Chapter 9 Probability (Q & A)

To find the probability of rolling a 6 on a fair six-sided die, we need to consider the number of favorable outcomes and the total number of possible outcomes.

A fair six-sided die has six possible outcomes when rolled: 1, 2, 3, 4, 5, and 6. Each of these outcomes is equally likely.

The total number of possible outcomes is 6.

The favorable outcome is rolling a 6. There is only one way to roll a 6.

The number of favorable outcomes is 1.

The probability of an event is calculated using the formula:

In this case, the probability of rolling a 6 is:

Therefore, the probability of rolling a 6 on a fair six-sided die is $1/6$.

Comparing this result with the given options:

(A) $1/6$

The correct option is (A).

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

Number of red balls = 5

Number of blue balls = 3

The total number of balls in the bag is the sum of the red balls and the blue balls:

Substituting the given values:

The number of favorable outcomes (drawing a blue ball) is the number of blue balls, which is 3.

The probability of an event is calculated as:

In this case, the probability of drawing a blue ball is:

Substituting the values:

Therefore, the probability of drawing a blue ball at random is $3/8$.

Comparing this with the given options:

(A) $3/8$

The correct option is (A).

The probability of an event is a measure of how likely that event is to occur. It is a fundamental concept in probability theory.

The probability of any event, denoted by $P(E)$, must satisfy certain conditions.

The lowest possible probability is 0. An event with a probability of 0 is considered an impossible event. For example, the probability of rolling a 7 on a standard six-sided die is 0.

The highest possible probability is 1. An event with a probability of 1 is considered a certain event. For example, the probability of rolling any number from 1 to 6 on a standard six-sided die is 1.

Probabilities cannot be negative, as you cannot have a negative chance of something happening. Similarly, probabilities cannot be greater than 1, as you cannot have more than a 100% chance of something happening.

Therefore, the range of possible values for the probability of an event $E$ is inclusive of 0 and 1. This can be expressed as:

This means that the probability of an event can be 0, 1, or any value in between 0 and 1.

Comparing this with the given options:

(A) $0 < P(E) < 1$ (This excludes the possibilities of an impossible or certain event.)

(B) $0 \leq P(E) \leq 1$ (This correctly includes the possibilities of impossible and certain events.)

(C) $P(E) \geq 0$ (This is true, but it doesn't set an upper bound.)

(D) $P(E) \leq 1$ (This is true, but it doesn't set a lower bound.)

The most comprehensive and correct range is $0 \leq P(E) \leq 1$.

The correct option is (B).

In probability theory, the value of the probability of an event provides information about its likelihood of occurring.

When the probability of an event is 0, it means that the event has absolutely no chance of happening under the given conditions.

Such an event is defined as an **impossible event**.

For example, rolling a 7 on a standard six-sided die is an impossible event, and its probability is 0.

Let's consider the other options to confirm why they are incorrect:

(A) "Sure" event is synonymous with a "certain" event.

(C) A "random" event is one whose outcome is uncertain and can vary.

(D) A "certain" event is an event that is guaranteed to happen, and its probability is 1.

Therefore, an event with a probability of 0 is called an impossible event.

Comparing this with the given options, the correct choice is (B).

In the study of probability, the value assigned to an event's likelihood ranges from 0 to 1.

When the probability of an event is 1, it signifies that the event is guaranteed to occur. There is no doubt that this event will happen.

An event that has a probability of 1 is known as a **sure event** or a **certain event**.

For instance, when rolling a standard six-sided die, the event of getting a number less than 7 is a certain event, as all possible outcomes (1, 2, 3, 4, 5, 6) satisfy this condition. The probability of this event is 1.

Let's review the provided options:

(A) An "impossible" event has a probability of 0.

(B) A "random" event is simply an event whose outcome is not predetermined.

(C) An "uncertain" event is a general term for an event whose outcome is not known, but doesn't specifically mean its probability is 1.

(D) A "sure or certain" event accurately describes an event with a probability of 1.

Therefore, an event with a probability of 1 is called a sure or certain event.

The correct option is (D).

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

Assertion (A): The probability of getting heads when flipping a coin is 0.5.

To determine the probability of getting heads when flipping a coin, we consider the basic principles of probability.

When a fair coin is flipped, there are two possible outcomes: Heads (H) or Tails (T). Each of these outcomes is equally likely.

The total number of possible outcomes is 2.

The number of favorable outcomes for getting heads is 1.

Using the probability formula $P(\text{Event}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$:

The value $1/2$ is equal to $0.5$. Therefore, Assertion (A) is true.

Reason (R): Assuming the coin is fair, there are two equally likely outcomes (Heads or Tails), and one of them is favourable (Heads).

This statement accurately describes the conditions for calculating the probability of getting heads when flipping a fair coin. A fair coin implies that each outcome has an equal chance of occurring. In this context, "favourable" refers to the specific outcome we are interested in, which is heads.

This statement correctly sets up the calculation for the probability of getting heads.

Therefore, Reason (R) is also true.

Now, we need to check if Reason (R) correctly explains Assertion (A).

The reason provided in (R) – that there are two equally likely outcomes (Heads or Tails) and one is favorable (Heads) – is precisely why the probability of getting heads is calculated as $1/2$ or $0.5$. The fairness of the coin and the identification of favorable and total outcomes are the basis for the assertion.

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

Based on this analysis, both A and R are true, and R is the correct explanation of A.

The correct option is (A).

In probability, the sum of the probabilities of all possible outcomes for an event must equal 1. In this case, the event is the weather tomorrow, and the two possible outcomes are that it rains or it does not rain.

Let $P(\text{Rain})$ be the probability that it will rain tomorrow.

Let $P(\text{No Rain})$ be the probability that it will NOT rain tomorrow.

According to the problem statement, the probability of rain tomorrow is 70%. To work with this in probability calculations, we convert the percentage to a decimal:

Since there are only two possible outcomes (rain or no rain), the sum of their probabilities must be 1:

We are given $P(\text{Rain}) = 0.7$. We need to find $P(\text{No Rain})$.

Substituting the known value into the equation:

To find $P(\text{No Rain})$, we subtract 0.7 from both sides of the equation:

$P(\text{No Rain}) = 0.3$

So, the probability that it will NOT rain tomorrow is 0.3.

Comparing this with the given options:

(A) 0.7 (This is the probability of rain, not no rain.)

(B) 0.3 (This is the correct probability of no rain.)

(C) 1 (This would mean it is certain not to rain, which is not indicated.)

(D) Cannot be determined (The probability can be determined from the given information.)

The correct option is (B).

In probability theory, the concept of complementary events is fundamental. A complementary event to an event E is an event that occurs if and only if E does not occur. This is often denoted as E' or $E^c$.

A key axiom of probability states that for any event E, the sum of the probability of E and the probability of its complement E' is always equal to 1. This can be expressed as:

This relationship means that the probability of an event not occurring is equal to 1 minus the probability of the event occurring.

Rearranging the formula, we get the probability of the complement:

In the given case study, the event E is rain, and its probability is $P(E) = 0.7$. The complement of E, denoted as E', is the event that it will NOT rain.

Using the formula for complementary events, the probability of not raining is:

Therefore, the probability of the complement of E (not raining), $P(E')$, is given by the formula $1 - P(E)$.

Let's evaluate the given options:

(A) $1 - P(E)$: This matches our derived formula for the probability of a complementary event.

(B) $P(E) + 1$: This would result in a probability greater than 1, which is impossible.

(C) $P(E) / 2$: This is incorrect; there is no general rule that the probability of a complement is half the probability of the event.

(D) $P(E')$ and $P(E)$ are independent: Complementary events are not independent; their probabilities are directly related.

The correct option is (A).

The probability of any event must fall within a specific range. Probabilities are used to quantify the likelihood of an event occurring, and by definition, this likelihood cannot be less than zero (impossible event) or greater than one (certain event).

The acceptable range for a probability value is $0 \leq P(E) \leq 1$. This means that any value between 0 and 1, inclusive, can be a valid probability.

Let's examine each of the given options:

(A) $0.25$: This value is between 0 and 1 ($0 \leq 0.25 \leq 1$). Therefore, $0.25$ can be a probability.

(B) $3/4$: This fraction is equivalent to $0.75$. Since $0 \leq 0.75 \leq 1$, $3/4$ can be a probability.

(C) $1.5$: This value is greater than 1 ($1.5 > 1$). Probabilities cannot exceed 1, as this would imply a certainty greater than 100% of an event occurring, which is not possible.

(D) $0$: This value is at the lower bound of the probability range ($0 \leq 0 \leq 1$). A probability of 0 indicates an impossible event.

Based on the fundamental rule that probabilities must be between 0 and 1, the value $1.5$ cannot be a probability.

The correct option is (C).

In probability theory, a **random experiment** is defined as a process or activity whose results cannot be predicted with certainty beforehand, even though all the possible outcomes of the experiment are known.

The key characteristics of a random experiment are:

1. Uncertainty of Outcome: It is impossible to know the specific outcome of a single trial of the experiment before it is conducted.
2. Known Set of Possible Outcomes: The set of all possible results of the experiment is known.

Let's consider some examples:

• Flipping a coin: We know the possible outcomes are heads or tails, but we cannot predict with certainty which one will occur on a single flip.
• Rolling a die: We know the possible outcomes are the numbers 1 through 6, but we cannot predict with certainty the number that will appear on a single roll.
• Drawing a card from a shuffled deck: We know all 52 cards in the deck, but we cannot predict with certainty which card will be drawn on a single draw.

The statement in the question perfectly matches this definition of a random experiment. The emphasis is on the unpredictability of a single outcome, coupled with the knowledge of all potential outcomes.

Let's evaluate the given options:

(A) True: This statement accurately defines a random experiment.

(B) False: This is incorrect because the statement correctly defines a random experiment.

(C) True, but only if the experiment is repeated many times.: While repeated trials help us estimate probabilities (as per the Law of Large Numbers), the definition of a random experiment itself does not depend on repetition. The uncertainty and known outcomes apply to a single trial.

(D) False, because the outcomes are unknown.: This is incorrect. The outcomes are known; it's the specific outcome of a single trial that is unknown.

Therefore, the statement provided is a correct definition of a random experiment.

The correct option is (A).

The **sample space** of a random experiment is the set of all possible outcomes.

In this case, the experiment is tossing a fair coin twice. We need to list all the possible sequences of outcomes for these two tosses.

Let 'H' represent the outcome of getting a Head, and 'T' represent the outcome of getting a Tail.

When a coin is tossed twice, the possible outcomes are:

• The first toss is Heads (H) and the second toss is Heads (H) $\rightarrow$ HH
• The first toss is Heads (H) and the second toss is Tails (T) $\rightarrow$ HT
• The first toss is Tails (T) and the second toss is Heads (H) $\rightarrow$ TH
• The first toss is Tails (T) and the second toss is Tails (T) $\rightarrow$ TT

Therefore, the sample space for tossing a fair coin twice is the set containing these four outcomes: $\{HH, HT, TH, TT\}$.

Let's look at the given options:

(A) $\{HH, TT\}$: This option only includes the outcomes where both tosses are the same, missing the mixed outcomes.

(B) $\{H, T\}$: This represents the sample space for tossing a coin only once, not twice.

(C) $\{HH, HT, TH, TT\}$: This option lists all four possible combinations of outcomes when tossing a coin twice.

(D) $\{HH, HT, TT\}$: This option is missing the outcome TH.

The correct sample space that includes all possible outcomes is $\{HH, HT, TH, TT\}$.

The correct option is (C).

The **sample space** of an experiment is the set of all possible outcomes. We are considering the experiment of rolling two fair six-sided dice simultaneously.

For a single fair six-sided die, there are 6 possible outcomes: {1, 2, 3, 4, 5, 6}.

When we roll two dice, the outcome of the first die is independent of the outcome of the second die. To find the total number of possible outcomes when two events occur independently, we multiply the number of outcomes for each event.

Number of outcomes for the first die = 6

Number of outcomes for the second die = 6

The total number of elements in the sample space (total number of possible outcomes) is the product of the number of outcomes for each die:

Substituting the values:

The sample space would consist of pairs of numbers $(d_1, d_2)$, where $d_1$ is the outcome of the first die and $d_2$ is the outcome of the second die. For example, some outcomes are (1,1), (1,2), (2,1), (6,6), etc. There are 36 such unique pairs.

Let's consider the given options:

(A) 12: This might be derived from adding the number of outcomes (6+6), but for independent events, we multiply.

(B) 36: This is the product of the number of outcomes for each die (6 x 6).

(C) 6: This is the number of outcomes for a single die, not two.

(D) 21: This number is not directly related to the sample space calculation in this scenario.

The number of elements in the sample space when two fair six-sided dice are rolled is 36.

The correct option is (B).

The **sample space** of a random experiment is the set of all possible outcomes of that experiment.

In this question, the experiment is drawing a single card from a well-shuffled deck of 52 playing cards.

A standard deck of 52 playing cards consists of 4 suits (Hearts, Diamonds, Clubs, Spades), and each suit has 13 ranks (Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King).

Since the deck is well-shuffled, each of the 52 cards has an equal chance of being drawn.

Therefore, the sample space consists of all the individual cards that can be drawn. The size of the sample space is simply the total number of distinct items that can be drawn.

In this case, the total number of cards in the deck is 52.

So, the size of the sample space is 52.

Let's examine the options:

(A) 13: This represents the number of ranks or the number of cards in each suit.

(B) 26: This represents the number of red cards (Hearts + Diamonds) or black cards (Clubs + Spades).

(C) 52: This is the total number of cards in a standard deck, which is the total number of possible outcomes.

(D) 4: This represents the number of suits.

The size of the sample space is the total number of distinct possible outcomes, which is 52.

The correct option is (C).

The **sample space** is the set of all possible outcomes of an experiment. In this case, the experiment involves two sequential events: flipping a coin and then rolling a fair six-sided die.

First, consider the coin flip. There are two possible outcomes:

• Heads (H)
• Tails (T)

Next, consider the roll of a fair six-sided die. There are six possible outcomes:

• 1
• 2
• 3
• 4
• 5
• 6

To find the total number of elements in the sample space for this combined experiment, we need to consider all possible combinations of the outcomes from each individual event. Since the coin flip and the die roll are independent events, we can find the total number of outcomes by multiplying the number of outcomes for each event.

Number of outcomes from flipping a coin = 2 (H, T)

Number of outcomes from rolling a die = 6 (1, 2, 3, 4, 5, 6)

The size of the sample space is calculated as:

Substituting the values:

The sample space would consist of ordered pairs, where the first element is the coin outcome and the second is the die outcome. For example, some outcomes are (H, 1), (H, 2), (T, 3), (T, 6), etc. There are 12 such unique pairs.

Let's evaluate the given options:

(A) 8: This is not the correct calculation.

(B) 12: This is the product of the number of outcomes for the coin flip (2) and the die roll (6).

(C) 36: This would be the sample space size for rolling two dice or flipping a coin four times ($2^4$).

(D) 6: This is the number of outcomes for rolling a die, not for the combined experiment.

The size of the sample space for flipping a coin and then rolling a die is 12.

The correct option is (B).

We need to evaluate both the Assertion (A) and the Reason (R) and then determine their relationship.

Assertion (A): The sample space for rolling a single die is $\{1, 2, 3, 4, 5, 6\}$.

A single die has six faces, numbered from 1 to 6. When a fair die is rolled, each of these numbers is a possible outcome. The sample space is defined as the set of all possible outcomes. Therefore, the sample space for rolling a single die is indeed $\{1, 2, 3, 4, 5, 6\}$. This assertion is true.

Reason (R): The sample space is the set of all possible outcomes of a random experiment.

This statement is the definition of a sample space in probability theory. It correctly states that the sample space encompasses every single outcome that can result from a particular random experiment. This reason is also true.

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

Assertion (A) provides a specific example of a sample space for a particular random experiment (rolling a single die). Reason (R) provides the general definition of what a sample space is.

The reason (R) explains *why* the set $\{1, 2, 3, 4, 5, 6\}$ is the sample space for rolling a single die. It's because this set contains all the possible outcomes of that random experiment, which is precisely what a sample space is defined to be.

Therefore, Reason (R) serves as the correct justification for Assertion (A).

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

The correct option is (A).

The problem describes a random experiment where two coins are tossed simultaneously. The sample space (S) for this experiment, which lists all possible outcomes, is given as $S = \{HH, HT, TH, TT\}$.

We are asked to find the outcomes that belong to event E, where E is defined as "getting at least one head".

Let's examine each outcome in the sample space to see if it satisfies the condition of having at least one head:

• HH: This outcome has two heads. Since it has more than one head, it certainly has "at least one head". So, HH is in E.
• HT: This outcome has one head and one tail. It satisfies the condition of having "at least one head". So, HT is in E.
• TH: This outcome has one tail and one head. It also satisfies the condition of having "at least one head". So, TH is in E.
• TT: This outcome has two tails. It does not have any heads, let alone "at least one head". So, TT is not in E.

Therefore, the outcomes that correspond to the event E (getting at least one head) are HH, HT, and TH.

The event E can be written as the set: $E = \{HH, HT, TH\}$.

Now, let's compare this with the given options:

(A) $\{HH, TT\}$: This includes TT, which does not have at least one head.

(B) $\{HH, HT, TH\}$: This set correctly includes all outcomes with at least one head.

(C) $\{HT, TH\}$: This set includes outcomes with exactly one head, but misses the outcome with two heads (HH).

(D) $\{HH\}$: This set only includes the outcome with two heads, missing the outcomes with exactly one head.

The correct set of outcomes for event E is $\{HH, HT, TH\}$.

The correct option is (B).

We are given the sample space for tossing two coins simultaneously as $S = \{HH, HT, TH, TT\}$. The total number of possible outcomes is the size of the sample space, denoted as $|S|$.

$|S| = 4$

From the previous question (Q16), we identified event E as "getting at least one head". The outcomes in event E are $E = \{HH, HT, TH\}$.

The number of outcomes favorable to event E is the size of event E, denoted as $|E|$.

$|E| = 3$

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

Substituting the values we found:

So, the probability of getting at least one head when tossing two coins is $3/4$.

Let's check the given options:

(A) $1/4$: This is the probability of getting no heads (TT).

(B) $1/2$: This would be the probability of getting exactly one head (HT or TH), but not including HH.

(C) $3/4$: This matches our calculated probability.

(D) $1$: This would mean it's a certain event, which it is not.

The correct option is (C).

In probability theory, the fundamental concepts are related as follows:

1. Experiment: A process or action that produces one or more outcomes.
2. Outcome: A single possible result of an experiment.
3. Sample Space (S): The set of *all* possible outcomes of an experiment.
4. Event (E): A collection of one or more outcomes from the sample space. In other words, an event is a subset of the sample space.

Consider an example: Rolling a single die.

• Experiment: Rolling a die.
• Possible Outcomes: 1, 2, 3, 4, 5, 6.
• Sample Space: $S = \{1, 2, 3, 4, 5, 6\}$.
• Event (e.g., getting an even number): $E = \{2, 4, 6\}$.

In this example, the event E ($ \{2, 4, 6\}$) is a subset of the sample space S ($ \{1, 2, 3, 4, 5, 6\}$).

Thus, an event is always defined as a subset of the sample space.

Let's look at the options:

(A) Probability: Probability is a numerical measure assigned to an event or outcome, not a set that an event is a subset of.

(B) Experiment: An event is not a subset of the experiment itself, but rather a collection of outcomes resulting from it.

(C) Sample space: This aligns with the definition and examples. An event is a collection of specific outcomes from the set of all possible outcomes (the sample space).

(D) Outcome: An event can contain one or more outcomes, but it is a collection of outcomes, not a subset of a single outcome.

Therefore, an event is a subset of the sample space.

The correct option is (C).

In probability theory, events are classified based on the number of outcomes they contain.

A **simple event** (also known as an elementary event) is defined as an event that consists of only one single outcome from the sample space.

For example, when rolling a fair six-sided die, the event of rolling a '3' is a simple event because it contains only one outcome: {3}. Similarly, the event of getting heads when flipping a coin is a simple event: {H}.

Let's consider the other classifications of events to understand why they are not simple events:

• Impossible Event: An event containing no outcomes (an empty set, $\emptyset$). Its probability is 0.
• Compound Event: An event containing more than one outcome. For example, rolling an even number on a die would be $\{2, 4, 6\}$.
• Certain Event: An event containing all possible outcomes in the sample space. Its probability is 1.

Based on the definition, a simple event contains exactly one outcome.

Evaluating the options:

(A) No outcomes: This describes an impossible event.

(B) Exactly one outcome: This accurately defines a simple event.

(C) More than one outcome: This describes a compound event.

(D) All possible outcomes: This describes a certain event.

Therefore, a simple event is an event containing exactly one outcome.

The correct option is (B).

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

A **simple event** is an event that consists of exactly one outcome from the sample space.

Let's analyze each option in the context of rolling a single die:

1. (A) Getting a 1: The event is $\{1\}$. This event contains exactly one outcome from the sample space. Therefore, it is a simple event.
2. (B) Getting a 6: The event is $\{6\}$. This event contains exactly one outcome from the sample space. Therefore, it is a simple event.
3. (C) Getting an odd number: The odd numbers in the sample space are 1, 3, and 5. So, the event is $\{1, 3, 5\}$. This event contains three outcomes. Since it contains more than one outcome, it is not a simple event; it is a compound event.
4. (D) Getting a 3: The event is $\{3\}$. This event contains exactly one outcome from the sample space. Therefore, it is a simple event.

