1.

For the first throw, there are 5 outcomes that are not 6 (1, 2, 3, 4, 5).

For the second throw, there are 5 outcomes that are not 6 (1, 2, 3, 4, 5).

The number of outcomes where no 6 appears is $5 \times 5 = 25$.

The probability of no 6 appearing is $\frac{25}{36}$.

The probability that at least one 6 appears is $1 - P(\text{no 6 appears})$.

Summary of Probabilities:

(i) Probability of an even number on the first throw and a prime number on the second throw = $\frac{1}{4}$

(ii) Probability that the sum of the numbers is less than or equal to 5 = $\frac{5}{18}$

(iii) Probability that at least one 6 appears = $\frac{11}{36}$

Statement of the Addition Theorem of Probability:

For any two events E and F, the probability of the occurrence of at least one of them (i.e., the probability of their union) is given by the sum of their individual probabilities minus the probability of their intersection.

Proof of the Addition Theorem of Probability:

Let S be the sample space with a finite number of equally likely outcomes. Let n(S) be the total number of outcomes in the sample space. Let n(E) be the number of outcomes in event E, n(F) be the number of outcomes in event F, and $n(E \cap F)$ be the number of outcomes in the intersection of E and F.

The probabilities are defined as:

We are interested in the probability of the union of E and F, $P(E \cup F)$. This corresponds to the number of outcomes in the union of E and F, $n(E \cup F)$.

Using the Principle of Inclusion-Exclusion for sets, we know that:

Now, we can find the probability of the union by dividing both sides of equation (5) by the total number of outcomes in the sample space, n(S):

We can split the right side of the equation:

Substituting the probability definitions from equations (2), (3), and (4) into equation (7):

This proves the Addition Theorem of Probability for any two events E and F.

Given:

Percentage of students studying Mathematics = 40%

Percentage of students studying Biology = 30%

Percentage of students studying both Mathematics and Biology = 10%

Let M be the event that a student studies Mathematics, and B be the event that a student studies Biology.

We are given the probabilities:

(i) Probability that the student studies Mathematics or Biology:

We need to find $P(M \cup B)$. Using the Addition Theorem of Probability:

So, the probability that the student studies Mathematics or Biology is 0.60 or 60%.

(ii) Probability that the student studies neither Mathematics nor Biology:

This corresponds to the complement of the event that the student studies Mathematics or Biology, i.e., $P((M \cup B)^c)$.

Using the rule for complements:

So, the probability that the student studies neither Mathematics nor Biology is 0.40 or 40%.

Summary of Probabilities:

(i) Probability that the student studies Mathematics or Biology = 0.60

(ii) Probability that the student studies neither Mathematics nor Biology = 0.40

Given:

Number of men = 7

Number of women = 4

Total number of people = $7 + 4 = 11$

A committee of 5 persons is to be formed.

Total number of ways to form the committee:

The total number of ways to choose 5 people from 11 is given by the combination formula $C(n, k) = \frac{n!}{k!(n-k)!}$.

(i) Probability that the committee consists of exactly 2 women:

To have exactly 2 women in the committee, we need to choose 2 women from 4 and 3 men from 7.

Number of ways to choose 2 women from 4 = $C(4, 2) = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} = \frac{4 \times 3}{2 \times 1} = 6$

Number of ways to choose 3 men from 7 = $C(7, 3) = \frac{7!}{3!(7-3)!} = \frac{7!}{3!4!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$

Number of ways to form the committee with exactly 2 women = $C(4, 2) \times C(7, 3) = 6 \times 35 = 210$

Probability (exactly 2 women) = $\frac{\text{Number of ways to form committee with 2 women}}{\text{Total number of ways to form the committee}}$

Simplify the fraction:

(ii) Probability that the committee consists of at least 3 men:

"At least 3 men" means either 3 men and 2 women, or 4 men and 1 woman, or 5 men and 0 women.