The question asks which of the following is NOT a simple event.

Based on our analysis, "Getting an odd number" is the event that is not a simple event because it includes multiple outcomes.

The correct option is (C).

In probability theory, two events are considered **mutually exclusive** if they cannot happen at the same time. This means that if one event occurs, the other cannot.

Let's analyze the given options:

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

This condition defines **independent events**. For independent events, the occurrence of one does not affect the probability of the other. Mutually exclusive events are generally not independent (unless both have a probability of 0).

(B) They cannot occur simultaneously.

This is the direct definition of mutually exclusive events. If two events cannot happen at the same time, their intersection is empty, meaning there are no common outcomes. This is a correct definition of mutually exclusive events.

(C) $P(A \cup B) = P(A) + P(B)$

This equation is derived from the general addition rule for probabilities, $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. If events A and B are mutually exclusive, then their intersection is empty, meaning $P(A \cap B) = 0$. In this specific case, the general addition rule simplifies to $P(A \cup B) = P(A) + P(B)$. Therefore, this equation is a consequence of mutual exclusivity and is also a way to identify if events are mutually exclusive.

(D) Both (B) and (C) are correct if they cover the sample space.

Option (B) is the definition of mutually exclusive events. Option (C) is a consequence of mutual exclusivity and can also be used to identify them. However, the condition that they "cover the sample space" refers to events being **exhaustive**. If two mutually exclusive events also cover the sample space, they form a partition of the sample space, and their probabilities sum to 1 ($P(A) + P(B) = 1$). While true in that specific scenario, it's not the definition of mutual exclusivity itself.

The most fundamental and direct definition of mutually exclusive events is that they cannot occur simultaneously.

Option (B) is the direct definition. Option (C) is a property that arises from mutual exclusivity, and if also exhaustive, their probabilities sum to 1.

The question asks what it *means* for two events to be mutually exclusive. The most accurate answer is that they cannot occur simultaneously.

The correct option is (B).

In probability theory, two events are considered **exhaustive** if at least one of them must occur. This means that the union of the two events covers the entire sample space, encompassing all possible outcomes of the experiment.

Let's analyze the given options:

(A) $A \cup B = S$ (the sample space)

This statement means that the combination of outcomes in event A and event B together includes every possible outcome of the experiment. This is the direct definition of exhaustive events. So, this option is correct.

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

This statement means that events A and B are mutually exclusive, i.e., they cannot occur at the same time. Mutual exclusivity is a different property from exhaustiveness, although mutually exclusive events can also be exhaustive if they cover the entire sample space.

(C) $P(A \cup B) = 1$

The probability of the union of events A and B represents the probability that either A or B or both occur. If the events are exhaustive, then their union accounts for all possible outcomes, and the probability of the union must be 1 (representing certainty). This statement is a consequence of exhaustiveness and is also a way to check for it.

(D) Both (A) and (C)

Since both statement (A) (the definition) and statement (C) (a derived property that can be used to check for exhaustiveness) are correct ways to describe exhaustive events, this option appears to be the most comprehensive answer.

If $A \cup B = S$, then it logically follows that $P(A \cup B) = P(S) = 1$. Conversely, if $P(A \cup B) = 1$, it implies that the union of A and B covers all possible outcomes, meaning $A \cup B = S$ (assuming A and B are indeed events derived from the sample space S).

Therefore, both (A) and (C) correctly describe exhaustive events.

The correct option is (D).

We are considering the experiment of rolling a single fair six-sided die. The sample space is $S = \{1, 2, 3, 4, 5, 6\}$.

Let A be the event of getting an even number. The outcomes for event A are $\{2, 4, 6\}$.

Let B be the event of getting an odd number. The outcomes for event B are $\{1, 3, 5\}$.

We need to determine if these events are mutually exclusive and exhaustive.

1. Mutually Exclusive:

Two events are mutually exclusive if they cannot occur at the same time, meaning their intersection is empty ($A \cap B = \emptyset$).

In this case, the outcomes in A are $\{2, 4, 6\}$ and the outcomes in B are $\{1, 3, 5\}$. There are no common outcomes between A and B. Therefore, $A \cap B = \emptyset$.

This means that events A and B are mutually exclusive.

2. Exhaustive:

Two events are exhaustive if their union covers the entire sample space ($A \cup B = S$).

Let's find the union of A and B: $A \cup B = \{2, 4, 6\} \cup \{1, 3, 5\} = \{1, 2, 3, 4, 5, 6\}$.

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

Since $A \cup B = S$, the events A and B are exhaustive.

Conclusion:

Events A and B are both mutually exclusive and exhaustive.

Let's review the options:

(A) Mutually exclusive, but not exhaustive. (Incorrect, they are exhaustive.)

(B) Exhaustive, but not mutually exclusive. (Incorrect, they are mutually exclusive.)

(C) Neither mutually exclusive nor exhaustive. (Incorrect, they are both.)

(D) Both mutually exclusive and exhaustive. (Correct.)

The correct option is (D).

The probability of the union of two events, $P(A \cup B)$, represents the probability that either event A occurs, or event B occurs, or both occur.

This is governed by the **General Addition Rule** in probability theory.

The general addition rule accounts for the possibility that both events A and B might occur simultaneously. The probability of both events occurring is denoted by the intersection of A and B, $P(A \cap B)$.

The formula is derived using the principle of inclusion-exclusion. If we simply add $P(A)$ and $P(B)$, we would be counting the outcomes that are common to both A and B twice (once in $P(A)$ and once in $P(B)$). Therefore, we must subtract the probability of their intersection, $P(A \cap B)$, to correct for this double counting.

The formula for the probability of the union of two events A and B is:

Let's examine the given options:

(A) $P(A) + P(B)$: This formula is only correct if events A and B are mutually exclusive (meaning $P(A \cap B) = 0$), which is not stated in the question.

(B) $P(A) + P(B) - P(A \cap B)$: This is the correct general formula for the probability of the union of two events.

(C) $P(A) \times P(B)$: This formula is used to calculate the probability of the intersection of two independent events, $P(A \cap B)$.

(D) $P(A) - P(B)$: This expression does not represent a standard probability formula for the union or intersection of events.

Therefore, the correct formula for the probability of A or B occurring is $P(A) + P(B) - P(A \cap B)$.

The correct option is (B).

We need to evaluate the Assertion (A) and the Reason (R) and their relationship.

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

The general addition rule for probabilities states that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. If events A and B are mutually exclusive, it means they cannot occur at the same time, so their intersection is the empty set, $A \cap B = \emptyset$. The probability of an empty set is 0, i.e., $P(A \cap B) = 0$. Substituting this into the general addition rule gives $P(A \cup B) = P(A) + P(B) - 0$, which simplifies to $P(A \cup B) = P(A) + P(B)$. Thus, Assertion (A) is true.

Reason (R): For mutually exclusive events, $A \cap B = \emptyset$, so $P(A \cap B) = 0$.

This statement correctly defines the condition for mutually exclusive events ($A \cap B = \emptyset$) and correctly states that the probability of an empty set is 0. This is a fundamental property used in probability theory. Thus, Reason (R) is true.

Now, we assess if Reason (R) explains Assertion (A).

Reason (R) provides the justification for why the formula $P(A \cup B) = P(A) + P(B)$ holds true for mutually exclusive events. It explains that the absence of common outcomes ($A \cap B = \emptyset$) leads to a zero probability of the intersection ($P(A \cap B) = 0$), which is the term that is removed from the general addition rule to arrive at the simplified formula for mutually exclusive events.

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

Both A and R are true, and R correctly explains A.

The correct option is (A).

We are drawing a single card from a standard deck of 52 playing cards. The sample space S consists of all 52 cards, so $|S| = 52$.

Let A be the event of drawing a King.

There are four Kings in a deck: King of Hearts, King of Diamonds, King of Clubs, and King of Spades. So, the number of outcomes in event A is $|A| = 4$.

Let B be the event of drawing a Spade.

There are 13 Spades in a deck (Ace of Spades, 2 of Spades, ..., King of Spades). So, the number of outcomes in event B is $|B| = 13$.

We need to find $P(A \cap B)$, which is the probability that the drawn card is both a King AND a Spade.

The event $A \cap B$ represents the outcomes that are common to both event A (drawing a King) and event B (drawing a Spade).

We need to identify the cards that are both Kings and Spades.

There is only one card in the deck that fits this description: the King of Spades.

So, the number of outcomes in the intersection of A and B is $|A \cap B| = 1$.

The probability of the intersection of A and B is calculated as:

Substituting the values:

Let's check the options:

(A) $4/52$: This is $P(A)$, the probability of drawing a King.

(B) $13/52$: This is $P(B)$, the probability of drawing a Spade.

(C) $1/52$: This is the probability of drawing a card that is both a King and a Spade.

(D) $17/52$: This would be the probability of drawing a King OR a Spade if they were mutually exclusive ($P(A)+P(B)$), which they are not.

The correct option is (C).

We are drawing a single card from a standard deck of 52 playing cards. The sample space S has $|S| = 52$.

Event A: Drawing a King. There are 4 Kings, so $P(A) = 4/52$.

Event B: Drawing a Spade. There are 13 Spades, so $P(B) = 13/52$.

From the previous question (Q26), we found that $A \cap B$ is the event of drawing the King of Spades, and $P(A \cap B) = 1/52$.

We need to find $P(A \cup B)$, which is the probability of drawing a King OR a Spade (or both).

We use the general addition rule for probabilities: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.

Let's substitute the probabilities we know:

Now, let's simplify the expression:

This fraction can be simplified by dividing both the numerator and denominator by their greatest common divisor, which is 4:

Now let's look at the options provided:

(A) $4/52 + 13/52$: This is $P(A) + P(B)$, which would be correct if A and B were mutually exclusive, but they are not.

(B) $4/52 \times 13/52$: This is the formula for $P(A) \times P(B)$, which would be used for the probability of independent events if we were calculating something else, not $P(A \cup B)$.

(C) $4/52 + 13/52 - 1/52$: This exactly matches the formula $P(A) + P(B) - P(A \cap B)$ with the specific probabilities for this case.

(D) $17/52$: This is the result of $P(A) + P(B)$ without subtracting $P(A \cap B)$, which is incorrect because the events are not mutually exclusive.

Therefore, the correct expression for $P(A \cup B)$ is $4/52 + 13/52 - 1/52$.

The correct option is (C).

Conditional probability, denoted as $P(A|B)$, quantifies the probability of an event A occurring, given that another event B has already occurred. This concept effectively narrows down the sample space to only those outcomes where event B has taken place.

To derive the formula for $P(A|B)$, we start by considering the intersection of events A and B, which is $A \cap B$. This intersection represents the outcomes where both A and B occur.

When we know that event B has occurred, our new sample space is effectively reduced to the outcomes within event B. The probability of event A occurring within this reduced sample space is the proportion of outcomes in $A \cap B$ relative to the total outcomes in B.

Therefore, the formula for conditional probability is:

Similarly, the probability of B occurring given that A has already occurred is $P(B|A) = P(A \cap B) / P(A)$ (for $P(A) > 0$).

Let's look at the given options:

(A) $P(A \cap B) / P(B)$: This is the correct formula for $P(A|B)$.

(B) $P(A \cup B) / P(B)$: This is not the standard formula for conditional probability.

(C) $P(A \cap B) / P(A)$: This is the formula for $P(B|A)$, the probability of B given A.

(D) $P(A) \times P(B)$: This is the formula for the probability of the intersection of two independent events, $P(A \cap B)$.

The correct formula for conditional probability $P(A|B)$ is $P(A \cap B) / P(B)$.

The correct option is (A).

Two events are considered **independent** if the occurrence or non-occurrence of one event does not change the probability of the other event occurring.

Let's analyze the given options based on this definition and the standard formulas in probability:

(A) The occurrence of one does not affect the probability of the other.

This is the fundamental definition of independent events. It accurately describes the relationship between two independent events.

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

This equation is incorrect. The formula for the union of two events is $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. The sum of probabilities $P(A) + P(B)$ doesn't typically represent the intersection unless specific conditions are met which are not related to independence.

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

This statement implies that the probability of A given B is the same as the probability of B given A. While this can be true in some specific cases, it is not the definition of independence. The definition of independence is related to how the occurrence of one event affects the probability of the other, not necessarily equal conditional probabilities.

(D) They are mutually exclusive.

Mutually exclusive events cannot occur at the same time, meaning $P(A \cap B) = 0$. Independent events, on the other hand, generally have a non-zero probability of intersection if $P(A) > 0$ and $P(B) > 0$. If two events are mutually exclusive and their probabilities are greater than zero, they cannot be independent.

The defining characteristic of independent events is that the probability of one event occurring is not influenced by whether the other event has occurred.

The correct option is (A).

A mathematically equivalent condition for independence is $P(A \cap B) = P(A) \times P(B)$, assuming $P(A) > 0$ and $P(B) > 0$. However, option (A) provides the conceptual definition.

Two events A and B are defined as **independent** if the occurrence of one event does not affect the probability of the other event occurring. Mathematically, this independence is expressed through a specific relationship between their probabilities.

The key formula that defines independence is that the probability of the intersection of two independent events (i.e., the probability that both A and B occur) is equal to the product of their individual probabilities.

This relationship is given by:

Let's review the provided options:

(A) $P(A) + P(B)$: This formula is related to the probability of the union of mutually exclusive events, $P(A \cup B)$, when $P(A \cap B) = 0$.

(B) $P(A) \times P(B)$: This is the correct formula for the probability of the intersection of two independent events.

(C) $P(A) - P(B)$: This expression does not represent a standard probability formula for the intersection of events.

(D) $P(A|B)$: This represents the conditional probability of A given B, which is $P(A \cap B) / P(B)$. While related to independence (if independent, $P(A|B) = P(A)$), it is not the formula for $P(A \cap B)$ itself.

Therefore, if events A and B are independent, the probability of both A and B occurring, $P(A \cap B)$, is equal to the product of their individual probabilities, $P(A) \times P(B)$.

The correct option is (B).

This problem involves calculating the probability of two dependent events occurring in sequence without replacement.

First, let's determine the initial state of the bag:

Number of red balls = 4

Number of blue balls = 6

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

We want to find the probability that the first ball drawn is red AND the second ball drawn is blue.

Step 1: Probability that the first ball is red.

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

Step 2: Probability that the second ball is blue, given the first was red and NOT replaced.

After drawing one red ball and not replacing it, the contents of the bag change:

Number of red balls remaining = $4 - 1 = 3$

Number of blue balls remaining = 6 (since a red ball was drawn)

Total number of balls remaining in the bag = $3 + 6 = 9$.

Now, the probability of drawing a blue ball as the second ball, given that the first ball was red, is:

Step 3: Probability of both events happening.

To find the probability that the first ball is red AND the second ball is blue, we multiply the probabilities of these two dependent events:

Substituting the probabilities calculated in Step 1 and Step 2:

Now, let's compare this result with the given options:

(A) $(4/10) \times (6/10)$: This would be correct if the first ball was replaced.

(B) $(4/10) + (6/9)$: This uses addition, which is for 'or' probabilities, not 'and' probabilities.

(C) $(4/10) \times (6/9)$: This matches our calculated probability.

(D) $(6/10) \times (4/9)$: This would be the probability of drawing a blue ball first, then a red ball.

The correct expression is $(4/10) \times (6/9)$.

The correct option is (C).

Let's examine the Assertion (A) and the Reason (R) individually and then their relationship.

Assertion (A): If $P(A|B) = P(A)$, then events A and B are independent.

The definition of conditional probability is $P(A|B) = \frac{P(A \cap B)}{P(B)}$ (assuming $P(B) > 0$).

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

Substituting this into the conditional probability formula:

If $P(B) > 0$, we can cancel out $P(B)$: $P(A|B) = P(A)$.

Conversely, if $P(A|B) = P(A)$, then $\frac{P(A \cap B)}{P(B)} = P(A)$, which implies $P(A \cap B) = P(A) \times P(B)$, the condition for independence. Thus, Assertion (A) is true.

Reason (R): The definition of independence is that the conditional probability of A given B is equal to the probability of A.

This statement is a direct consequence of the definition of independence. If A and B are independent, the knowledge that B has occurred does not change the probability of A occurring, so $P(A|B)$ should be equal to $P(A)$. This is a valid interpretation and often used as an alternative definition or test for independence. Thus, Reason (R) is also true.

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

Reason (R) states that the equality $P(A|B) = P(A)$ *is* the definition of independence. Assertion (A) states that if this equality holds, then A and B are independent. Therefore, Reason (R) provides the definition that underpins Assertion (A).

In essence, Assertion (A) presents a condition, and Reason (R) explains what that condition means in terms of the definition of independence.

Both A and R are true, and R correctly explains A.

The correct option is (A).

We are given the following information from a survey of 100 students:

Total number of students = 100.

Number of students who like Mathematics, $|M| = 60$.

Number of students who like Physics, $|P| = 40$.

Number of students who like both Mathematics and Physics, $|M \cap P| = 20$.

We need to find the conditional probability $P(P|M)$, which is the probability that a randomly selected student likes Physics given that they already like Mathematics.

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

First, let's calculate $P(M)$ and $P(M \cap P)$ from the given data:

$P(M) = \frac{\text{Number of students who like Mathematics}}{\text{Total number of students}} = \frac{|M|}{\text{Total}} = \frac{60}{100}$

$P(M \cap P) = \frac{\text{Number of students who like both M and P}}{\text{Total number of students}} = \frac{|M \cap P|}{\text{Total}} = \frac{20}{100}$

Now, we can substitute these values into the conditional probability formula:

To simplify this complex fraction, we can multiply the numerator by the reciprocal of the denominator:

Alternatively, we can think of this problem by directly considering the reduced sample space. Given that a student likes Mathematics, we are only concerned with the 60 students who like Mathematics. Out of these 60 students, we want to know how many also like Physics.

Number of students who like Mathematics (our new sample space) = 60.

Number of students who like Physics AND Mathematics = 20.

So, the probability of liking Physics given they like Mathematics is:

Now let's look at the options:

(A) $20/60$: This matches our calculated probability.

(B) $20/40$: This would be $P(M|P)$, the probability of liking Mathematics given they like Physics.

(C) $60/100$: This is $P(M)$, the probability of liking Mathematics.

(D) $40/100$: This is $P(P)$, the probability of liking Physics.

The correct option is (A).

We are given the following information:

Total students = 100

$|M| = 60$ (likes Mathematics)

$|P| = 40$ (likes Physics)

$|M \cap P| = 20$ (likes both)

To determine if the events "liking Mathematics" (M) and "liking Physics" (P) are independent, we need to check if the condition for independence holds true. The primary condition for independence is $P(M \cap P) = P(M) \times P(P)$.

Let's calculate the probabilities:

$P(M) = \frac{|M|}{\text{Total}} = \frac{60}{100}$

$P(P) = \frac{|P|}{\text{Total}} = \frac{40}{100}$

$P(M \cap P) = \frac{|M \cap P|}{\text{Total}} = \frac{20}{100}$

Now, let's check the independence condition $P(M \cap P) = P(M) \times P(P)$.

Calculate the product of the individual probabilities:

Now, compare this product with the actual probability of the intersection:

$P(M \cap P) = \frac{20}{100}$

$P(M) \times P(P) = \frac{24}{100}$

Since $\frac{20}{100} \neq \frac{24}{100}$, we have $P(M \cap P) \neq P(M) \times P(P)$.

Therefore, the events "liking Mathematics" and "liking Physics" are not independent.

Let's evaluate the given options:

(A) Yes, because $P(M \cap P) = P(M) \times P(P)$. (Incorrect, the equality does not hold.)

(B) No, because $P(M \cap P) \neq P(M) \times P(P)$. (Correct, this is why they are not independent.)

(C) Yes, because some students like both. (Incorrect. The fact that there's an intersection ($P(M \cap P) > 0$) does not imply independence.)

(D) No, because $P(M|P) \neq P(M)$. Let's check this condition: $P(M|P) = \frac{P(M \cap P)}{P(P)} = \frac{20/100}{40/100} = \frac{20}{40} = \frac{1}{2} = \frac{50}{100}$. Since $P(M) = 60/100$, $P(M|P) \neq P(M)$. This is also a valid reason for them not being independent. However, option (B) directly tests the primary definition $P(M \cap P) = P(M) \times P(P)$. Option (B) is the most direct test of independence using the product rule.

Option (B) directly uses the definition of independence involving the product of probabilities.

The correct option is (B).

The Law of Total Probability provides a way to calculate the probability of an event B by considering all the possible ways B can occur through a set of mutually exclusive and exhaustive events. The formula stated is $P(B) = \sum\limits_{i=1}^n P(B|A_i) P(A_i)$.

Let's break down the conditions required for this formula:

1. $P(B|A_i)$: This is the conditional probability of event B occurring given that event $A_i$ has occurred. This term is well-defined only if $P(A_i) > 0$ for all $i$.
2. $P(A_i)$: This is the probability of event $A_i$.
3. $\sum\limits_{i=1}^n P(A_i) = 1$: The sum of the probabilities of all events $A_i$ must equal 1. This implies that the $A_i$ are exhaustive, meaning they cover all possibilities in the sample space.
4. $A_i \cap A_j = \emptyset$ for $i \neq j$: The events $A_i$ must be mutually exclusive, meaning that no two events $A_i$ and $A_j$ can occur at the same time.

A set of events $\{A_1, A_2, \dots, A_n\}$ that are mutually exclusive and exhaustive is called a **partition** of the sample space.

Now let's examine the given options in light of these requirements:

(A) $A_i$ must be independent events. Independence of $A_i$ from each other is not a requirement for the Law of Total Probability. In fact, for the law to apply, the $A_i$ must be mutually exclusive, which means they are generally not independent (unless their probabilities are 0).

(B) $P(A_i)$ must be zero for all $i$. This is incorrect. For the conditional probability $P(B|A_i)$ to be defined, $P(A_i)$ must be greater than zero. Also, if all $P(A_i)$ were zero, their sum could not be 1.

(C) $A_i$ must form a partition of the sample space. This means the events $A_i$ must be mutually exclusive (no overlap) and exhaustive (cover all possibilities). This is exactly the condition required for the Law of Total Probability to be applied correctly.

(D) $P(B|A_i)$ must be 1 for all $i$. This would imply that if any $A_i$ occurs, B is certain to occur. This is a very specific scenario and not a general requirement for the law. The $P(B|A_i)$ values can vary.

The core requirement for the Law of Total Probability is that the events $A_i$ form a partition of the sample space.

The correct option is (C).

This problem can be solved using the Law of Total Probability. We have two bags, and a bag is chosen at random. After choosing a bag, a ball is drawn from it.

Let $B_1$ be the event that Bag 1 is chosen.

Let $B_2$ be the event that Bag 2 is chosen.

Let R be the event of drawing a red ball.

Since a bag is chosen at random, the probability of choosing either bag is equal:

$P(B_1) = 1/2$

$P(B_2) = 1/2$

Now, let's find the probability of drawing a red ball from each bag:

Bag 1: Contains 3 red balls and 2 blue balls. Total balls = $3 + 2 = 5$.

The probability of drawing a red ball from Bag 1 is:

Bag 2: Contains 2 red balls and 4 blue balls. Total balls = $2 + 4 = 6$.

The probability of drawing a red ball from Bag 2 is:

According to the Law of Total Probability, the probability of event R (drawing a red ball) is the sum of the probabilities of R occurring through each of the mutually exclusive and exhaustive events $B_1$ and $B_2$.

The formula is: $P(R) = P(R|B_1)P(B_1) + P(R|B_2)P(B_2)$

Substituting the values we calculated:

This can also be written as:

Now, let's compare this with the given options:

(A) $(1/2) \times (3/5) + (1/2) \times (2/6)$: This exactly matches the formula derived from the Law of Total Probability.

(B) $(3/5) + (2/6)$: This simply adds the probabilities of drawing a red ball from each bag, ignoring the fact that a bag is chosen at random.

(C) $(3/5) \times (2/6)$: This multiplies the probabilities of drawing a red ball from each bag, which is not the correct operation for this scenario.

(D) $1/2 \times (3/5 + 2/6)$: This incorrectly adds the conditional probabilities first and then multiplies by 1/2. The multiplication by $P(B_i)$ should be applied to each conditional probability $P(R|B_i)$ separately.

The correct expression for the probability of drawing a red ball is $(1/2) \times (3/5) + (1/2) \times (2/6)$.

The correct option is (A).

This problem involves using the Law of Total Probability. We want to find the overall probability that an item is flagged as defective. An item can be flagged as defective in two mutually exclusive ways: it is actually defective and flagged correctly, or it is not defective and flagged incorrectly.

Let D be the event that an item is defective.

Let ND be the event that an item is non-defective (the complement of D).

Let F be the event that an item is flagged as defective.

From the problem statement:

Defect rate = 2%, so $P(D) = 0.02$.

This means the probability of an item being non-defective is $P(ND) = 1 - P(D) = 1 - 0.02 = 0.98$.

The inspector correctly identifies 95% of defective items as defective. This is the conditional probability of being flagged given it's defective: $P(F|D) = 0.95$.

The inspector incorrectly flags 1% of non-defective items as defective. This is the conditional probability of being flagged given it's non-defective: $P(F|ND) = 0.01$.

We want to find the probability that an item is flagged as defective, $P(F)$. We can use the Law of Total Probability, where the sample space is partitioned by the events D (defective) and ND (non-defective).

The Law of Total Probability states: $P(F) = P(F|D)P(D) + P(F|ND)P(ND)$.

Now let's analyze the options:

(A) $P(\text{Defective and Flagged}) + P(\text{Non-defective and Flagged})$

The probability of "Defective and Flagged" is $P(F \cap D)$. Using the formula for conditional probability, $P(F \cap D) = P(F|D)P(D)$.

The probability of "Non-defective and Flagged" is $P(F \cap ND)$. Using the formula for conditional probability, $P(F \cap ND) = P(F|ND)P(ND)$.

The event of being flagged as defective (F) is the union of two mutually exclusive events: being defective and flagged ($F \cap D$), and being non-defective and flagged ($F \cap ND$). Therefore, $P(F) = P(F \cap D) + P(F \cap ND)$.

So, option (A) is a correct formulation for $P(F)$.

(B) $P(\text{Flagged}|\text{Defective})P(\text{Defective}) + P(\text{Flagged}|\text{Non-defective})P(\text{Non-defective})$

Substituting the event names with our notation:

$P(F|D)P(D) + P(F|ND)P(ND)$

This is exactly the Law of Total Probability for event F based on the partition {D, ND}. So, option (B) is also a correct formulation.

(C) $0.95 \times 0.02 + 0.01 \times 0.98$

This option directly substitutes the values from the problem into the formula from option (B):

$P(F|D) = 0.95$, $P(D) = 0.02$

$P(F|ND) = 0.01$, $P(ND) = 0.98$

So, $0.95 \times 0.02 + 0.01 \times 0.98$ is the numerical calculation based on the correct formula.

This option represents the calculation of $P(F \cap D) + P(F \cap ND)$, which equals $P(F)$. So, option (C) is also a correct formulation.

(D) All of the above that are correct formulations.

Since options (A), (B), and (C) are all correct formulations or calculations of the probability that a randomly selected item is flagged as defective, this option is the most appropriate answer.

The correct option is (D).

Let's evaluate Assertion (A) and Reason (R) separately and then consider their relationship.

Assertion (A): The sum of probabilities of mutually exclusive and exhaustive events is 1.

If a set of events $\{E_1, E_2, \dots, E_n\}$ are mutually exclusive, it means that $P(E_i \cap E_j) = 0$ for all $i \neq j$. If they are also exhaustive, it means their union covers the entire sample space, $E_1 \cup E_2 \cup \dots \cup E_n = S$.

The probability of the union of mutually exclusive events is $P(E_1 \cup E_2 \cup \dots \cup E_n) = P(E_1) + P(E_2) + \dots + P(E_n)$.

Since these events are exhaustive, their union is the sample space S, and $P(S) = 1$. Therefore, the sum of their probabilities must be 1: $P(E_1) + P(E_2) + \dots + P(E_n) = 1$.

Thus, Assertion (A) is true.

Reason (R): The union of mutually exclusive and exhaustive events covers the entire sample space, and the probability of the sample space is 1.

This statement correctly explains why the sum of probabilities of mutually exclusive and exhaustive events is 1. The first part, "The union of mutually exclusive and exhaustive events covers the entire sample space," is the definition of exhaustive events. The second part, "and the probability of the sample space is 1," is a fundamental axiom of probability.

These two facts together directly lead to the conclusion that the sum of the probabilities of such events must be 1.

Thus, Reason (R) is true and provides a correct explanation.

Since Assertion (A) is true and Reason (R) correctly explains why Assertion (A) is true, the relationship between them is that R explains A.

The correct option is (A).

The problem states that there are three bags, and one of them is selected at random. This means that each bag has an equal chance of being chosen.

Let $B_1$ be the event of selecting Bag 1.

Let $B_2$ be the event of selecting Bag 2.

Let $B_3$ be the event of selecting Bag 3.

Since the selection of the bag is random and there are three bags, the probability of selecting any specific bag is equal.

The total number of bags to choose from is 3.

The number of favorable outcomes for selecting Bag 1 is 1.

The probability of selecting Bag 1 is calculated as:

Substituting the values:

The information about the contents of the bags (number of red and black balls) is not needed to answer this specific question about the probability of selecting a particular bag.

Let's examine the options:

(A) $1/3$: This is the correct probability of selecting one bag out of three at random.

(B) $1/5$: This might relate to the number of balls in Bag 1, but not the probability of selecting the bag.

(C) $2/5$: This might relate to the number of red balls in Bag 1 out of the total balls in Bag 1.

(D) $1/2$: This would be correct if there were only two bags.

The probability of selecting Bag 1 is $1/3$.

The correct option is (A).

We are given information about three bags:

Bag 1: 2 red balls, 3 black balls (Total = 5 balls)

Bag 2: 4 red balls, 1 black ball (Total = 5 balls)

Bag 3: 3 red balls, 4 black balls (Total = 7 balls)

We are asked to find the probability of drawing a red ball given that Bag 2 was selected. This is a conditional probability.

Let R be the event of drawing a red ball.

Let $B_1$, $B_2$, and $B_3$ be the events of selecting Bag 1, Bag 2, and Bag 3, respectively.

We want to find $P(R|B_2)$. This means we are only considering the scenario where Bag 2 has been selected.

In Bag 2, there are:

Number of red balls = 4

Number of black balls = 1

Total number of balls in Bag 2 = $4 + 1 = 5$.

The probability of drawing a red ball from Bag 2 is the number of red balls in Bag 2 divided by the total number of balls in Bag 2:

Substituting the values:

Now, let's compare this with the given options:

(A) $1/3$: This is the probability of selecting Bag 1.

(B) $4/5$: This is the probability of drawing a red ball from Bag 2.

(C) $4/10$: This is not relevant to the given question.

(D) $1/5$: This is not relevant to the given question.

The probability of drawing a red ball given that Bag 2 was selected is $4/5$.

The correct option is (B).

Bayes' Theorem is a fundamental concept in probability and statistics that describes the probability of an event based on prior knowledge of conditions that might be related to the event. It is particularly useful for updating probabilities when new evidence becomes available.

Bayes' Theorem allows us to relate conditional probabilities in different directions. If we have the probability of event A given event B ($P(A|B)$) and the probability of event B given event A ($P(B|A)$), along with the marginal probabilities $P(A)$ and $P(B)$, Bayes' Theorem provides a way to calculate the "inverse" or "posterior" probability.

Specifically, Bayes' Theorem is typically stated as:

In this formula:

• $P(A|B)$ is the **posterior probability**: the probability of A given the new evidence B.
• $P(B|A)$ is the **likelihood**: the probability of the evidence B given A.
• $P(A)$ is the **prior probability**: the initial probability of A before observing the evidence B.
• $P(B)$ is the **marginal probability of the evidence**.

The theorem's primary use is to update our beliefs (prior probabilities) in light of new data to arrive at updated beliefs (posterior probabilities).

Let's review the options:

(A) Marginal probability: Marginal probability ($P(A)$ or $P(B)$) is often an input to Bayes' Theorem, not what it calculates.

(B) Joint probability: Joint probability ($P(A \cap B)$) can be calculated using Bayes' Theorem indirectly ($P(A \cap B) = P(A|B)P(B)$ or $P(A \cap B) = P(B|A)P(A)$), but it's not the direct output or the primary focus.

(C) Prior probability: Prior probability is the starting point, the initial belief before new evidence is considered.

(D) Posterior or inverse probability: Bayes' Theorem is used to calculate the posterior probability, which is the updated probability of an event after considering new evidence. It's often referred to as "inverse" probability because it allows us to infer from effect to cause or update beliefs in an inverse direction of the direct conditional probability.

Therefore, Bayes' Theorem is used to calculate posterior or inverse probability.

The correct option is (D).

Bayes' Theorem provides a way to calculate a conditional probability, specifically the probability of an event A given that event B has occurred ($P(A|B)$), by relating it to other probabilities.

The theorem is derived from the definition of conditional probability and the multiplication rule.

We know the definition of conditional probability:

We also know that the probability of the intersection $P(A \cap B)$ can be expressed in two ways using the multiplication rule:

and

By equating equations (2) and (3), we get $P(A|B) P(B) = P(B|A) P(A)$.

Now, we can rearrange this equation to solve for $P(A|B)$:

This is the formula for Bayes' Theorem.

Let's examine the given options:

(A) $\frac{P(B|A) P(A)}{P(B)}$: This is the correct formula for Bayes' Theorem.

(B) $\frac{P(A|B) P(B)}{P(A)}$: This formula is derived by rearranging Bayes' Theorem to solve for $P(B|A)$ or $P(A)$, not $P(A|B)$.

(C) $\frac{P(A \cap B)}{P(A)}$: This is the formula for $P(B|A)$, the probability of B given A.

(D) $P(A) \times P(B|A)$: This is a part of the derivation, $P(A \cap B)$, but not the full formula for $P(A|B)$.

The correct formula for Bayes' Theorem for calculating $P(A|B)$ is $\frac{P(B|A) P(A)}{P(B)}$.

The correct option is (A).

Bayes' Theorem is used to update the probability for a hypothesis (or event) based on new evidence. It relates the conditional probabilities of events. The standard formula for Bayes' Theorem is:

In this formula, each term has a specific name reflecting its role in the Bayesian inference process:

• $P(A|B)$: This is the **posterior probability**. It is the updated probability of event A occurring after observing the evidence B.
• $P(B|A)$: This is the **likelihood**. It represents the probability of observing the evidence B given that event A is true.
• $P(A)$: This is the **prior probability**. It is the initial probability of event A before any new evidence (B) is considered. It represents our belief in event A before the experiment or observation.
• $P(B)$: This is the **marginal probability of the evidence**. It serves as a normalizing constant.

The question asks for the term associated with $P(A)$ in Bayes' Theorem.

Based on the definitions above, $P(A)$ is referred to as the prior probability.

Let's review the options:

(A) prior: This correctly identifies $P(A)$ as the initial probability or belief.

(B) posterior: This refers to $P(A|B)$, the updated probability after observing evidence.

(C) likelihood: This refers to $P(B|A)$, the probability of the evidence given the event.

(D) marginal: While $P(A)$ is a marginal probability, in the context of Bayes' Theorem and its inferential role, "prior" is the more specific and descriptive term.

Therefore, $P(A)$ in Bayes' Theorem is called the prior probability.

The correct option is (A).

This problem requires the application of Bayes' Theorem to calculate a posterior probability.

Let D be the event that a person has the disease.

Let ND be the event that a person does not have the disease (complement of D).

Let Pos be the event that the test result is positive.

From the problem statement:

Prevalence of the disease = 1% $\implies P(D) = 0.01$.

Probability of not having the disease = $P(ND) = 1 - P(D) = 1 - 0.01 = 0.99$.

Test accuracy (sensitivity) = 90% (correctly identifies disease if present) $\implies P(\text{Pos}|D) = 0.90$.

False positive rate = 5% (incorrectly flags disease as present when it's not) $\implies P(\text{Pos}|ND) = 0.05$.

We want to find the probability that a person actually has the disease given that the test is positive, which is $P(D|\text{Pos})$.

Using Bayes' Theorem:

The term $P(\text{Pos})$ in the denominator is the overall probability of testing positive, which can be calculated using the Law of Total Probability:

$P(\text{Pos}) = P(\text{Pos}|D) P(D) + P(\text{Pos}|ND) P(ND)$

Now let's analyze the options:

(A) $\frac{P(\text{Positive}|\text{Disease}) P(\text{Disease})}{P(\text{Positive}|\text{Disease}) P(\text{Disease}) + P(\text{Positive}|\text{No Disease}) P(\text{No Disease})}$

Substituting our event notations: $\frac{P(\text{Pos}|D) P(D)}{P(\text{Pos}|D) P(D) + P(\text{Pos}|ND) P(ND)}$. This is the direct application of Bayes' Theorem using the Law of Total Probability for the denominator. Thus, option (A) is a correct formulation.

(B) $\frac{0.90 \times 0.01}{0.90 \times 0.01 + 0.05 \times 0.99}$

This option substitutes the numerical values from the problem into the formula presented in option (A). It correctly represents the calculation for $P(D|\text{Pos})$. Thus, option (B) is also a correct formulation.

(C) $\frac{0.90 \times 0.01}{P(\text{Positive})}$

The numerator $0.90 \times 0.01$ is $P(\text{Pos}|D)P(D)$, which is $P(\text{Pos} \cap D)$. The formula for $P(D|\text{Pos})$ is $\frac{P(\text{Pos} \cap D)}{P(\text{Pos})}$. So, this option correctly states the numerator and the denominator as $P(\text{Positive})$. Thus, option (C) is also a correct formulation.

(D) All of the above are correct formulations.

Since options (A), (B), and (C) all correctly represent the application of Bayes' Theorem or the calculation of the required probability using the given information, this option is the most fitting.

The correct option is (D).

Let's evaluate Assertion (A) and Reason (R) individually and then examine their relationship.

Assertion (A): Bayes' Theorem is useful for updating the probability of a hypothesis based on new evidence.

This is the core purpose and utility of Bayes' Theorem. It mathematically describes how to revise an initial probability (prior probability) based on new data (evidence) to obtain an updated probability (posterior probability). This process is fundamental to Bayesian inference and learning from data. Thus, Assertion (A) is true.

Reason (R): It provides a way to calculate the conditional probability $P(A|B)$ from $P(B|A)$, $P(A)$, and $P(B)$.

Bayes' Theorem is indeed stated as $P(A|B) = \frac{P(B|A) P(A)}{P(B)}$. This formula explicitly shows how to calculate the conditional probability $P(A|B)$ (the posterior probability) using the likelihood $P(B|A)$, the prior probability $P(A)$, and the marginal probability of the evidence $P(B)$. Thus, Reason (R) is also true.

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

Assertion (A) states that Bayes' Theorem is useful for updating probabilities based on evidence. Reason (R) explains *how* it does this: by using the formula that relates the posterior probability ($P(A|B)$) to the prior probability ($P(A)$) and the likelihood of the evidence ($P(B|A)$). The ability to calculate $P(A|B)$ from these components is precisely what allows for the updating of probabilities, moving from the prior to the posterior.

Therefore, Reason (R) provides the mechanism or the "how" behind the utility described in Assertion (A). It explains *why* Bayes' Theorem is useful for updating probabilities.

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

The correct option is (A).

This problem involves applying the Law of Total Probability to find $P(R)$, which is then used in Bayes' Theorem.

We are given three bags:

Bag 1 ($B_1$): 2 red balls, 3 black balls. Total = 5 balls. $P(R|B_1) = 2/5$.

Bag 2 ($B_2$): 4 red balls, 1 black ball. Total = 5 balls. $P(R|B_2) = 4/5$.

Bag 3 ($B_3$): 3 red balls, 4 black balls. Total = 7 balls. $P(R|B_3) = 3/7$.

Since a bag is selected at random, the probability of selecting each bag is equal:

$P(B_1) = 1/3$

$P(B_2) = 1/3$

$P(B_3) = 1/3$

The Law of Total Probability states that for a sample space partitioned by mutually exclusive and exhaustive events $B_1, B_2, B_3$, the probability of an event R can be calculated as:

$P(R) = P(R|B_1)P(B_1) + P(R|B_2)P(B_2) + P(R|B_3)P(B_3)$

Let's analyze the options:

(A) $P(R|B_1)P(B_1) + P(R|B_2)P(B_2) + P(R|B_3)P(B_3)$

This option correctly states the general formula for $P(R)$ using the Law of Total Probability. It uses the notation for the probabilities involved.

(B) $(2/5)(1/3) + (4/5)(1/3) + (3/7)(1/3)$

This option substitutes the specific numerical values for $P(R|B_i)$ and $P(B_i)$ into the formula from option (A). It represents the correct calculation for $P(R)$.

(C) $(2/5) + (4/5) + (3/7)$

This option incorrectly adds the conditional probabilities without multiplying them by the probabilities of the respective bags, $P(B_i)$.

(D) Both (A) and (B) are correct formulations.

Option (A) provides the formula in terms of symbolic probabilities, while option (B) provides the same formula with the specific values substituted. Both are correct ways to represent $P(R)$ in this context.

The correct option is (D) because both (A) and (B) are correct ways to express $P(R)$.

We are continuing from the previous case study. We want to find $P(B_1|R)$, the probability that Bag 1 was selected given that a red ball was drawn.

We will use Bayes' Theorem, which states: $P(B_1|R) = \frac{P(R|B_1) P(B_1)}{P(R)}$.

From the previous questions and calculations:

$P(R|B_1) = 2/5$ (Probability of drawing red from Bag 1)

$P(B_1) = 1/3$ (Probability of selecting Bag 1)

$P(R) = (2/5)(1/3) + (4/5)(1/3) + (3/7)(1/3)$ (Total probability of drawing a red ball)

Now, substitute these values into Bayes' Theorem for $P(B_1|R)$:

Substitute the expressions:

Let's examine the given options:

(A) $\frac{(2/5)(1/3)}{(2/5)(1/3) + (4/5)(1/3) + (3/7)(1/3)}$: This option exactly matches the formula derived using Bayes' Theorem with the specific probabilities.

(B) $\frac{2/5}{(2/5) + (4/5) + (3/7)} \times 1/3$: This is incorrect. The $1/3$ factor should be applied to each term in the numerator and denominator separately, not just to the entire denominator or as a separate multiplier.

(C) $\frac{(2/5)}{(2/5) + (4/5) + (3/7)} \times 1/3$: This is also incorrect for the same reason as (B). The $P(B_1)$ term should be multiplied by $P(R|B_1)$ in the numerator.

(D) $\frac{P(R|B_1)}{P(R)}$: This is a partial application of Bayes' Theorem; it's missing the multiplication by $P(B_1)$ in the numerator.