Case 1: 3 men and 2 women

Number of ways = $C(7, 3) \times C(4, 2) = 35 \times 6 = 210$ (calculated above)

Case 2: 4 men and 1 woman

Number of ways = $C(7, 4) \times C(4, 1)$

$C(7, 4) = C(7, 3) = 35$

$C(4, 1) = 4$

Number of ways = $35 \times 4 = 140$

Case 3: 5 men and 0 women

Number of ways = $C(7, 5) \times C(4, 0)$

$C(7, 5) = C(7, 2) = \frac{7 \times 6}{2 \times 1} = 21$

$C(4, 0) = 1$

Number of ways = $21 \times 1 = 21$

Total number of ways to have at least 3 men = $210 + 140 + 21 = 371$

Probability (at least 3 men) = $\frac{\text{Number of ways to form committee with at least 3 men}}{\text{Total number of ways to form the committee}}$

Simplify the fraction:

Check for common factors. 462 is divisible by 7 ($462 = 7 \times 66$). 371 is also divisible by 7 ($371 = 7 \times 53$).

Summary of Probabilities:

(i) Probability that the committee consists of exactly 2 women = $\frac{5}{11}$

(ii) Probability that the committee consists of at least 3 men = $\frac{53}{66}$

Given:

A standard pack of 52 cards.

Total number of outcomes: 52

(i) Probability that the card drawn is a king or a queen or a jack:

There are 4 kings, 4 queens, and 4 jacks in a deck of 52 cards.

These are mutually exclusive events (a card cannot be a king and a queen simultaneously).

Number of kings = 4

Number of queens = 4

Number of jacks = 4

Total number of favourable outcomes (king or queen or jack) = $4 + 4 + 4 = 12$

Probability (king or queen or jack) = $\frac{\text{Number of kings, queens, or jacks}}{\text{Total number of cards}}$

Simplify the fraction:

(ii) Probability that the card drawn is a red card or a face card:

Let R be the event of drawing a red card, and F be the event of drawing a face card.

Number of red cards = 26 (13 hearts and 13 diamonds)

Number of face cards = 12 (Jack, Queen, King for each of the 4 suits)

Number of cards that are both red and face cards: These are the Jack, Queen, and King of Hearts, and the Jack, Queen, and King of Diamonds. So, there are $3 + 3 = 6$ red face cards.

Using the Addition Theorem of Probability: $P(R \cup F) = P(R) + P(F) - P(R \cap F)$

$P(R) = \frac{26}{52}$

$P(F) = \frac{12}{52}$

$P(R \cap F) = \frac{6}{52}$

Simplify the fraction:

(iii) Probability that the card drawn is a diamond card or a king:

Let D be the event of drawing a diamond card, and K be the event of drawing a king.

Number of diamond cards = 13

Number of kings = 4

Number of cards that are both a diamond and a king: This is the King of Diamonds. So, there is 1 such card.

Using the Addition Theorem of Probability: $P(D \cup K) = P(D) + P(K) - P(D \cap K)$

$P(D) = \frac{13}{52}$

$P(K) = \frac{4}{52}$

$P(D \cap K) = \frac{1}{52}$

Simplify the fraction:

Summary of Probabilities:

(i) Probability that the card drawn is a king or a queen or a jack = $\frac{3}{13}$

(ii) Probability that the card drawn is a red card or a face card = $\frac{8}{13}$

(iii) Probability that the card drawn is a diamond card or a king = $\frac{4}{13}$

Property to Prove: $P(\text{not } E) = 1 - P(E)$

Proof:

Let S be the sample space and E be an event. The occurrence of E and the non-occurrence of E (not E, denoted as $E^c$) are complementary events. This means that every outcome in the sample space S either belongs to E or to $E^c$, and no outcome belongs to both.

Mathematically, we can express this relationship as:

And the intersection of E and $E^c$ is the empty set:

Now, let's consider the axioms of probability:

1. Axiom 1 (Non-negativity): For any event E, $P(E) \geq 0$.
2. Axiom 2 (Normalization): The probability of the sample space is 1, i.e., $P(S) = 1$.
3. Axiom 3 (Additivity for Mutually Exclusive Events): If $E_1, E_2, \dots, E_n$ are mutually exclusive events (i.e., $E_i \cap E_j = \emptyset$ for $i \neq j$), then $P(E_1 \cup E_2 \cup \dots \cup E_n) = P(E_1) + P(E_2) + \dots + P(E_n)$.

From Axiom 3, since E and $E^c$ are mutually exclusive (from equation (2)), we can write:

From equation (1), we know that $E \cup E^c = S$. Therefore, we can substitute S into equation (3):

From Axiom 2, we know that $P(S) = 1$. Substituting this into equation (4):

Rearranging equation (5) to solve for $P(E^c)$, which is $P(\text{not } E)$:

This proves the property $P(\text{not } E) = 1 - P(E)$ using the axioms of probability and the concept of complementary events.