The correct formulation for $P(B_1|R)$ is the one that correctly applies Bayes' Theorem with the Law of Total Probability for the denominator.

The correct option is (A).

We are considering the experiment of rolling a single fair six-sided die. The sample space for this experiment is $S = \{1, 2, 3, 4, 5, 6\}$.

We need to find the probability of getting a prime number.

First, let's identify which numbers in the sample space are prime numbers.

A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.

Let's examine each number in the sample space:

• 1: By definition, 1 is not a prime number.
• 2: The divisors of 2 are 1 and 2. It is greater than 1 and has only two divisors. So, 2 is a prime number.
• 3: The divisors of 3 are 1 and 3. It is greater than 1 and has only two divisors. So, 3 is a prime number.
• 4: The divisors of 4 are 1, 2, and 4. It has more than two divisors. So, 4 is not a prime number.
• 5: The divisors of 5 are 1 and 5. It is greater than 1 and has only two divisors. So, 5 is a prime number.
• 6: The divisors of 6 are 1, 2, 3, and 6. It has more than two divisors. So, 6 is not a prime number.

The prime numbers in the sample space $\{1, 2, 3, 4, 5, 6\}$ are $\{2, 3, 5\}$.

Let E be the event of getting a prime number. The outcomes in event E are $\{2, 3, 5\}$.

The number of favorable outcomes (getting a prime number) is $|E| = 3$.

The total number of possible outcomes (the size of the sample space) is $|S| = 6$.

The probability of event E is calculated as:

Substituting the values:

This probability can be simplified to $1/2$, but the options are given in terms of a denominator of 6.

Let's check the options:

(A) $1/6$: This would mean only one outcome is prime.

(B) $2/6$: This would mean two outcomes are prime.

(C) $3/6$: This correctly represents that there are three prime numbers (2, 3, 5) out of six possible outcomes.

(D) $4/6$: This would mean four outcomes are prime.

The probability of getting a prime number when rolling a single die is $3/6$.

The correct option is (C).

We are given the probabilities of two events A and B, and the probability of their intersection:

$P(A) = 0.6$

$P(B) = 0.4$

$P(A \cap B) = 0.2$

We need to find the probability of the union of A and B, denoted as $P(A \cup B)$.

The formula for the probability of the union of two events is given by the General Addition Rule:

Now, we substitute the given values into the formula:

Perform the addition and subtraction:

$P(A \cup B) = 1.0 - 0.2$

$P(A \cup B) = 0.8$

Let's check the given options:

(A) 1: This would be correct if the events were mutually exclusive and their probabilities summed to 1, or if they covered the entire sample space and accounted for overlap.

(B) 0.8: This matches our calculated value.

(C) 0.24: This might be obtained by incorrectly multiplying probabilities ($0.6 \times 0.4$), which is for independent events' intersection.

(D) 0.6: This is $P(A)$, not $P(A \cup B)$.

The probability of A or B occurring is 0.8.

The correct option is (B).

We are given the following probabilities:

$P(A) = 0.6$

$P(B) = 0.4$

$P(A \cap B) = 0.2$

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

The formula for conditional probability is:

Now, substitute the given values into the formula:

Let's examine the options:

(A) $0.2/0.6$: This would be $P(B|A) = P(A \cap B) / P(A)$.

(B) $0.2/0.4$: This matches our calculation for $P(A|B)$.

(C) $0.6/0.2$: This is $P(A) / P(A \cap B)$, which is not a standard probability formula.

(D) $0.4/0.2$: This is $P(B) / P(A \cap B)$, which is not a standard probability formula.

The correct expression for $P(A|B)$ is $0.2/0.4$.

The correct option is (B).

We have a bag with 10 tokens numbered from 1 to 10. The sample space S consists of all possible outcomes when drawing a token:

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

The total number of possible outcomes is $|S| = 10$.

We are interested in the event of drawing a token with a number divisible by 3. Let's identify these numbers within the sample space.

A number is divisible by 3 if it leaves no remainder when divided by 3.

Let's check each number from 1 to 10:

• 1 $\div$ 3 = 0 remainder 1
• 2 $\div$ 3 = 0 remainder 2
• 3 $\div$ 3 = 1 remainder 0 $\implies$ 3 is divisible by 3
• 4 $\div$ 3 = 1 remainder 1
• 5 $\div$ 3 = 1 remainder 2
• 6 $\div$ 3 = 2 remainder 0 $\implies$ 6 is divisible by 3
• 7 $\div$ 3 = 2 remainder 1
• 8 $\div$ 3 = 2 remainder 2
• 9 $\div$ 3 = 3 remainder 0 $\implies$ 9 is divisible by 3
• 10 $\div$ 3 = 3 remainder 1

The numbers divisible by 3 in the sample space are $\{3, 6, 9\}$.

Let E be the event of drawing a token with a number divisible by 3. The outcomes in E are $\{3, 6, 9\}$.

The number of favorable outcomes is $|E| = 3$.

The probability of event E is calculated as:

Substituting the values:

Now, let's compare this with the given options:

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

(B) $4/10$: This would imply there are 4 numbers divisible by 3.

(C) $1/3$: This is the simplified form of $3/9$ or $4/12$, not relevant here.

(D) $1/10$: This would imply only one number is divisible by 3.

The probability of drawing a token with a number divisible by 3 is $3/10$.

The correct option is (A).

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

An **impossible event** is an event that cannot occur under any circumstances within the given sample space. The probability of an impossible event is 0.

Let's analyze each option:

1. (A) Getting a number less than 7: The numbers in the sample space that are less than 7 are $\{1, 2, 3, 4, 5, 6\}$. This event includes all outcomes in the sample space. This is a certain event, not an impossible event.
2. (B) Getting a number greater than 6: The numbers in the sample space that are greater than 6 are none. There is no outcome in $\{1, 2, 3, 4, 5, 6\}$ that is greater than 6. Therefore, this event cannot occur. This is an impossible event.
3. (C) Getting an even number: The even numbers in the sample space are $\{2, 4, 6\}$. This event can occur.
4. (D) Getting a number divisible by 1: All integers are divisible by 1. The numbers in the sample space that are divisible by 1 are $\{1, 2, 3, 4, 5, 6\}$. This event includes all outcomes in the sample space. This is a certain event.

The event that cannot occur when rolling a single die is getting a number greater than 6.

The correct option is (B).

We are given the probabilities of two independent events A and B:

$P(A) = 0.5$

$P(B) = 0.3$

Since events A and B are independent, we know that the probability of their intersection is the product of their individual probabilities: $P(A \cap B) = P(A) \times P(B)$.

We need to find $P(A \cup B)$, the probability that either A or B or both occur. We use the General Addition Rule for probabilities:

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

Now, substitute this value and the given probabilities into the addition rule:

Calculate the result:

$P(A \cup B) = 0.8 - 0.15 = 0.65$

Now let's look at the options provided:

(A) 0.8: This is $P(A) + P(B)$, which would be correct only if the events were mutually exclusive and their intersection probability was 0.

(B) 0.15: This is $P(A \cap B)$, the probability of both events occurring, not the probability of either occurring.

(C) $0.5 + 0.3 - (0.5 \times 0.3)$: This expression correctly represents $P(A) + P(B) - P(A \cap B)$ for independent events, which evaluates to $0.5 + 0.3 - 0.15 = 0.65$.

(D) Both (A) and (C) are correct. Option (C) is a correct formulation of the calculation. If option (C) evaluates to 0.8, then both would be correct. However, we calculated $P(A \cup B) = 0.65$. Option (A) states 0.8, which is incorrect.

Let's re-evaluate the options based on our calculation of $P(A \cup B) = 0.65$.

Option (A) is incorrect because $P(A \cup B)$ is not 0.8.

Option (B) is incorrect because it's $P(A \cap B)$.

Option (C) is a correct formulation for the calculation, which results in 0.65.

Option (D) suggests both (A) and (C) are correct. Since (A) is incorrect, (D) cannot be correct.

There might be a misunderstanding in the question options or my interpretation. Let's re-check the calculation and options.

Calculation: $P(A \cup B) = 0.5 + 0.3 - (0.5 \times 0.3) = 0.8 - 0.15 = 0.65$.

Option (A) is 0.8.

Option (C) is the correct expression $0.5 + 0.3 - (0.5 \times 0.3)$, which correctly evaluates to 0.65.

If the question implies that one of the options IS the value of $P(A \cup B)$, then none of the options represent 0.65. However, if the question asks for a *correct formulation*, then (C) is correct.

Let's assume the question asks for the correct *formulation* and also provides the *value*. In that case, (C) is the correct formulation, and its value is 0.65. Option (A) is just a value. It seems option (D) might be intended if option (A) was also 0.65. Given the options, option (C) is the most precise correct statement.

If the question is asking for the correct formulation, and option (A) is just a value, then (C) is the only correct formulation.

Let's consider if there's a scenario where (A) could be correct. If the events were mutually exclusive, $P(A \cup B) = P(A) + P(B) = 0.5 + 0.3 = 0.8$. But the events are stated to be independent, and their intersection is not zero ($0.5 \times 0.3 = 0.15 \neq 0$), so they are not mutually exclusive.

It seems there might be an error in the provided options if the calculated value 0.65 is not present. However, option (C) is the correct expression for calculating $P(A \cup B)$ given the information.

Let's proceed with the assumption that the question is asking for the correct *formulation* or calculation process.

Option (C) is the correct calculation process. Option (A) is an incorrect value based on our calculation.

If we strictly interpret "find $P(A \cup B)$", we should look for the value. Since 0.65 is not an option, let's re-read carefully.

Option (C) is a correct *statement* of how to calculate $P(A \cup B)$. It is a correct formulation. Option (A) is an incorrect *value*. Option (D) states both (A) and (C) are correct. Since (A) is an incorrect value, (D) cannot be correct.

Therefore, (C) is the only option that is a correct statement about finding $P(A \cup B)$.

The correct option is (C).

We have a box containing 5 defective bulbs and 15 non-defective bulbs. The total number of bulbs is $5 + 15 = 20$.

Two bulbs are drawn at random without replacement. We want to find the probability that both bulbs are defective.

This involves two sequential events, and since the draws are without replacement, the outcome of the first draw affects the probabilities for the second draw.

Step 1: Probability that the first bulb drawn is defective.

There are 5 defective bulbs out of a total of 20 bulbs.

Step 2: Probability that the second bulb drawn is defective, given that the first was defective and not replaced.

After drawing one defective bulb and not replacing it, the conditions change:

Number of defective bulbs remaining = $5 - 1 = 4$.

Total number of bulbs remaining = $20 - 1 = 19$.

The probability of drawing a second defective bulb, given the first was defective, is:

Step 3: Probability that both bulbs are defective.

To find the probability that both events occur, we multiply the probabilities from Step 1 and Step 2:

Substituting the calculated probabilities:

Now, let's compare this with the given options:

(A) $(5/20) \times (4/19)$: This matches our calculated probability.

(B) $(5/20) \times (5/20)$: This would be the probability if the draws were with replacement.

(C) $(5/20) + (4/19)$: This uses addition, which is for 'or' probabilities, not 'and' probabilities.

(D) $(5/20) \times (15/19)$: This would be the probability of drawing one defective and one non-defective bulb in a specific order (or related to it).

The correct probability that both bulbs are defective is $(5/20) \times (4/19)$.

The correct option is (A).

We need to match each term with its correct description:

(i) Sample Space

The sample space is the set of all possible outcomes of a random experiment. This matches description (d).

Match: (i) - (d)

(ii) Event

An event is a subset of the sample space, meaning it is a collection of one or more outcomes from the set of all possible outcomes. This matches description (a).

Match: (ii) - (a)

(iii) Mutually Exclusive Events

Mutually exclusive events are events that cannot occur at the same time. If one happens, the other cannot. This matches description (b).

Match: (iii) - (b)

(iv) Conditional Probability

Conditional probability, denoted as $P(A|B)$, is the probability of an event A occurring given that event B has already occurred. This matches description (c).

Match: (iv) - (c)

Combining these matches, we get:

(i) - (d)

(ii) - (a)

(iii) - (b)

(iv) - (c)

Now let's compare this with the given options:

(A) (i)-(a), (ii)-(d), (iii)-(b), (iv)-(c) (Incorrect pairing for (i) and (ii))

(B) (i)-(d), (ii)-(a), (iii)-(b), (iv)-(c) (Matches all our findings)

(C) (i)-(d), (ii)-(a), (iii)-(c), (iv)-(b) (Incorrect pairing for (iii) and (iv))

(D) (i)-(a), (ii)-(d), (iii)-(c), (iv)-(b) (Incorrect pairing for (i), (ii), (iii), and (iv))

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

The correct option is (B).

We have a standard deck of 52 playing cards. The sample space S consists of all 52 cards, so $|S| = 52$.

We are interested in the event of drawing a face card. The face cards in a deck are Kings, Queens, and Jacks.

In a standard deck, there are 4 suits (Hearts, Diamonds, Clubs, Spades).

For each suit, there is one King, one Queen, and one Jack.

Number of Kings = 4

Number of Queens = 4

Number of Jacks = 4

The total number of face cards is the sum of the number of Kings, Queens, and Jacks:

Total face cards = $4 + 4 + 4 = 12$.

Let F be the event of drawing a face card. The number of favorable outcomes is $|F| = 12$.

The probability of drawing a face card is calculated as:

Substituting the values:

Now, let's simplify this fraction. The greatest common divisor of 12 and 52 is 4.

Divide both the numerator and the denominator by 4:

So, the probability of drawing a face card is $12/52$, which simplifies to $3/13$.

Let's examine the options:

(A) $3/13$: This is the simplified form of the probability.

(B) $1/13$: This would be the probability of drawing a specific rank (like any specific King) from a suit.

(C) $12/52$: This is the unsimplified form of the probability.

(D) Both (A) and (C) are correct. Since both the unsimplified and simplified forms of the correct probability are provided as options, this option is the most appropriate.

The probability of drawing a face card is indeed both $12/52$ and its simplified form $3/13$.

The correct option is (D).

In probability theory, $A'$ denotes the complement of event A. The complement of an event A includes all outcomes in the sample space that are not in A.

A fundamental property of probability is that the sum of the probability of an event and the probability of its complement is equal to 1. This can be expressed as:

We are given that $P(A) = 0.7$. We need to find $P(A')$.

Using the property mentioned above, we can rearrange the formula to solve for $P(A')$:

Substitute the given value of $P(A)$:

Calculate the result:

$P(A') = 0.3$

Now, let's check the given options:

(A) 0.7: This is $P(A)$ itself.

(B) 0.3: This matches our calculated value for $P(A')$.

(C) 1: This is the probability of the entire sample space or a certain event.

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

Therefore, if $P(A) = 0.7$, then $P(A') = 0.3$.

The correct option is (B).

When rolling two fair six-sided dice, the total number of possible outcomes is $6 \times 6 = 36$. Each outcome is a pair $(d_1, d_2)$, where $d_1$ is the result of the first die and $d_2$ is the result of the second die.

We want to find the probability that the sum of the numbers on the two dice is 7.

Let's list the combinations of outcomes from the two dice that add up to 7:

• (1, 6) - First die shows 1, second die shows 6.
• (2, 5) - First die shows 2, second die shows 5.
• (3, 4) - First die shows 3, second die shows 4.
• (4, 3) - First die shows 4, second die shows 3.
• (5, 2) - First die shows 5, second die shows 2.
• (6, 1) - First die shows 6, second die shows 1.

There are 6 favorable outcomes where the sum is 7.

The total number of possible outcomes when rolling two dice is 36.

The probability of getting a sum of 7 is the number of favorable outcomes divided by the total number of possible outcomes:

This fraction can be simplified. Dividing both the numerator and the denominator by their greatest common divisor, which is 6:

Now let's check the options:

(A) $6/36$: This is the unsimplified probability.

(B) $1/6$: This is the simplified probability.

(C) $7/36$: This might be confusion with the sum itself being 7.

(D) Both (A) and (B): Since both $6/36$ and its simplified form $1/6$ are correct representations of the probability, this option is the most accurate.

The probability of getting a sum of 7 when rolling two dice is both $6/36$ and $1/6$.

The correct option is (D).

If two events A and B are independent, it means that the occurrence of one does not affect the probability of the other. Mathematically, this is expressed as $P(A \cap B) = P(A)P(B)$.

A key property of independent events is that this independence extends to their complements as well.

Let's prove these statements:

1. Are $A'$ and $B'$ independent?

We need to check if $P(A' \cap B') = P(A')P(B')$.

We know that $P(A') = 1 - P(A)$ and $P(B') = 1 - P(B)$.

Also, from De Morgan's laws, $A' \cap B' = (A \cup B)'$.

So, $P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B)$.

Using the addition rule, $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.

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

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

Now, $P(A' \cap B') = 1 - (P(A) + P(B) - P(A)P(B)) = 1 - P(A) - P(B) + P(A)P(B)$.

We also know that $P(A')P(B') = (1 - P(A))(1 - P(B)) = 1 - P(B) - P(A) + P(A)P(B)$.

Since $P(A' \cap B') = P(A')P(B')$, $A'$ and $B'$ are independent. Thus, statement (A) is true.

2. Are A and $B'$ independent?

We need to check if $P(A \cap B') = P(A)P(B')$.

We know that $P(A) = P(A \cap B) + P(A \cap B')$.

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

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

Therefore, $P(A \cap B') = P(A) - P(A)P(B) = P(A)(1 - P(B))$.

And $P(A)P(B') = P(A)(1 - P(B))$.

Since $P(A \cap B') = P(A)P(B')$, A and $B'$ are independent. Thus, statement (B) is true.

3. Are $A'$ and B independent?

We need to check if $P(A' \cap B) = P(A')P(B)$.

We know that $P(B) = P(A \cap B) + P(A' \cap B)$.

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

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

Therefore, $P(A' \cap B) = P(B) - P(A)P(B) = P(B)(1 - P(A))$.

And $P(A')P(B) = (1 - P(A))P(B)$.

Since $P(A' \cap B) = P(A')P(B)$, $A'$ and B are independent. Thus, statement (C) is true.

Since statements (A), (B), and (C) are all true, option (D) is the correct answer.

The correct option is (D).

This problem involves using Bayes' Theorem to calculate a posterior probability.

Let A be the event that an item is produced by Machine A.

Let B be the event that an item is produced by Machine B.

Let D be the event that an item is defective.

From the problem statement:

Machine A produces 60% of items $\implies P(A) = 0.60$.

Machine B produces 40% of items $\implies P(B) = 0.40$.

Machine A has a 3% defect rate $\implies P(D|A) = 0.03$.

Machine B has a 5% defect rate $\implies P(D|B) = 0.05$.

We want to find the probability that an item came from Machine A given that it is defective. This is $P(A|D)$.

Bayes' Theorem states: $P(A|D) = \frac{P(D|A) P(A)}{P(D)}$.

The term $P(D)$ in the denominator is the overall probability of an item being defective, which can be found using the Law of Total Probability:

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

Let's analyze the options:

(A) $P(\text{Defective}|\text{Machine A})P(\text{Machine A}) / P(\text{Defective})$

Substituting our notation: $\frac{P(D|A) P(A)}{P(D)}$. This is the correct formula for Bayes' Theorem applied to this problem.

(B) $\frac{0.03 \times 0.60}{0.03 \times 0.60 + 0.05 \times 0.40}$

This option substitutes the numerical values into the Bayes' Theorem formula. The numerator is $P(D|A)P(A)$. The denominator is $P(D|A)P(A) + P(D|B)P(B)$, which is the Law of Total Probability for $P(D)$. This is a correct formulation.

(C) $\frac{0.018}{0.018 + 0.020}$

The numerator $0.018$ is the result of $0.03 \times 0.60$.

The first term in the denominator $0.018$ is also $0.03 \times 0.60$.

The second term in the denominator $0.020$ is the result of $0.05 \times 0.40$.

So, this option represents the calculation of $\frac{P(D|A)P(A)}{P(D|A)P(A) + P(D|B)P(B)}$. This is a numerically correct representation of the probability $P(A|D)$. This is also a correct formulation.

(D) All of the above that are correct formulations.

Since options (A), (B), and (C) are all correct ways to formulate or calculate the probability that the defective item came from Machine A, this option is the most appropriate answer.

The correct option is (D).

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

In this case, the experiment is tossing a coin 3 times. Each toss can result in either Heads (H) or Tails (T).

We need to list all possible sequences of outcomes for three tosses:

• First toss H, Second toss H, Third toss H $\implies$ HHH
• First toss H, Second toss H, Third toss T $\implies$ HHT
• First toss H, Second toss T, Third toss H $\implies$ HTH
• First toss H, Second toss T, Third toss T $\implies$ HTT
• First toss T, Second toss H, Third toss H $\implies$ THH
• First toss T, Second toss H, Third toss T $\implies$ THT
• First toss T, Second toss T, Third toss H $\implies$ TTH
• First toss T, Second toss T, Third toss T $\implies$ TTT

So, the sample space is the set containing all these 8 possible outcomes: $\{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$.

Let's check the given options:

(A) $\{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$: This option lists all 8 possible outcomes when tossing a coin three times.

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

(C) $\{HH, HT, TH, TT\}$: This is the sample space for tossing a coin twice.

(D) $\{HHH, TTT\}$: This only includes the outcomes where all tosses are the same, missing many possibilities.

The correct sample space for tossing a coin 3 times is the one that includes all 8 possible sequences of Heads and Tails.

The correct option is (A).

When a coin is tossed 3 times, the sample space S consists of 8 equally likely outcomes, as determined in the previous question: $S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$.

The total number of possible outcomes is $|S| = 8$.

We are interested in the event of getting exactly two heads. Let's identify the outcomes from the sample space that have exactly two heads:

• HHT: Two heads, one tail.
• HTH: Two heads, one tail.
• THH: Two heads, one tail.

The outcomes with exactly two heads are $\{HHT, HTH, THH\}$.

Let E be the event of getting exactly two heads. The number of favorable outcomes is $|E| = 3$.

The probability of event E is calculated as:

Substituting the values:

Now, let's check the given options:

(A) $2/8$: This would imply 2 favorable outcomes.

(B) $3/8$: This matches our calculated probability.

(C) $1/8$: This would imply only 1 favorable outcome.

(D) $4/8$: This would imply 4 favorable outcomes (e.g., getting at least two heads).

The probability of getting exactly two heads when tossing a coin 3 times is $3/8$.

The correct option is (B).

In the context of probability theory, an "experiment" has a specific and broad definition. It's not limited to formal scientific experiments but encompasses any activity or process where the outcome is uncertain but can be clearly defined.

Let's break down the meaning of an experiment in probability:

• Process: It's an action, procedure, or test.
• Well-defined outcomes: The set of all possible results of the process must be known beforehand. Even if we can't predict the exact outcome of a single trial, we must know what the potential outcomes are.
• Uncertainty: The outcome of a single trial of the experiment cannot be predicted with certainty.

Consider common examples:

• Tossing a coin: The process is tossing the coin. The outcomes are Heads or Tails. The outcome of a single toss is uncertain.
• Rolling a die: The process is rolling the die. The outcomes are numbers 1 through 6. The outcome of a single roll is uncertain.
• Drawing a card from a deck: The process is drawing a card. The outcomes are the 52 cards in the deck. The outcome of a single draw is uncertain.

Now let's evaluate the given options:

(A) Only scientific experiments. This is too narrow. While scientific experiments are a type of probability experiment, the term is much broader. Tossing a coin or rolling a die are not typically called "scientific experiments" but are fundamental probability experiments.

(B) Any process that generates a set of well-defined outcomes. This definition aligns perfectly with the meaning of an experiment in probability. The key elements of a process, well-defined outcomes, and inherent uncertainty are captured here.

(C) An event whose outcome is certain. This describes a certain event, not an experiment. An experiment is the process itself, and its outcome is uncertain.

(D) The calculation of probability. The calculation of probability is what we do *after* defining the experiment and its outcomes; it is not the experiment itself.

Therefore, the term "experiment" in probability refers to any process that generates a set of well-defined outcomes.

The correct option is (B).

We are given the probabilities of two events A and B:

$P(A) = 0.3$

$P(B) = 0.5$

We are also told that events A and B are **mutually exclusive**. Mutually exclusive events are events that cannot occur at the same time.

By definition, if two events A and B are mutually exclusive, their intersection is the empty set ($\emptyset$). This means there are no common outcomes between event A and event B.

Therefore, the probability of their intersection, $P(A \cap B)$, is 0.

The values of $P(A)$ and $P(B)$ are provided but are not needed to determine $P(A \cap B)$ when mutual exclusivity is stated.

Let's check the options:

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

(B) 0.15: This might be obtained by multiplying probabilities ($0.3 \times 0.5$), which is for independent events' intersection, not mutually exclusive.

(C) 0.8: This is $P(A) + P(B)$, which would be $P(A \cup B)$ if A and B were mutually exclusive.

(D) Cannot be determined: This is incorrect, as the definition of mutually exclusive events directly gives us the answer.

The probability of the intersection of mutually exclusive events is always 0.

The correct option is (A).

We are given the probabilities of two events A and B:

$P(A) = 0.3$

$P(B) = 0.5$

We are also told that events A and B are **mutually exclusive**. This means that events A and B cannot happen at the same time. Therefore, the probability of their intersection is 0:

We need to find the probability of the union of A and B, $P(A \cup B)$. The 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:

So, for mutually exclusive events, $P(A \cup B) = P(A) + P(B)$.

Now, substitute the given values:

Calculate the result:

$P(A \cup B) = 0.8$

Let's review the options:

(A) 0.15: This might be obtained by multiplying $P(A) \times P(B)$, which is for independent events' intersection.

(B) 0.8: This matches our calculated value for $P(A \cup B)$.

(C) 0.2: This is the difference $P(B) - P(A)$ or potentially $P(A \cap B)$ if they were independent and calculated differently.

(D) 1: This is the probability of the entire sample space or a certain event.

The probability of A or B occurring, given they are mutually exclusive, is 0.8.

The correct option is (B).

The fundamental definition of probability, often called the classical definition, states that the probability of an event is the ratio of the number of outcomes favorable to that event to the total number of possible outcomes in the sample space.

This definition relies on a crucial assumption about the nature of these outcomes:

The **Equally Likely Assumption**: For this formula to be valid, each individual outcome in the sample space must have an equal chance of occurring. If outcomes are not equally likely, this simple ratio method does not apply directly, and we would need to use empirical probability or other methods.

Let's analyze the options:

(A) random: While the outcomes of a random experiment are uncertain, "random" describes the process, not a property of individual outcomes that allows for simple probability calculation by counting.

(B) simple: A "simple event" is an event with only one outcome. This describes the nature of a single outcome contributing to an event, but not a property of all outcomes in the sample space necessary for the basic probability formula.

(C) equally likely: This directly addresses the assumption needed for the ratio of favorable outcomes to total outcomes to be a valid measure of probability. If outcomes are equally likely, each has the same probability, simplifying the calculation.

(D) definite: "Definite" implies certainty. Outcomes in a probability experiment are typically not definite; their occurrence is uncertain.

Therefore, the probability calculation based on counting favorable and total outcomes assumes that all outcomes are equally likely.

The correct option is (C).

We are given the following probabilities:

$P(A) = 0.6$

$P(B|A) = 0.5$ (This is the conditional probability of B given A)

We need to find the probability of the intersection of A and B, denoted as $P(A \cap B)$.

The formula for conditional probability relates $P(A \cap B)$, $P(B|A)$, and $P(A)$. The multiplication rule for conditional probability states:

Now, we substitute the given values into this formula:

Calculate the product:

$P(A \cap B) = 0.30$

Let's check the given options:

(A) 0.3: This matches our calculated value for $P(A \cap B)$.

(B) 1.1: This is not a valid probability, as probabilities cannot exceed 1.

(C) 0.1: This is not the result of the calculation.

(D) 0.8: This might be $P(A) + P(B)$ if they were mutually exclusive, but it's not $P(A \cap B)$.

The probability of $A \cap B$ is 0.3.

The correct option is (A).

Bayes' Theorem is used to update the probability of a hypothesis (event A) based on new evidence (event B). The theorem is typically expressed as:

In this formula:

• $P(A|B)$: This is the **posterior probability** – the updated probability of A after observing B.
• $P(A)$: This is the **prior probability** – the initial probability of A before observing B.
• $P(B)$: This is the **marginal probability of the evidence**.
• $P(B|A)$: This term represents the probability of observing the evidence B given that the hypothesis A is true. This is known as the **likelihood**.

The question asks for the term that refers to $P(B|A)$ in Bayes' Theorem.

Based on the roles of each term in the theorem, $P(B|A)$ is called the likelihood.

Let's examine the options:

(A) prior probability: This refers to $P(A)$.

(B) posterior probability: This refers to $P(A|B)$.

(C) likelihood: This refers to $P(B|A)$.

(D) marginal probability: This refers to $P(B)$.

The correct option is (C).

We have a bag with 5 red balls and 5 black balls. The total number of balls is $5 + 5 = 10$.

A ball is drawn, its color is noted, and crucially, it is returned to the bag. This means the draws are independent, and the probabilities for the second draw are the same as for the first draw.

We want to find the probability that the first ball is red AND the second ball is black.

Step 1: Probability that the first ball drawn is red.

There are 5 red balls out of a total of 10 balls.

Step 2: Probability that the second ball drawn is black.

Since the first ball was returned to the bag, the total number of balls and the number of black balls remain unchanged.

There are 5 black balls out of a total of 10 balls.

Step 3: Probability of both events happening.

Since the draws are independent (due to replacement), we multiply the probabilities of each event:

Substituting the calculated probabilities:

Now, let's compare this with the given options:

(A) $(5/10) \times (4/9)$: This is for dependent events without replacement.

(B) $(5/10) \times (5/9)$: This mixes probabilities from replacement and no-replacement.

(C) $(5/10) \times (5/10)$: This matches our calculated probability for independent events with replacement.

(D) $(5/10) + (5/10)$: This uses addition, which is for 'or' probabilities, not 'and' probabilities.

The correct probability is $(5/10) \times (5/10)$.

The correct option is (C).

We are considering the experiment of rolling a single fair six-sided die. The sample space is $S = \{1, 2, 3, 4, 5, 6\}$.

Let E be the event of getting a number greater than 4. The outcomes that satisfy this condition are those numbers in the sample space that are strictly greater than 4.

The numbers in S greater than 4 are 5 and 6.

So, the event E is $\{5, 6\}$.

The complement of an event E, denoted as $E'$, consists of all outcomes in the sample space S that are not in E.

In other words, $E' = S - E$.

Let's find the outcomes in S that are NOT in E:

The outcomes in S are $\{1, 2, 3, 4, 5, 6\}$.

The outcomes in E are $\{5, 6\}$.

Removing the outcomes of E from S, we get: $\{1, 2, 3, 4\}$.

So, the complement of E, $E'$, consists of the outcomes $\{1, 2, 3, 4\}$.

Now let's describe this set of outcomes in words:

The numbers 1, 2, 3, and 4 are all numbers that are less than or equal to 4.

Let's evaluate the options:

(A) Getting a number less than 4: This would be the set $\{1, 2, 3\}$. This is not the complement because it excludes 4.

(B) Getting a number less than or equal to 4: This corresponds to the set $\{1, 2, 3, 4\}$, which is exactly our complement $E'$.

(C) Getting a number equal to 4: This would be the set $\{4\}$, which is only one outcome from the complement.

(D) Getting an odd number: This would be the set $\{1, 3, 5\}$. This is not the complement because it includes 5 and excludes 1, 2, 3, 4.

The complement of getting a number greater than 4 is getting a number that is not greater than 4, which means getting a number less than or equal to 4.

The correct option is (B).

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

$P(A) = 0.5$

$P(B) = 0.6$

$P(A \cup B) = 0.8$

We need to find the probability of the intersection of A and B, $P(A \cap B)$.

The relationship between the probabilities of two events, their union, and their intersection is given by the General Addition Rule:

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

Now, substitute the given values into the formula:

Perform the calculation:

$P(A \cap B) = 1.1 - 0.8$

$P(A \cap B) = 0.3$

Let's examine the options:

(A) 0.3: This matches our calculated value for $P(A \cap B)$.

(B) 0.1: This is not the correct result.

(C) 0.2: This is not the correct result.

(D) 1.1: This is not a valid probability, as probabilities cannot exceed 1.

The probability of $A \cap B$ is 0.3.

The correct option is (A).

We are given the following probabilities:

$P(A) = 0.5$

$P(B) = 0.6$

$P(A \cup B) = 0.8$

To determine if events A and B are independent, we need to check if the condition $P(A \cap B) = P(A) \times P(B)$ holds true.

First, let's calculate $P(A \cap B)$ using the given information and the Addition Rule: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.

Rearranging the formula to solve for $P(A \cap B)$: $P(A \cap B) = P(A) + P(B) - P(A \cup B)$.

$P(A \cap B) = 1.1 - 0.8 = 0.3$

Now, let's calculate the product of the individual probabilities:

Now we compare $P(A \cap B)$ with $P(A) \times P(B)$:

$P(A \cap B) = 0.3$

$P(A) \times P(B) = 0.30$

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

Let's evaluate the options:

(A) Yes, because $P(A \cap B) = P(A) \times P(B)$. (This is correct; the condition for independence is met.)

(B) No, because $P(A \cap B) \neq P(A) \times P(B)$. (Incorrect, as the equality holds.)

(C) Yes, because $P(A) + P(B) > 1$. (The sum of probabilities being greater than 1 does not directly prove independence, although it implies there must be an overlap.)

(D) Cannot be determined. (Incorrect, we were able to determine independence.)

The events are independent because the probability of their intersection is equal to the product of their individual probabilities.

The correct option is (A).

We have a bag containing balls of different colors:

Number of red balls = 4

Number of blue balls = 3

Number of green balls = 5

The total number of balls in the bag is the sum of the balls of all colors:

Total balls = $4 + 3 + 5 = 12$.

We want to find the probability that the ball drawn is not green.

There are two ways to approach this:

Method 1: Directly count favorable outcomes

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

Number of non-green balls = Number of red balls + Number of blue balls

Number of non-green balls = $4 + 3 = 7$.

The probability of drawing a non-green ball is:

Method 2: Using the complement rule

Let G be the event of drawing a green ball. The probability of drawing a green ball is:

The event "not green" is the complement of event G, denoted as $G'$. According to the complement rule, $P(G') = 1 - P(G)$.

$P(\text{Not Green}) = 1 - P(G) = 1 - \frac{5}{12}$

Both methods yield the same result.

Now, let's check the options:

(A) $5/12$: This is the probability of drawing a green ball.

(B) $7/12$: This matches our calculated probability of drawing a ball that is not green.

(C) $4/12$: This is the probability of drawing a red ball.

(D) $3/12$: This is the probability of drawing a blue ball.

The probability that the ball drawn is not green is $7/12$.

The correct option is (B).

When rolling two fair six-sided dice, there are $6 \times 6 = 36$ possible outcomes in the sample space. Each outcome is equally likely.

We want to find the probability of getting a sum of at most 5. "At most 5" means the sum can be 2, 3, 4, or 5.

Let's list the combinations of outcomes $(d_1, d_2)$ from the two dice that result in these sums:

Sum = 2:

• (1, 1) - 1 combination

Sum = 3:

• (1, 2)
• (2, 1) - 2 combinations

Sum = 4:

• (1, 3)
• (2, 2)
• (3, 1) - 3 combinations

Sum = 5:

• (1, 4)
• (2, 3)
• (3, 2)
• (4, 1) - 4 combinations

The total number of favorable outcomes (where the sum is at most 5) is the sum of the combinations for each sum:

Total favorable outcomes = (combinations for sum 2) + (combinations for sum 3) + (combinations for sum 4) + (combinations for sum 5)

Total favorable outcomes = $1 + 2 + 3 + 4 = 10$.

The total number of possible outcomes is 36.

The probability of getting a sum of at most 5 is:

Now, let's check the options:

(A) $10/36$: This matches our calculated probability.

(B) $1/6$: This is the probability of getting a sum of 7.

(C) $5/36$: This would imply 5 favorable outcomes.

(D) $15/36$: This would imply 15 favorable outcomes (e.g., sum is at most 7).

The probability of getting a sum of at most 5 when rolling two dice is $10/36$.

The correct option is (A).

We are given that events A and B are independent. This means that the occurrence of one event does not affect the probability of the other event.

A fundamental property of independent events is that this independence extends to their complements as well. If A and B are independent, then:

• A and B' are independent.
• A' and B are independent.
• A' and B' are independent.

We want to find $P(A|B')$.

The definition of conditional probability is $P(A|B') = \frac{P(A \cap B')}{P(B')}$.

Since A and B are independent, it follows that A and B' are also independent. Therefore, the probability of their intersection is the product of their individual probabilities:

Substituting this into the formula for conditional probability:

Assuming $P(B') \neq 0$ (which is generally true unless B is a certain event), we can cancel out $P(B')$ from the numerator and denominator:

This result makes sense intuitively: if A is independent of B, then the occurrence or non-occurrence of B (represented by B') should not change the probability of A.

Let's check the options:

(A) $P(A|B)$: This would only be equal to $P(A)$ if A and B were independent, but it's not necessarily equal to $P(A|B')$.

(B) $P(A)$: This is what we derived.

(C) $P(B)$: This is the probability of event B, not relevant here.

(D) $P(A)P(B')$: This is the probability of the intersection $P(A \cap B')$, not $P(A|B')$.

Therefore, if A and B are independent events, $P(A|B')$ is equal to $P(A)$.

The correct option is (B).

Let E be the event that a student passed in English.

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

We are given the following probabilities:

$P(E) = 0.7$ (70% passed in English)

$P(M) = 0.6$ (60% passed in Mathematics)

$P(E \cap M) = 0.4$ (40% passed in both subjects)

We need to find the probability that the student passed in at least one of the subjects. This corresponds to the probability of the union of the two events, $P(E \cup M)$.

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

Now, let's look at the options to see which one represents this formula and calculation:

(A) $0.7 + 0.6$: This is $P(E) + P(M)$. This would be correct if the events were mutually exclusive, but they are not, as $P(E \cap M) = 0.4$, not 0.

(B) $0.7 + 0.6 - 0.4$: This directly applies the Addition Rule with the given probabilities: $P(E) + P(M) - P(E \cap M)$. This is the correct way to calculate $P(E \cup M)$.

(C) 0.4: This is $P(E \cap M)$, the probability of passing both subjects, not at least one.

(D) $1 - 0.4$: This is $1 - P(E \cap M)$. This would represent the probability that a student did NOT pass both subjects, which is different from passing at least one subject.

Option (B) is the correct formulation for finding the probability that the student passed in at least one of the subjects.

The correct option is (B).

We have a group of 5 men and 3 women. The total number of people is $5 + 3 = 8$.

A committee of 2 people is to be selected from this group.

We need to find the probability that the committee consists of exactly one man and one woman.

First, let's find the total number of ways to select a committee of 2 people from 8 people. This is a combination problem, as the order of selection does not matter.

The total number of ways to choose 2 people from 8 is given by the combination formula ${}_nC_k = \frac{n!}{k!(n-k)!}$, where $n$ is the total number of items and $k$ is the number of items to choose.

Total number of possible committees = ${}_8C_2 = \frac{8!}{2!(8-2)!} = \frac{8!}{2!6!} = \frac{8 \times 7}{2 \times 1} = 28$.

Next, let's find the number of ways to select a committee with exactly one man and one woman.

Number of ways to choose 1 man from 5 men = ${}_5C_1 = \frac{5!}{1!(5-1)!} = \frac{5!}{1!4!} = 5$.

Number of ways to choose 1 woman from 3 women = ${}_3C_1 = \frac{3!}{1!(3-1)!} = \frac{3!}{1!2!} = 3$.

The number of ways to form a committee with exactly one man and one woman is the product of these two numbers (by the multiplication principle):

Number of favorable committees = ${}_5C_1 \times {}_3C_1 = 5 \times 3 = 15$.

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

$P(\text{1 man and 1 woman}) = \frac{\text{Number of favorable committees}}{\text{Total number of possible committees}} = \frac{15}{28}$.

Now let's examine the given options to see how they relate to this calculation:

(A) $\frac{{}_5C_1 \times {}_3C_1}{{}_8C_2}$

This option directly represents our calculation: $\frac{5 \times 3}{28} = \frac{15}{28}$. This is a correct formulation.

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

This expression also calculates the probability. Let's see why:

The numerator $5 \times 3$ represents the number of ways to pick one man and one woman (which is ${}_5C_1 \times {}_3C_1$).

The denominator $8 \times 7$ represents the total number of ways to pick 2 people in order (permutations), $P(8,2) = \frac{8!}{(8-2)!} = 8 \times 7 = 56$.

However, the formula ${}_nC_k = \frac{P(n,k)}{k!}$. So, ${}_8C_2 = \frac{P(8,2)}{2!} = \frac{8 \times 7}{2 \times 1}$.

The expression in option (B) is $\frac{5 \times 3}{8 \times 7} = \frac{15}{56}$. This is half of our correct probability (15/28). This option seems to be calculating something else, possibly related to ordered selection, or it's an incorrect simplification of the combination calculation.

Let's re-evaluate (B) carefully. If we consider the selections sequentially: Probability of picking a man first is $5/8$. Probability of picking a woman second (given a man was picked first) is $3/7$. Probability of picking a man first and then a woman is $(5/8) \times (3/7) = 15/56$. Probability of picking a woman first is $3/8$. Probability of picking a man second (given a woman was picked first) is $5/7$. Probability of picking a woman first and then a man is $(3/8) \times (5/7) = 15/56$. The probability of getting one man and one woman (order doesn't matter) is the sum of these two probabilities: $15/56 + 15/56 = 30/56 = 15/28$.

The expression $\frac{5 \times 3}{8 \times 7}$ is $\frac{15}{56}$. This is not the correct probability of getting exactly one man and one woman.

Let's re-examine (A). ${}_5C_1 = 5$, ${}_3C_1 = 3$, ${}_8C_2 = 28$. So, (A) is $\frac{5 \times 3}{28} = \frac{15}{28}$. This is correct.

Let's reconsider if (B) could be interpreted differently. The denominator $8 \times 7$ is $P(8,2)$. The numerator $5 \times 3$ is the number of favorable pairs if order mattered. If order mattered, the total number of ordered pairs would be $P(8,2) = 56$. The number of ordered pairs with one man and one woman would be $(5 \times 3) + (3 \times 5) = 15 + 15 = 30$. So the probability would be $30/56 = 15/28$. The expression in (B) is $15/56$, which is only part of the ordered outcome.

It seems Option (A) is the correct formulation using combinations.

(C) $\frac{{}_5C_1 + {}_3C_1}{{}_8C_2}$

This option incorrectly uses addition in the numerator instead of multiplication. It would mean adding the number of ways to pick one man to the number of ways to pick one woman, which doesn't form the committee of one man and one woman.

(D) Both (A) and (B) are correct ways to calculate it.

Since (A) is correct and (B) seems incorrect as a complete representation of the probability, this option is likely incorrect.

Based on the standard approach using combinations, option (A) is the correct formulation.

The correct option is (A).

We are given two conditions about events A and B:

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

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

Let's recall the definitions of these terms:

• Conditional Probability: $P(A|B) = \frac{P(A \cap B)}{P(B)}$ (assuming $P(B) > 0$)
• Independence: Events A and B are independent if $P(A \cap B) = P(A)P(B)$. An equivalent condition is $P(A|B) = P(A)$ (if $P(B) > 0$) and $P(B|A) = P(B)$ (if $P(A) > 0$).
• Mutually Exclusive Events: Events A and B are mutually exclusive if $A \cap B = \emptyset$, which means $P(A \cap B) = 0$.
• Exhaustive Events: Events A and B are exhaustive if $A \cup B = S$ (the sample space), meaning $P(A \cup B) = 1$.
• Subset: Event A is a subset of B if all outcomes in A are also in B.

Let's analyze the given conditions:

The condition $P(A|B) = P(A)$ is one of the definitions of independence for events A and B (provided $P(B) > 0$). If the probability of A occurring is the same whether B has occurred or not, then A and B are independent.

Similarly, the condition $P(B|A) = P(B)$ is another definition of independence for events A and B (provided $P(A) > 0$). If the probability of B occurring is the same whether A has occurred or not, then B and A are independent.

Since both given conditions are direct definitions of independence, it necessarily follows that events A and B are independent.

Let's consider why the other options are not necessarily true:

• (A) Mutually exclusive: If A and B were mutually exclusive, then $P(A \cap B) = 0$. For independence, $P(A \cap B) = P(A)P(B) = 0.5 \times 0.6 = 0.3$. Since $0 \neq 0.3$, they are not mutually exclusive.
• (C) Exhaustive: For A and B to be exhaustive, $P(A \cup B) = 1$. We know $P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.5 + 0.6 - 0.3 = 0.8$. Since $0.8 \neq 1$, they are not exhaustive.
• (D) A is a subset of B: This is not necessarily true. Independence does not imply a subset relationship.

The given conditions directly imply that A and B are independent.

The correct option is (B).

We have a standard deck of 52 cards. We are drawing two cards without replacement, and we want to find the probability that both cards are Kings.

There are 4 Kings in a deck of 52 cards.

We can solve this problem using two methods:

Method 1: Using sequential probabilities

Step 1: Probability of drawing a King on the first draw.

There are 4 Kings in 52 cards.

Step 2: Probability of drawing a King on the second draw, given the first was a King and not replaced.

After drawing one King, there are 3 Kings left in the deck, and the total number of cards remaining is 51.

Step 3: Probability of both events occurring.

Multiply the probabilities from Step 1 and Step 2:

This matches option (B).

Method 2: Using combinations

The total number of ways to draw 2 cards from a deck of 52 is given by the combination formula ${}_nC_k = \frac{n!}{k!(n-k)!}$.

Total number of ways to choose 2 cards from 52 = ${}_{52}C_2$.

The number of ways to draw 2 Kings from the 4 Kings available in the deck is ${}_4C_2$.

The probability of drawing two Kings is the ratio of the number of ways to draw 2 Kings to the total number of ways to draw 2 cards:

Let's verify that this is equivalent to Method 1:

${}_4C_2 = \frac{4!}{2!2!} = \frac{4 \times 3}{2 \times 1} = 6$

${}_{52}C_2 = \frac{52!}{2!50!} = \frac{52 \times 51}{2 \times 1} = 26 \times 51 = 1326$

So, $P(\text{Both Kings}) = \frac{6}{1326}$.

Now, let's simplify $\frac{4}{52} \times \frac{3}{51} = \frac{1}{13} \times \frac{3}{51} = \frac{1}{13} \times \frac{1}{17} = \frac{1}{221}$.

And $\frac{6}{1326} = \frac{3}{663} = \frac{1}{221}$.

Both methods give the same result, $1/221$. The expressions themselves are correct representations.

Let's re-examine the options:

(A) $\frac{4}{52} \times \frac{3}{52}$: Incorrect because the second probability should account for no replacement ($3/51$).

(B) $\frac{4}{52} \times \frac{3}{51}$: This correctly represents the sequential probability calculation for drawing two Kings without replacement.

(C) $\frac{{}_4C_2}{{}_{52}C_2}$: This correctly represents the probability using combinations.

(D) Both (B) and (C) are correct. Since both expressions represent the correct probability, this option is the most fitting.

The correct option is (D).

This problem involves a combined experiment: tossing a coin and rolling a die. We need to find the probability of a specific event E, which is getting a Head AND an even number.

Let H be the event of getting a Head when tossing a coin. The probability is $P(H) = 1/2$.

Let Even be the event of getting an even number when rolling a die. The possible outcomes for a die are $\{1, 2, 3, 4, 5, 6\}$. The even numbers are $\{2, 4, 6\}$. So, there are 3 even numbers out of 6 possible outcomes.

The probability of getting an even number is $P(\text{Even}) = 3/6 = 1/2$.

The event E is that both of these occur: getting a Head AND an even number. Since tossing a coin and rolling a die are independent events, the probability of both occurring is the product of their individual probabilities:

Substitute the probabilities:

Calculate the result:

$P(E) = \frac{3}{12} = \frac{1}{4}$

Now, let's look at the options:

(A) $1/2 \times 3/6$: This correctly represents the calculation $P(H) \times P(\text{Even})$.

(B) $1/2 + 3/6$: This uses addition, which is for 'or' probabilities, not 'and' probabilities.

(C) $1/12$: This is the result of $1/2 \times 1/6$. It would be correct if the probability of an even number was $1/6$ (e.g., if it was a 12-sided die with only one even number out of 12), or if it was $1/2 \times 1/12$ which is not the case. Alternatively, $1/12$ is the result if you calculated $1/2 \times 1/6$, which is wrong.

(D) $3/12$: This is the simplified numerical result of $1/2 \times 3/6$. Since option (A) correctly shows the calculation and (D) shows the numerical result, and the question asks "What is P(E)?", both might be considered correct in different ways.

However, option (A) explicitly shows the calculation using the probabilities of the independent events.

Option (D) shows the numerical result of that calculation.

If the question asks for the probability, the numerical value is often preferred. $3/12$ simplifies to $1/4$.

Let's reconsider option (C). $1/12$ would imply $1/2 \times 1/6$. The probability of even is $3/6$, not $1/6$. So (C) is incorrect.

Option (A) shows the correct formula application. Option (D) shows the numerical result. In multiple-choice questions like this, if both the correct calculation setup and the correct numerical answer are present, and there isn't an option for "Both (A) and (D)", we need to choose the best fit. The question asks "What is P(E)?", which typically expects a value.

Let's evaluate (A) numerically: $1/2 \times 3/6 = 3/12$.

Let's evaluate (D) numerically: $3/12$.

So, both (A) and (D) represent the same probability. However, (A) shows the reasoning, and (D) shows the result. Often, questions asking "What is the probability" expect the numerical value.

If the intent is to show the calculation, (A) is best. If the intent is to show the final value, (D) is best.

Given the options, option (A) is a correct representation of the probability calculation. Option (D) is the correct numerical result.

Let's assume the question is asking for the most complete answer that shows how the probability is derived.

The probability is indeed calculated as $(1/2) \times (3/6)$, which equals $3/12$.

Both (A) and (D) express the probability correctly. However, option (D) says "Both (A) and (D) are correct formulations." This is self-referential and confusing.

Let's assume the options meant: (A) Correct Calculation, (B) Incorrect Operation, (C) Incorrect Numerical Value, (D) Correct Numerical Value. In that case, (A) and (D) represent the same correct probability. If the option was "Both (A) and (D) are correct", then it would be clear.

Let's interpret option (D) as "Both (A) and the calculation represented by (D) are correct formulations."

The probability is $3/12$. Option (A) is $1/2 \times 3/6$, which equals $3/12$. Option (D) is $3/12$. Thus, both (A) and (D) represent the same correct probability.

Therefore, the most accurate answer is that both (A) and (D) are correct representations.

The correct option is (D).

We are given that A and B are independent events, with $P(A) = 0.4$ and $P(B) = 0.5$. We need to find $P(A' \cap B')$.

A key property of independent events is that if A and B are independent, then their complements ($A'$ and $B'$) are also independent.

Property of independence: If A and B are independent, then $P(A \cap B) = P(A)P(B)$.

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

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

This matches option (B).

We can also express $P(A')$ and $P(B')$ in terms of $P(A)$ and $P(B)$ using the complement rule:

$P(A') = 1 - P(A)$

$P(B') = 1 - P(B)$

Substituting these into the expression for $P(A' \cap B')$:

$P(A' \cap B') = (1 - P(A)) \times (1 - P(B))$

This matches option (C).

Let's evaluate the options:

(A) $P(A' \cup B')$: This is the probability of the union of the complements, not the intersection.

(B) $P(A') \times P(B')$: This is the correct formula for the intersection of independent events $A'$ and $B'$.

(C) $(1-P(A)) \times (1-P(B))$: This is the correct expression using the probabilities of A and B to calculate $P(A') \times P(B')$.

(D) Both (B) and (C). Since both (B) and (C) are correct formulations of $P(A' \cap B')$, this option is the most appropriate.

Therefore, both (B) and (C) are correct formulations for $P(A' \cap B')$.

The correct option is (D).

When rolling two fair six-sided dice, there are a total of $6 \times 6 = 36$ possible outcomes, each equally likely.

We want to find the probability of getting a sum of 9.

Let's list the combinations of outcomes $(d_1, d_2)$ from the two dice that result in a sum of 9:

• (3, 6) - First die shows 3, second die shows 6.
• (4, 5) - First die shows 4, second die shows 5.
• (5, 4) - First die shows 5, second die shows 4.
• (6, 3) - First die shows 6, second die shows 3.

There are 4 favorable outcomes where the sum is 9.

The total number of possible outcomes is 36.

The probability of getting a sum of 9 is the number of favorable outcomes divided by the total number of possible outcomes:

Now, let's check the options:

(A) $4/36$: This matches our calculated probability.

(B) $5/36$: This would imply 5 favorable outcomes.

(C) $6/36$: This is the probability of getting a sum of 7.

(D) $9/36$: This would imply 9 favorable outcomes.

The probability of getting a sum of 9 when rolling two dice is $4/36$.

The correct option is (A).

In this scenario, the process is a student randomly guessing the answer to a multiple-choice question.

The question has 4 options.

Only one of these options is correct.

The student guesses randomly, meaning each option has an equal chance of being chosen.

The sample space consists of the 4 options the student can choose from. Each option is a possible outcome of the guessing process.

Total number of possible outcomes = 4 (the four options).

The event we are interested in is that the student guesses the correct answer.

There is only one correct option.

Number of favorable outcomes = 1 (the single correct option).

The probability of an event is calculated as the ratio of the number of favorable outcomes to the total number of possible outcomes, assuming all outcomes are equally likely (which they are in random guessing).

Substituting the values:

Let's review the options:

(A) $1/4$: This matches our calculated probability.

(B) $3/4$: This is the probability of guessing the wrong answer.

(C) $1/2$: This would be the probability if there were 2 options.

(D) $1$: This would mean it's a certain event.

The probability that the student guesses the correct answer is $1/4$.

The correct option is (A).

We have a standard deck of 52 cards. We are drawing two cards with replacement, and we want to find the probability that both cards are Kings.

There are 4 Kings in a deck of 52 cards.

Since the card is replaced after the first draw, the two draws are independent events. The probability of drawing a King remains the same for both draws.

Step 1: Probability of drawing a King on the first draw.

There are 4 Kings in 52 cards.

Step 2: Probability of drawing a King on the second draw (with replacement).

Since the first card was replaced, the deck is back to its original state: 4 Kings in 52 cards.

Step 3: Probability of both events happening.

Since the draws are independent, we multiply the probabilities:

Substituting the probabilities:

Now, let's examine the options:

(A) $\frac{4}{52} \times \frac{3}{51}$: This is for drawing without replacement.

(B) $\frac{4}{52} \times \frac{4}{52}$: This correctly represents the probability for independent events with replacement.

(C) $\frac{{}_4C_2}{{}_{52}C_2}$: This is for drawing two Kings without replacement using combinations.

(D) $\frac{{}_4P_2}{{}_{52}P_2}$: This uses permutations, which would be for ordered selections. If order matters, the number of ways to pick 2 Kings in order is $P(4,2) = 4 \times 3 = 12$, and the total number of ordered pairs is $P(52,2) = 52 \times 51$. So, $\frac{12}{52 \times 51}$. This is not the same as $\frac{4}{52} \times \frac{4}{52}$.

The correct representation for drawing two Kings with replacement is $\frac{4}{52} \times \frac{4}{52}$.

The correct option is (B).

Random Experiment: A random experiment is an experiment whose outcome cannot be predicted with certainty, even though all possible outcomes are known. The outcome depends on chance.

Example of a Random Experiment: Tossing a fair coin.

In this experiment, the possible outcomes are getting a Head (H) or a Tail (T). We cannot predict with certainty which outcome will occur before the toss.

Sample Space: The sample space of a random experiment is the set of all possible outcomes of that experiment.

Example of a Sample Space: For the random experiment of tossing a fair coin, the sample space (denoted by S) is the set containing all possible outcomes.

S = {H, T}

Here, H represents the outcome of getting a Head, and T represents the outcome of getting a Tail.

When a coin is tossed three times, each toss can result in either a Head (H) or a Tail (T). To find the sample space, we list all possible combinations of outcomes for the three tosses.

The possible outcomes are:

First toss: H or T

Second toss: H or T

Third toss: H or T

The sample space (S) is the set of all possible sequences of these outcomes:

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

There are $2^3 = 8$ possible outcomes in the sample space.

When two dice are rolled simultaneously, each die can land on any number from 1 to 6. The sample space consists of all possible pairs of outcomes, where the first element of the pair represents the outcome of the first die and the second element represents the outcome of the second die.

Let the outcome of the first die be $d_1$ and the outcome of the second die be $d_2$. The sample space $S$ can be represented as:

$S = \{ (d_1, d_2) \mid d_1 \in \{1, 2, 3, 4, 5, 6\} \text{ and } d_2 \in \{1, 2, 3, 4, 5, 6\} \}$

Listing all the possible pairs:

The total number of outcomes in the sample space is $6 \times 6 = 36$.

In this experiment, we are drawing a single ball at random from a bag that contains balls of two different colors: red and black.

The possible outcomes are drawing a red ball or drawing a black ball. We can represent these outcomes.

Let R denote the event of drawing a red ball.

Let B denote the event of drawing a black ball.

The sample space $S$ for this experiment is the set of all possible outcomes:

$S = \{R, B\}$

Here, R signifies that the drawn ball is red, and B signifies that the drawn ball is black.

Event: An event is a specific outcome or a set of outcomes of a random experiment. It is a subset of the sample space.

Example related to rolling a single die:

Consider the random experiment of rolling a single die. The sample space $S$ for this experiment is:

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

An event is a particular result or a collection of results from this sample space. Here are a few examples of events:

1. Event A: Getting an even number. The outcomes that satisfy this event are 2, 4, and 6. So, $A = \{2, 4, 6\}$.

2. Event B: Getting a number greater than 4. The outcomes that satisfy this event are 5 and 6. So, $B = \{5, 6\}$.

3. Event C: Getting a number less than 2. The only outcome that satisfies this event is 1. So, $C = \{1\}$.

4. Event D: Getting the number 7. There is no outcome in the sample space that satisfies this. So, $D = \{\}$ (the empty set, also called an impossible event).

When two coins are tossed, each coin can land on either a Head (H) or a Tail (T). The possible outcomes are the combinations of these results for both coins.

The sample space $S$ for tossing two coins is:

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

Here, the first letter in each outcome represents the result of the first coin, and the second letter represents the result of the second coin.

Let E be the event of getting at least one head. This means that in the outcomes of the two coin tosses, there should be one head or two heads.

We examine the sample space to identify the outcomes that satisfy this condition:

• HH: Has two heads (satisfies "at least one head").
• HT: Has one head (satisfies "at least one head").
• TH: Has one head (satisfies "at least one head").
• TT: Has no heads (does not satisfy "at least one head").

Therefore, the elements of event E are:

$E = \{HH, HT, TH\}$

Mutually Exclusive Events: Mutually exclusive events are events that cannot occur at the same time. If one event happens, the other cannot. In terms of sets, their intersection is empty.

Example using drawing a single card from a deck:

Consider the random experiment of drawing a single card from a standard deck of 52 playing cards. The sample space $S$ consists of all 52 cards.

Let's define two events:

• Event A: Drawing a King. There are 4 Kings in a deck (King of Hearts, King of Diamonds, King of Clubs, King of Spades).
• Event B: Drawing a Queen. There are 4 Queens in a deck (Queen of Hearts, Queen of Diamonds, Queen of Clubs, Queen of Spades).

These two events, A and B, are mutually exclusive because a single card drawn from the deck cannot be both a King and a Queen simultaneously. The set of outcomes for Event A and the set of outcomes for Event B have no common elements; their intersection is empty.

Mathematically, if $A \cap B = \emptyset$, then events A and B are mutually exclusive.

Exhaustive Events: Exhaustive events are a set of events such that at least one of them must occur. In other words, the union of exhaustive events is equal to the entire sample space.

Example using rolling a single die:

Consider the random experiment of rolling a single die. The sample space $S$ for this experiment is:

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

Let's consider the following events:

• Event A: Getting an even number. The outcomes are $\{2, 4, 6\}$.
• Event B: Getting an odd number. The outcomes are $\{1, 3, 5\}$.

Events A and B are exhaustive events because every possible outcome when rolling a die is either an even number or an odd number. There are no other possibilities. The union of these events covers the entire sample space:

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

Impossible Event: An impossible event is an event that cannot occur in a random experiment. Its probability is 0. In set theory terms, it is represented by the empty set ($\emptyset$).

Sure Event: A sure event (or certain event) is an event that is certain to occur in a random experiment. Its probability is 1. In set theory terms, it is equal to the entire sample space ($S$).

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

Given:

$P(A) = 0.5$

$P(B) = 0.6$

$P(A \cap B) = 0.2$

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

Substitute the given values into the formula:

Therefore, the probability of $A \cup B$ is 0.9.

We are given that events $A$ and $B$ are mutually exclusive. This means that they cannot occur at the same time, so the probability of their intersection is 0, i.e., $P(A \cap B) = 0$.

We are also given:

$P(A) = 0.3$

$P(B) = 0.4$

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

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

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

Therefore, the probability of $A \cup B$ is 0.7.

When a single 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 $n(S) = 6$.

We are interested in the event of getting a prime number. A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.

Let $E$ be the event of getting a prime number. The prime numbers in the sample space are:

$E = \{2, 3, 5\}$

The number of favorable outcomes for event $E$ is $n(E) = 3$.

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

Substituting the values:

Simplifying the fraction:

Therefore, the probability of getting a prime number when a single die is rolled is $\frac{1}{2}$.

A standard deck of 52 cards has four suits: Hearts, Diamonds, Clubs, and Spades. Each suit has 13 cards.

The total number of possible outcomes when drawing a single card from a well-shuffled deck is the total number of cards in the deck.

Total number of cards = 52.

So, the total number of possible outcomes, $n(S) = 52$.

The deck is divided into two colors: red and black.

• Hearts and Diamonds are red suits.
• Clubs and Spades are black suits.

Each suit has 13 cards. Therefore, the number of red cards is the sum of cards in Hearts and Diamonds:

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

Number of red cards = 13 + 13 = 26.

Let $E$ be the event of drawing a red card. The number of favorable outcomes for event $E$ is the number of red cards.

Number of favorable outcomes, $n(E) = 26$.

The probability of an event is given by the formula:

Substituting the values:

Simplifying the fraction:

Thus, the probability of drawing a red card is $\frac{1}{2}$.

Conditional Probability: Conditional probability is the probability of an event occurring given that another event has already occurred. It quantifies how the likelihood of one event changes when we know that a related event has taken place.

The formula for the conditional probability of event $A$ given event $B$ (denoted as $P(A|B)$) is:

This formula is valid only if $P(B) > 0$.

We are given the following probabilities:

$P(A) = 0.8$

$P(B) = 0.5$

$P(B|A) = 0.4$

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

We can use the definition of conditional probability, $P(B|A) = \frac{P(A \cap B)}{P(A)}$.

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

Now, substitute the given values into the formula:

Therefore, the probability of $A \cap B$ is 0.32.

We are given the following probabilities:

$P(A) = 0.6$

$P(A \cap B) = 0.3$

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

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

Substitute the given values into the formula:

To simplify the fraction:

Therefore, $P(B|A) = 0.5$.

First, let's determine the sample space when two coins are tossed. Let H represent Heads and T represent Tails.

The sample space $S$ is:

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

The total number of possible outcomes is $n(S) = 4$.

Let $A$ be the event of getting two heads.

$A = \{HH\}$

The number of outcomes for event $A$ is $n(A) = 1$.

Let $B$ be the event that at least one head is obtained. This means we can get one head or two heads.

$B = \{HH, HT, TH\}$

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

We need to find the probability of getting two heads given that at least one head is obtained. This is a conditional probability, $P(A|B)$.

The formula for conditional probability is:

First, let's find $P(A \cap B)$. The intersection of $A$ and $B$ is the event where both "getting two heads" and "getting at least one head" occur.

$A \cap B = \{HH\} \cap \{HH, HT, TH\} = \{HH\}$

The number of outcomes for $A \cap B$ is $n(A \cap B) = 1$.

Now we can calculate the probabilities:

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

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

Substitute these values into the conditional probability formula:

To simplify:

Therefore, the probability of getting two heads given that at least one head is obtained is $\frac{1}{3}$.

Independent Events: Two events, $A$ and $B$, are considered independent if the occurrence or non-occurrence of one event does not affect the probability of the other event occurring.

If events $A$ and $B$ are independent, the relationship between their probabilities is given by the multiplication rule for independent events:

This means that the probability of both events $A$ and $B$ occurring is simply the product of their individual probabilities.

We are given that events $A$ and $B$ are independent. This means that the probability of their intersection is the product of their individual probabilities:

We are given:

$P(A) = 0.4$

$P(B) = 0.7$

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

Now we need to find $P(A \cup B)$. The general formula for the union of two events is:

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

Therefore, $P(A \cup B) = 0.82$.

The problem involves two sequential events: drawing the first ball and drawing the second ball. Since the first ball is replaced after its color is noted, the two draws are independent events.

First, let's determine the total number of balls in the bag:

Number of red balls = 5

Number of blue balls = 3

Total number of balls = 5 + 3 = 8

Probability of drawing a red ball in the first draw:

Let $R_1$ be the event that the first ball drawn is red.

$P(R_1) = \frac{\text{Number of red balls}}{\text{Total number of balls}} = \frac{5}{8}$

Since the first ball is replaced, the composition of the bag remains the same for the second draw. The events are independent.

Probability of drawing a red ball in the second draw:

Let $R_2$ be the event that the second ball drawn is red.

$P(R_2) = \frac{\text{Number of red balls}}{\text{Total number of balls}} = \frac{5}{8}$

We need to find the probability that both balls are red. Since the draws are independent events, the probability of both events occurring is the product of their individual probabilities:

Substitute the probabilities:

Therefore, the probability that both balls are red is $\frac{25}{64}$.

This problem involves conditional probability because the outcome of the second draw depends on the outcome of the first draw, as the ball is not replaced.

Initially, the bag contains:

Number of red balls = 5

Number of blue balls = 3

Total number of balls = 5 + 3 = 8

We are given that the first ball drawn was red. Let $R_1$ be the event that the first ball drawn is red.

After the first ball (which was red) is drawn and not replaced, the contents of the bag change:

Number of red balls remaining = 5 - 1 = 4

Number of blue balls remaining = 3

Total number of balls remaining = 4 + 3 = 7

We need to find the probability that the second ball drawn is blue, given that the first ball was red. Let $B_2$ be the event that the second ball drawn is blue.

The conditional probability $P(B_2 | R_1)$ is calculated based on the new composition of the bag after the first draw:

Substituting the values:

Therefore, the probability that the second ball is blue given that the first ball was red is $\frac{3}{7}$.

To determine if two events $A$ and $B$ are independent, we need to check if the probability of their intersection is equal to the product of their individual probabilities. That is, we check if $P(A \cap B) = P(A) \times P(B)$.

We are given:

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

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

$P(A \cap B) = \frac{1}{8}$

Let's calculate the product of $P(A)$ and $P(B)$:

Now, we compare this product with the given $P(A \cap B)$:

$P(A \cap B) = \frac{1}{8}$

Since $P(A \cap B) = \frac{1}{8}$ and $P(A) \times P(B) = \frac{1}{8}$, we can conclude that:

Therefore, events $A$ and $B$ are independent events.

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 $n(S) = 6$.

We are interested in the event of getting a number greater than 4. Let $E$ be this event.

The numbers in the sample space that are greater than 4 are 5 and 6.

So, the favorable outcomes for event $E$ are:

$E = \{5, 6\}$

The number of favorable outcomes for event $E$ is $n(E) = 2$.

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

Substituting the values:

Simplifying the fraction:

Therefore, the probability of getting a number greater than 4 when rolling a fair die is $\frac{1}{3}$.

When two dice are rolled, the total number of possible outcomes is $6 \times 6 = 36$. The sample space consists of pairs of numbers $(d_1, d_2)$, where $d_1$ is the result of the first die and $d_2$ is the result of the second die.

The total number of possible outcomes is $n(S) = 36$.

We want to find the probability that the sum of the numbers on the two dice is 7.

Let $E$ be the event that the sum of the numbers is 7.

We need to list all the pairs of outcomes $(d_1, d_2)$ such that $d_1 + d_2 = 7$:

• If $d_1 = 1$, then $d_2 = 6$, so $(1, 6)$.
• If $d_1 = 2$, then $d_2 = 5$, so $(2, 5)$.
• If $d_1 = 3$, then $d_2 = 4$, so $(3, 4)$.
• If $d_1 = 4$, then $d_2 = 3$, so $(4, 3)$.
• If $d_1 = 5$, then $d_2 = 2$, so $(5, 2)$.
• If $d_1 = 6$, then $d_2 = 1$, so $(6, 1)$.

The favorable outcomes for event $E$ are: $E = \{(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)\}$.

The number of favorable outcomes for event $E$ is $n(E) = 6$.

The probability of event $E$ is calculated as:

Substituting the values:

Simplifying the fraction:

Therefore, the probability that the sum of the numbers is 7 when two dice are rolled is $\frac{1}{6}$.

This problem involves sequential events without replacement, so we will use conditional probability.

Total number of apples in the box = 10

Number of rotten apples = 3

Number of not rotten apples = 10 - 3 = 7

Let $R_1$ be the event that the first apple selected is rotten.

Let $NR_2$ be the event that the second apple selected is not rotten.

We want to find the probability of both events occurring, $P(R_1 \cap NR_2)$. This can be calculated as $P(R_1) \times P(NR_2 | R_1)$.

Step 1: Probability that the first apple is rotten ($P(R_1)$).

The probability of selecting a rotten apple first is the number of rotten apples divided by the total number of apples:

Step 2: Probability that the second apple is not rotten given the first was rotten ($P(NR_2 | R_1)$).

After the first apple (which was rotten) is drawn and not replaced, the number of apples in the box changes:

Total number of apples remaining = 10 - 1 = 9

Number of rotten apples remaining = 3 - 1 = 2

Number of not rotten apples remaining = 7

The probability of selecting a not rotten apple as the second apple, given that the first was rotten, is:

Step 3: Probability that the first apple is rotten AND the second is not rotten.

To find the probability of both events happening in sequence, we multiply the probabilities from Step 1 and Step 2:

Now, simplify the expression:

Therefore, the probability that the first apple is rotten and the second is not rotten is $\frac{7}{30}$.

A set of events $E_1, E_2, \dots, E_n$ forms a partition of the sample space $S$ if the following two conditions are met:

1. Mutually Exclusive: The events must be mutually exclusive, meaning that no two events can occur at the same time. In other words, the intersection of any two distinct events is empty.

   $E_i \cap E_j = \emptyset$ for all $i \neq j$, where $i, j \in \{1, 2, \dots, n\}$

   …(i)

2. Exhaustive: The union of all the events must be equal to the entire sample space. This means that at least one of the events must occur.

   $E_1 \cup E_2 \cup \dots \cup E_n = S$

   …(ii)

In simpler terms, a partition divides the sample space into non-overlapping subsets that together cover the entire sample space.

Given:

Two mutually exclusive and exhaustive events $E_1$ and $E_2$, and any event $A$.

To State:

The formula for the Law of Total Probability for these events.

Law of Total Probability:

The Law of Total Probability states that if $E_1, E_2, \ldots, E_n$ is a set of mutually exclusive and exhaustive events (meaning that exactly one of these events must occur), then for any event $A$, the probability of $A$ can be calculated as:

In this specific case, where we have only two mutually exclusive and exhaustive events, $E_1$ and $E_2$, the formula simplifies to:

Where:

• $P(A)$ is the probability of event $A$.
• $P(A|E_1)$ is the conditional probability of event $A$ occurring given that event $E_1$ has occurred.
• $P(E_1)$ is the probability of event $E_1$.
• $P(A|E_2)$ is the conditional probability of event $A$ occurring given that event $E_2$ has occurred.
• $P(E_2)$ is the probability of event $E_2$.

Given:

Two mutually exclusive and exhaustive events $E_1$ and $E_2$, and any event $A$ such that $P(A) > 0$.

To State:

The formula for Bayes' Theorem for these conditions.

Bayes' Theorem:

Bayes' Theorem allows us to update the probability of a hypothesis (event $E_1$ or $E_2$) given new evidence (event $A$). For two mutually exclusive and exhaustive events $E_1$ and $E_2$, and an event $A$, Bayes' Theorem states:

The probability of $E_1$ given $A$ is:

And the probability of $E_2$ given $A$ is:

It is important to note that the denominator in both formulas, $P(A)$, is calculated using the Law of Total Probability:

Where:

• $P(E_1|A)$ is the posterior probability of $E_1$ given event $A$.
• $P(E_2|A)$ is the posterior probability of $E_2$ given event $A$.
• $P(A|E_1)$ is the likelihood of event $A$ given $E_1$.
• $P(E_1)$ is the prior probability of $E_1$.
• $P(A|E_2)$ is the likelihood of event $A$ given $E_2$.
• $P(E_2)$ is the prior probability of $E_2$.

Given:

Bag A contains 3 red balls and 4 black balls.

Bag B contains 5 red balls and 6 black balls.

One bag is chosen at random, and a ball is drawn.

To Find:

The probability that a ball is drawn from Bag A.

Solution:

Let $E_A$ be the event that Bag A is chosen.

Let $E_B$ be the event that Bag B is chosen.

Since one bag is chosen at random, the probability of choosing either bag is equal.

The question asks for the probability that a ball is drawn from Bag A. This is simply the probability of choosing Bag A, as stated in the problem setup.

Therefore, the probability that a ball is drawn from Bag A is $\frac{1}{2}$.

Given:

Bag A contains 3 red balls and 4 black balls. Total balls in Bag A = $3 + 4 = 7$.

Bag B contains 5 red balls and 6 black balls. Total balls in Bag B = $5 + 6 = 11$.

One bag is chosen at random.

To Find:

The probability of drawing a red ball.

Solution:

Let $E_A$ be the event that Bag A is chosen.

Let $E_B$ be the event that Bag B is chosen.

Let $R$ be the event of drawing a red ball.

Since one bag is chosen at random:

The probability of drawing a red ball from Bag A is:

The probability of drawing a red ball from Bag B is:

We can use the Law of Total Probability to find the probability of drawing a red ball:

Substituting the values:

Now, we calculate the terms:

To add these fractions, we find a common denominator, which is 154 ($14 \times 11$):

We can simplify this fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 2:

Therefore, the probability of drawing a red ball is $\frac{34}{77}$.

Given:

The conditional probability of event A given event B is equal to the probability of event A.

Implication:

We need to determine the relationship between events A and B when the condition $P(A|B) = P(A)$ holds true.

Solution:

We know the formula for conditional probability:

Provided that $P(B) \neq 0$.

Now, we equate the given condition (i) with the formula for conditional probability (ii):

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

This relationship, $P(A \cap B) = P(A)P(B)$, is the definition of statistical independence between two events A and B.

Therefore, if $P(A|B) = P(A)$, it implies that events A and B are independent.

This means that the occurrence of event B does not affect the probability of event A occurring, and vice versa.

Given:

A coin is biased such that heads (H) is twice as likely as tails (T).

To Find:

The probability of getting heads when the coin is tossed once.

Solution:

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

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

According to the problem statement, heads is twice as likely as tails:

Since there are only two possible outcomes for a coin toss (heads or tails), and these outcomes are mutually exclusive and exhaustive, the sum of their probabilities must be 1:

Now we can substitute equation (i) into equation (ii):

Combining the terms:

Solving for $P(T)$:

Now, we can find the probability of getting heads using equation (i):

Therefore, the probability of getting heads is $\frac{2}{3}$.

Given:

A group consists of 5 men and 4 women.

A committee of 3 is to be formed.

To Find:

The probability that the committee consists of exactly 2 men.

Solution:

First, we need to find the total number of ways to form a committee of 3 from the group of $5 + 4 = 9$ people. This is a combination problem, as the order in which members are chosen for the committee does not matter.

Total number of ways to choose a committee of 3 from 9 people is given by $\binom{9}{3}$:

Next, we need to find the number of ways to form a committee with exactly 2 men. This means we need to choose 2 men from the 5 men AND 1 woman from the 4 women (to complete the committee of 3).

Number of ways to choose 2 men from 5 is $\binom{5}{2}$:

Number of ways to choose 1 woman from 4 is $\binom{4}{1}$:

The number of committees with exactly 2 men and 1 woman is the product of these two numbers:

The probability of the committee consisting of exactly 2 men is the ratio of the number of favorable committees to the total number of possible committees:

Simplifying the fraction by dividing the numerator and denominator by their greatest common divisor, which is 4:

Therefore, the probability that the committee consists of exactly 2 men is $\frac{10}{21}$.

Given:

Probability of event A, $P(A) = 0.3$.

Probability of event B, $P(B) = 0.4$.

Probability of both A and B occurring, $P(A \cap B) = 0.1$.

To Find:

The probability of neither A nor B occurring, denoted as $P(\text{neither A nor B})$.

Solution:

The event "neither A nor B" can be expressed in terms of set notation as $(A \cup B)^c$, which means the complement of the union of A and B.

We know that the probability of the complement of an event is 1 minus the probability of the event:

To find $P(A \cup B)$, we use the formula for the probability of the union of two events:

Substitute the given values into equation (ii):

Now, substitute the value of $P(A \cup B)$ into equation (i) to find the probability of neither A nor B:

Therefore, the probability of neither A nor B is 0.4.

Given:

A standard deck of 52 playing cards.

To Find:

The probability that a card drawn is a face card (King, Queen, or Jack).

Solution:

A standard deck of 52 cards has 4 suits: Hearts, Diamonds, Clubs, and Spades.

Each suit contains the following cards:

• Ace
• 2, 3, 4, 5, 6, 7, 8, 9, 10
• Jack (J)
• Queen (Q)
• King (K)

The face cards are the King, Queen, and Jack.

In each suit, there are 3 face cards (Jack, Queen, King).

Since there are 4 suits, the total number of face cards in a deck is:

The total number of possible outcomes (drawing any card from the deck) is 52.

The probability of an event is calculated as:

In this case, the favorable outcomes are drawing a face card.

Probability of drawing a face card = $\frac{\text{Number of face cards}}{\text{Total number of cards}}$

We can simplify this fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 4:

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

Given:

A standard deck of 52 playing cards.

To Find:

The probability that a card drawn is a face card (King, Queen, or Jack).

Solution:

A standard deck of 52 cards has 4 suits: Hearts, Diamonds, Clubs, and Spades.

Each suit contains the following cards:

• Ace
• 2, 3, 4, 5, 6, 7, 8, 9, 10
• Jack (J)
• Queen (Q)
• King (K)

The face cards are the King, Queen, and Jack.

In each suit, there are 3 face cards (Jack, Queen, King).

Since there are 4 suits, the total number of face cards in a deck is:

The total number of possible outcomes (drawing any card from the deck) is 52.

The probability of an event is calculated as:

In this case, the favorable outcomes are drawing a face card.

Probability of drawing a face card = $\frac{\text{Number of face cards}}{\text{Total number of cards}}$

We can simplify this fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 4:

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

Given:

The probability of the complement of event A, $P(A') = 0.6$.

The probability of event B, $P(B) = 0.5$.

The probability of the intersection of A and B, $P(A \cap B) = 0.2$.

To Find:

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

Solution:

We are given $P(A') = 0.6$. We know that $P(A') = 1 - P(A)$.

Therefore, we can find $P(A)$:

Now we have $P(A) = 0.4$, $P(B) = 0.5$, and $P(A \cap B) = 0.2$.

To find $P(A \cup B)$, we use the formula for the probability of the union of two events:

Substitute the values we have:

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

Given:

The probabilities of solving a mathematics problem by students A, B, and C are:

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

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

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

The students work independently.

To Find:

The probability that the problem is solved.

Solution:

The problem is solved if at least one of the students solves it. It's easier to calculate the probability that none of them solve the problem and then subtract that from 1.

First, let's find the probabilities that each student does NOT solve the problem:

Probability that A does not solve it, $P(A') = 1 - P(A) = 1 - \frac{1}{2} = \frac{1}{2}$.

Probability that B does not solve it, $P(B') = 1 - P(B) = 1 - \frac{1}{3} = \frac{2}{3}$.

Probability that C does not solve it, $P(C') = 1 - P(C) = 1 - \frac{1}{4} = \frac{3}{4}$.

Since the students work independently, the probability that none of them solve the problem is the product of their individual probabilities of not solving it:

Now, we multiply these fractions:

Simplifying the fraction:

The probability that the problem is solved is the complement of the probability that none of them solve it:

Therefore, the probability that the problem is solved is $\frac{3}{4}$.

Given:

A discrete random variable X with the following probability distribution:

To Find:

The value of k.

Solution:

For any probability distribution of a discrete random variable, the sum of all probabilities must equal 1.

Therefore, we have:

Substitute the given probabilities into the equation:

Combine the known probability values:

Now, solve for k by subtracting 0.6 from both sides of the equation:

Therefore, the value of k is 0.4.

Given:

A pair of dice is rolled.

To Find:

The probability that the sum of the numbers shown on the two dice is greater than 9.

Solution:

When a pair of dice is rolled, the total number of possible outcomes is $6 \times 6 = 36$. Each outcome is equally likely.

We need to find the outcomes where the sum of the numbers is greater than 9. This means the sum can be 10, 11, or 12.

Let's list the combinations that result in these sums:

• Sum of 10: (4, 6), (5, 5), (6, 4) - 3 outcomes
• Sum of 11: (5, 6), (6, 5) - 2 outcomes
• Sum of 12: (6, 6) - 1 outcome

The total number of favorable outcomes (where the sum is greater than 9) is the sum of the outcomes for sums 10, 11, and 12:

The probability of an event is calculated as:

Therefore, the probability that the sum is greater than 9 is:

Simplifying the fraction:

Therefore, the probability that the sum is greater than 9 is $\frac{1}{6}$.

Given:

Two events, A and B.

To Express:

The event "exactly one of A or B occurs" in two ways:

1. In set notation.
2. In terms of probabilities using $P(A), P(B), P(A \cap B)$.

Solution:

a) In set notation:

The event "exactly one of A or B occurs" means that either A occurs and B does not, OR B occurs and A does not.

The event "A occurs and B does not" can be written as $A \cap B'$.

The event "B occurs and A does not" can be written as $B \cap A'$.

Since these two possibilities are mutually exclusive (they cannot happen at the same time), we can express "exactly one of A or B occurs" as the union of these two events:

This can also be expressed using the symmetric difference operator ($\Delta$):

Alternatively, it is the union of A and B, excluding the intersection of A and B:

b) In terms of probabilities using $P(A), P(B), P(A \cap B)$:

Using the first set notation $(A \cap B') \cup (B \cap A')$, since these two events are mutually exclusive, the probability is the sum of their individual probabilities:

$P(\text{exactly one of A or B occurs}) = P(A \cap B') + P(B \cap A')$

We know that $P(A \cap B') = P(A) - P(A \cap B)$ and $P(B \cap A') = P(B) - P(A \cap B)$.

Substituting these into the equation:

Simplifying this expression:

Alternatively, using the second set notation $(A \cup B) \setminus (A \cap B)$, the probability is:

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

We know that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.

Substituting this into the equation:

Simplifying this expression also leads to:

Therefore, the probability that exactly one of A or B occurs is $P(A) + P(B) - 2P(A \cap B)$.

Given:

The probability that a student passes test I, $P(I) = 0.8$.

The probability that a student passes test II, $P(II) = 0.7$.

The probability that the student passes both tests, $P(I \cap II) = 0.6$.

To Find:

The probability that the student passes at least one test.

Solution:

The event "passes at least one test" means the student passes test I, or passes test II, or passes both tests. This is represented by the union of the two events, $P(I \cup II)$.

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

Substitute the given values into the formula:

Perform the addition and subtraction:

Therefore, the probability that the student passes at least one test is 0.9.

Given:

Probability of event A, $P(A) = \frac{1}{3}$.

Probability of event B, $P(B) = \frac{1}{4}$.

Probability of the intersection of A and B, $P(A \cap B) = \frac{1}{12}$.

To Find:

The conditional probability of A given B, $P(A|B)$.

The conditional probability of B given A, $P(B|A)$.

Solution:

1. To find $P(A|B)$ (Probability of A given B):

We use the formula for conditional probability:

Substitute the given values:

To divide fractions, we multiply the numerator by the reciprocal of the denominator:

Simplify the fraction:

2. To find $P(B|A)$ (Probability of B given A):

We use the formula for conditional probability:

Substitute the given values:

To divide fractions, we multiply the numerator by the reciprocal of the denominator:

Simplify the fraction:

Therefore:

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

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

Given:

A coin is tossed and a die is rolled.

To Write:

The sample space for this joint experiment.

Solution:

The experiment involves two independent events:

• Coin Toss: The possible outcomes are Heads (H) and Tails (T).
• Die Roll: The possible outcomes are the numbers 1, 2, 3, 4, 5, 6.

The sample space for a joint experiment consists of all possible pairs of outcomes from each individual experiment.

We can list all possible combinations by pairing each outcome of the coin toss with each outcome of the die roll.

The sample space, denoted by S, can be written as:

The total number of outcomes in the sample space is the product of the number of outcomes for each event: 2 (for the coin) $\times$ 6 (for the die) = 12 outcomes.

Given:

A and B are two independent events.

To Prove:

A and $B'$ are also independent events.

Proof:

By definition, two events A and B are independent if $P(A \cap B) = P(A)P(B)$.

We need to show that $P(A \cap B') = P(A)P(B')$.

Consider the event A. It can be partitioned into two mutually exclusive events:

1. A occurs and B occurs ($A \cap B$).
2. A occurs and B does not occur ($A \cap B'$).

Therefore, we can write:

We want to find $P(A \cap B')$. Rearranging equation (1), we get:

Since A and B are independent, we can substitute $P(A \cap B) = P(A)P(B)$ into equation (2):

Factor out $P(A)$ from the right side:

We know that $1 - P(B) = P(B')$. So, we can write:

This is the definition of independence for events A and $B'$.

Thus, if A and B are independent, then A and $B'$ are also independent.

Hence proved.

Given:

A standard deck of 52 playing cards.

To Find:

The probability of drawing a King or a Spade.

Solution:

Let K be the event of drawing a King.

Let S be the event of drawing a Spade.

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

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

Now, let's determine the individual probabilities:

1. Probability of drawing a King ($P(K)$):

There are 4 Kings in a deck of 52 cards (King of Hearts, King of Diamonds, King of Clubs, King of Spades).

2. Probability of drawing a Spade ($P(S)$):

There are 13 Spades in a deck of 52 cards (Ace of Spades through King of Spades).

3. Probability of drawing a King AND a Spade ($P(K \cap S)$):

There is only one card that is both a King and a Spade: the King of Spades.

Now, substitute these values back into the union formula:

Perform the addition and subtraction:

Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 4:

Therefore, the probability of drawing a King or a Spade is $\frac{4}{13}$.

Given:

A deck of 52 playing cards.

Two cards are drawn simultaneously.

To Find:

The probability that one card is a King and the other is a Queen.

Solution:

First, we need to find the total number of ways to draw two cards from a deck of 52 cards. Since the cards are drawn simultaneously, the order does not matter, so we use combinations:

Next, we need to find the number of ways to draw one King and one Queen.

There are 4 Kings in the deck, and we need to choose 1 King:

There are 4 Queens in the deck, and we need to choose 1 Queen:

To get one King AND one Queen, we multiply the number of ways to choose each:

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

Now, we simplify the fraction. Both 16 and 1326 are divisible by 2:

Therefore, the probability that one card is a King and the other is a Queen is $\frac{8}{663}$.

Definition of the Complement of an Event:

Let $S$ be the sample space of a random experiment, which is the set of all possible outcomes.

Let $A$ be an event, which is a subset of the sample space $S$ ($A \subseteq S$).

The complement of the event $A$, denoted by $A'$ (or sometimes $\bar{A}$ or $A^c$), is the set of all outcomes in the sample space $S$ that are not in $A$.

In set notation, the complement of event $A$ is given by:

Essentially, $A'$ represents the event that "$A$ does not occur".

Relationship between $P(A)$ and $P(A')$:

Since the event $A$ and its complement $A'$ comprise all possible outcomes in the sample space $S$ and have no outcomes in common, they satisfy two conditions:

1. $A$ and $A'$ are mutually exclusive events, meaning their intersection is empty: $A \cap A' = \emptyset$.

2. $A$ and $A'$ are exhaustive events with respect to $S$, meaning their union is the entire sample space: $A \cup A' = S$.

For any event $A$ and its complement $A'$ in a sample space $S$, the sum of their probabilities is equal to 1. This is because the event $A \cup A'$ is the certain event ($S$), and the probability of the certain event is 1.

Using the axiom of probability for mutually exclusive events $P(A \cup A') = P(A) + P(A')$, and since $A \cup A' = S$, we have $P(A \cup A') = P(S) = 1$.

Therefore, the relationship between $P(A)$ and $P(A')$ is:

This relationship can also be expressed as:

or

Equation (2) is the standard formula used to calculate the probability of the complement of an event.

Problem Analysis:

We have a bag with 10 balls numbered from 1 to 10. We are interested in the probability of drawing a ball whose number satisfies two conditions simultaneously: it must be an even number and it must be a multiple of 3.

Solution:

First, let's identify the sample space ($S$), which is the set of all possible outcomes when drawing a ball from the bag.

The total number of possible outcomes is the number of elements in the sample space, denoted by $n(S)$.

Next, let's define the event we are interested in. Let $E$ be the event that the number drawn is even.

Let $M$ be the event that the number drawn is a multiple of 3.

We are looking for the probability that the number is even AND a multiple of 3. This corresponds to the intersection of events $E$ and $M$, denoted by $E \cap M$.

We need to find the numbers that are present in both sets $E$ and $M$.

Looking at the sets $E$ and $M$, the only common number is 6.

The number of outcomes favorable to the event ($E \cap M$) is $n(E \cap M)$.

The probability of an event is given by the formula:

In this case, the event is $E \cap M$. So, the probability is:

Substituting the values we found:

The probability that the number is even and a multiple of 3 is $\boldsymbol{\frac{1}{10}}$.

Problem Analysis:

We are given the probabilities of two events, $A$ and $B$, occurring, $P(A) = 0.5$ and $P(B) = 0.3$. We are also told that events $A$ and $B$ are mutually exclusive. We need to find the probability that both events $A$ and $B$ occur, which is denoted as $P(A \text{ and } B)$ or $P(A \cap B)$.

Solution:

Two events $A$ and $B$ are said to be mutually exclusive if they cannot occur at the same time. This means that the occurrence of one event prevents the occurrence of the other.

In terms of set theory, the intersection of two mutually exclusive events is the empty set ($\emptyset$).

The probability of the intersection of two mutually exclusive events is 0, because there are no outcomes common to both events.

Given that $A$ and $B$ are mutually exclusive, the probability of both $A$ and $B$ occurring is 0, regardless of their individual probabilities $P(A)$ and $P(B)$ (as long as they are valid probabilities, i.e., between 0 and 1 inclusive).

The probability $P(A \text{ and } B)$ is $\boldsymbol{0}$.

Problem Analysis:

We are given the percentage of students who passed in Physics, the percentage who passed in Mathematics, and the percentage who failed in both subjects. We need to find the percentage of students who passed in both Physics and Mathematics.

Let's denote the events:

$P$: A student passes in Physics.

$M$: A student passes in Mathematics.

We are given the following probabilities (expressed as decimals):

The percentage of students who fail in both subjects means they fail in Physics AND fail in Mathematics. Failing in Physics is the complement of passing Physics ($P'$), and failing in Mathematics is the complement of passing Mathematics ($M'$).

We want to find the percentage of students who pass in both subjects, which is the probability of the intersection of events $P$ and $M$, i.e., $P(P \cap M)$.

Solution:

We are given $P(P' \cap M') = 0.20$.

According to De Morgan's Law, the complement of the union of two events is the intersection of their complements:

Therefore, the probability of the complement of the union of $P$ and $M$ is equal to the probability of the intersection of their complements:

So, we have:

Using the complement rule, $P(A') = 1 - P(A)$, we can find the probability of the union $P \cup M$:

Substitute the value of $P((P \cup M)')$ into equation (1):

This means that $80\%$ of students pass in Physics OR Mathematics (or both).

Now, we use the Addition Rule for Probabilities, which states that for any two events $A$ and $B$:

Applying this to events $P$ and $M$:

We know $P(P \cup M) = 0.80$, $P(P) = 0.60$, and $P(M) = 0.70$. Substitute these values into equation (3):

Simplify the equation:

Rearrange the equation to solve for $P(P \cap M)$:

To express this probability as a percentage, multiply by 100:

The percentage of students who pass in both subjects is $\boldsymbol{50\%}$.

Problem Analysis:

We have a bag containing 5 red balls and 7 blue balls. The total number of balls is $5 + 7 = 12$.

We are drawing two balls from the bag without replacement. This means 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.

We need to find the probability of three different events:

i) Drawing two red balls.

ii) Drawing two blue balls.

iii) Drawing one red and one blue ball.

Solution:

Total number of balls in the bag = $5 \text{ (Red)} + 7 \text{ (Blue)} = 12$ balls.

i) Probability of drawing two red balls:

This means drawing a red ball first, and then drawing another red ball from the remaining balls.

Probability of drawing a red ball on the first draw:

After drawing one red ball, there are $5 - 1 = 4$ red balls left, and the total number of balls remaining is $12 - 1 = 11$.

Probability of drawing a red ball on the second draw, given that the first ball was red:

The probability of drawing two red balls is the product of the probabilities of these two consecutive events:

Simplifying the fraction:

ii) Probability of drawing two blue balls:

This means drawing a blue ball first, and then drawing another blue ball from the remaining balls.

Probability of drawing a blue ball on the first draw:

After drawing one blue ball, there are $7 - 1 = 6$ blue balls left, and the total number of balls remaining is $12 - 1 = 11$.

Probability of drawing a blue ball on the second draw, given that the first ball was blue:

The probability of drawing two blue balls is the product of the probabilities of these two consecutive events:

Simplifying the fraction:

iii) Probability of drawing one red and one blue ball:

This event can occur in two mutually exclusive ways:

Case 1: Drawing a Red ball first, then a Blue ball ($R_1, B_2$).

Case 2: Drawing a Blue ball first, then a Red ball ($B_1, R_2$).

Calculate the probability of Case 1 (R then B):

We know $P(\text{1st ball is Red}) = \frac{5}{12}$ from (1).

After drawing one red ball, there are still 7 blue balls left, and the total number of balls remaining is $12 - 1 = 11$.

Calculate the probability of Case 2 (B then R):

We know $P(\text{1st ball is Blue}) = \frac{7}{12}$ from (3).

After drawing one blue ball, there are still 5 red balls left, and the total number of balls remaining is $12 - 1 = 11$.

Since these two cases are mutually exclusive (either you draw R then B, or B then R), the probability of drawing one red and one blue ball is the sum of their probabilities:

Simplifying the fraction:

Summary of Probabilities:

i) Probability of drawing two red balls = $\boldsymbol{\frac{5}{33}}$

ii) Probability of drawing two blue balls = $\boldsymbol{\frac{7}{22}}$

iii) Probability of drawing one red and one blue ball = $\boldsymbol{\frac{35}{66}}$

Problem Analysis:

We are given the probabilities of two events A and B, and the probability of their intersection: $P(A) = 1/2$, $P(B) = 1/3$, and $P(A \cap B) = 1/4$. We need to calculate several related probabilities, including conditional probabilities and the probability of the union.

Solution:

We are given:

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

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

Substitute the given values into equation (i):

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

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

Substitute the given values into equation (ii):

Simplify the fraction:

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

The formula for this conditional probability is:

First, we find $P(B')$ using the complement rule $P(B') = 1 - P(B)$:

Next, we find $P(A' \cap B')$. Using De Morgan's Law, $A' \cap B' = (A \cup B)'$.

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

To find $P((A \cup B)')$, we first need to find $P(A \cup B)$ using the Addition Rule:

Substitute the given values into equation (v):

Find a common denominator (LCM of 2, 3, and 4 is 12):

Now, find $P(A' \cap B') = P((A \cup B)')$ using the complement rule:

Substitute the value from equation (vi) into equation (vii):

Finally, substitute the values from equations (iv) and (viii) into equation (iii):

Simplify the fraction:

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

We already calculated this probability in step iii) using the Addition Rule (equation v).

Summary of Results:

i) $P(A|B) = \boldsymbol{\frac{3}{4}}$

ii) $P(B|A) = \boldsymbol{\frac{1}{2}}$

iii) $P(A'|B') = \boldsymbol{\frac{5}{8}}$

iv) $P(A \cup B) = \boldsymbol{\frac{7}{12}}$

Problem Analysis:

We are given information about the proportion of items produced by two machines (A and B) and the probability of an item being defective based on which machine produced it. We are asked to find the probability that a randomly selected item, which is known to be defective, was produced by Machine A.

This is a classic application of Bayes' Theorem.

Let's define the events:

$A$: The selected item was produced by Machine A.

$B$: The selected item was produced by Machine B.

$D$: The selected item is defective.

We are given the following probabilities:

We want to find the probability that the item was produced by Machine A given that it is defective, which is $P(A|D)$.

Solution:

According to Bayes' Theorem, the probability of event A given event D is:

To use this formula, we first need to calculate the overall probability of an item being defective, $P(D)$. We can find this using the Law of Total Probability. Since an item must be produced by either Machine A or Machine B (and not both), events A and B form a partition of the sample space of production sources.

Substitute the given values into equation (2):

Now, substitute the values of $P(D|A)$, $P(A)$, and $P(D)$ from equation (3) into Bayes' Theorem (equation 1):

To simplify, we can multiply the numerator and denominator by 1000:

Simplify the fraction:

The probability can also be expressed as a decimal or percentage:

The probability that the defective item was produced by Machine A is $\boldsymbol{\frac{1}{2}}$ or $\boldsymbol{0.5}$ or $\boldsymbol{50\%}$.

Problem Analysis:

We are given the percentages of people who own a car, a motorcycle, and both. We need to find the probabilities of having a car or a motorcycle, having neither, and having a car given they have a motorcycle.

Let's define the events:

$C$: A person has a car.

$M$: A person has a motorcycle.

We are given the following probabilities:

Solution:

i) Probability that the person has a car or a motorcycle:

This is the probability of the union of events C and M, $P(C \cup M)$. We use the Addition Rule for Probabilities:

Substitute the given probabilities into equation (1):

This means $60\%$ of people have a car or a motorcycle (or both).

ii) Probability that the person has neither a car nor a motorcycle:

This means the person does NOT have a car (event $C'$) AND does NOT have a motorcycle (event $M'$). This is the probability of the intersection of the complements, $P(C' \cap M')$.

Using De Morgan's Law, we know that $C' \cap M' = (C \cup M)'$.

So, $P(C' \cap M') = P((C \cup M)')$.

Using the Complement Rule, $P(E') = 1 - P(E)$, we have:

Substitute the value of $P(C \cup M)$ from equation (2) into equation (3):

This means $40\%$ of people have neither a car nor a motorcycle.

iii) Probability that the person has a car given that he has a motorcycle:

This is the conditional probability $P(C|M)$. The formula for conditional probability is:

Substitute the given probabilities into equation (4):

This means that among the people who have a motorcycle, $25\%$ also have a car.

Summary of Results:

i) Probability of having a car or a motorcycle = $\boldsymbol{0.60}$ or $\boldsymbol{60\%}$

ii) Probability of having neither a car nor a motorcycle = $\boldsymbol{0.40}$ or $\boldsymbol{40\%}$

iii) Probability of having a car given that he has a motorcycle = $\boldsymbol{0.25}$ or $\boldsymbol{\frac{1}{4}}$ or $\boldsymbol{25\%}$

Problem Analysis:

We are drawing a single card from a standard deck of 52 cards. We have two defined events: drawing a spade and drawing an ace. We need to determine if these two events are independent and provide a mathematical justification.

Solution:

A standard deck of cards has 52 cards.

Let the sample space $S$ be the set of all possible outcomes when drawing a single card. The total number of outcomes is $n(S) = 52$.

Let event $E$ be 'the card is a spade'.

There are 13 spades in a deck (Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King of spades).

The probability of event E is:

Let event $F$ be 'the card is an ace'.

There are 4 aces in a deck (Ace of Spades, Ace of Hearts, Ace of Diamonds, Ace of Clubs).

The probability of event F is:

The intersection of events E and F, denoted by $E \cap F$, is the event that the card is a spade AND an ace.

The only card that is both a spade and an ace is the Ace of Spades.

The probability of the intersection is:

Two events E and F are considered independent if the probability of their intersection is equal to the product of their individual probabilities. Mathematically, this condition is:

Let's check if this condition holds using the probabilities we calculated:

From (1), $P(E) = \frac{1}{4}$.

From (2), $P(F) = \frac{1}{13}$.

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

Now, compare the value of $P(E \cap F)$ from (3) with the value of $P(E) \times P(F)$ from (5).

Since $\frac{1}{52} = \frac{1}{52}$, the condition $P(E \cap F) = P(E) \times P(F)$ is satisfied.

Justification:

Yes, events E ('the card is a spade') and F ('the card is an ace') are independent events because the probability of drawing a card that is both a spade and an ace ($P(E \cap F) = \frac{1}{52}$) is equal to the product of the individual probabilities of drawing a spade ($P(E) = \frac{1}{4}$) and drawing an ace ($P(F) = \frac{1}{13}$). $P(E \cap F) = P(E) \times P(F)$.

Problem Analysis:

This problem describes a two-stage experiment. The first stage is throwing a die, which determines the second stage, which is tossing a coin a certain number of times. We need to find the overall probability of getting exactly one head, considering both possible outcomes of the first stage.

Solution:

Let's define the events involved:

Let $D_1$ be the event that the die shows a 5 or 6.

Let $D_2$ be the event that the die shows a 1, 2, 3, or 4.

Let $H$ be the event that he gets exactly one head in the coin tosses.

First, calculate the probabilities of the events related to the die throw:

Total number of outcomes when throwing a die = 6 (i.e., 1, 2, 3, 4, 5, 6).

Number of outcomes for $D_1$ (5 or 6) = 2.

Number of outcomes for $D_2$ (1, 2, 3, or 4) = 4.

Note that $P(D_1) + P(D_2) = \frac{1}{3} + \frac{2}{3} = 1$, as $D_1$ and $D_2$ cover all possible outcomes of the die roll and are mutually exclusive.

Next, consider the probabilities of getting exactly one head for each case of the die throw:

Case 1: Die is 5 or 6 (Event $D_1$)

The coin is tossed three times. The possible outcomes are {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}. Total number of outcomes = $2^3 = 8$.

Outcomes with exactly one head = {HTT, THT, TTH}. Number of favorable outcomes = 3.

The conditional probability of getting exactly one head given $D_1$ occurred is $P(H|D_1)$:

Case 2: Die is 1, 2, 3, or 4 (Event $D_2$)

The coin is tossed two times. The possible outcomes are {HH, HT, TH, TT}. Total number of outcomes = $2^2 = 4$.

Outcomes with exactly one head = {HT, TH}. Number of favorable outcomes = 2.

The conditional probability of getting exactly one head given $D_2$ occurred is $P(H|D_2)$:

The event of getting exactly one head ($H$) can occur if either $D_1$ happens AND $H$ happens (in 3 tosses), OR if $D_2$ happens AND $H$ happens (in 2 tosses).

Using the Law of Total Probability, since $D_1$ and $D_2$ are mutually exclusive and exhaustive, the total probability of event H is:

Using the definition of conditional probability, $P(A \cap B) = P(A|B) P(B)$:

Substitute these into equation (5):

Substitute the probabilities from (1), (2), (3), and (4):

Simplify the fractions and find a common denominator (24):

The probability that he gets exactly one head is $\boldsymbol{\frac{11}{24}}$.

Problem Analysis:

We are given the probabilities of two independent events A and B: $P(A) = 0.4$ and $P(B) = 0.5$. We need to calculate the probabilities of three specific combinations of these events occurring.

The key information is that events A and B are independent.

Solution:

We are given:

Since A and B are independent events, their complements A' and B' are also independent, A and B' are independent, and A' and B are independent.

The probabilities of the complements are:

i) Probability that both A and B occur:

This is the probability of the intersection of A and B, $P(A \cap B)$. Since A and B are independent, we use the multiplication rule for independent events:

Substitute the given probabilities into equation (3):

ii) Probability that exactly one of A or B occurs:

This event means either A occurs and B does not ($A \cap B'$) OR B occurs and A does not ($B \cap A'$). These two outcomes are mutually exclusive.

So, the probability is $P((A \cap B') \cup (B \cap A')) = P(A \cap B') + P(B \cap A')$.

Since A and B are independent, A and B' are independent, and B and A' are independent. Therefore:

Substitute the probabilities from (1), (2), and the given values:

The probability of exactly one of A or B occurring is the sum of these probabilities:

iii) Probability that neither A nor B occurs:

This means that event A does not occur ($A'$) AND event B does not occur ($B'$). This is the probability of the intersection of the complements, $P(A' \cap B')$.

Since A and B are independent, their complements A' and B' are also independent. Therefore, we can use the multiplication rule for independent events:

Substitute the probabilities from (1) and (2) into equation (6):

Alternate Solution for iii):

Using De Morgan's Law, $A' \cap B' = (A \cup B)'$. So, $P(A' \cap B') = P((A \cup B)')$.

Using the complement rule, $P(E') = 1 - P(E)$, we have $P((A \cup B)') = 1 - P(A \cup B)$.

First, find $P(A \cup B)$ using the Addition Rule:

Substitute the given probabilities and the result from part i) ($P(A \cap B) = 0.20$):

Now, find $P(A' \cap B')$:

Both methods yield the same result.

Summary of Results:

i) Probability that both A and B occur = $\boldsymbol{0.20}$ or $\boldsymbol{20\%}$

ii) Probability that exactly one of A or B occurs = $\boldsymbol{0.50}$ or $\boldsymbol{50\%}$

iii) Probability that neither A nor B occurs = $\boldsymbol{0.30}$ or $\boldsymbol{30\%}$

Problem Analysis:

We have a bag containing 4 red balls and 5 black balls. The total number of balls is $4 + 5 = 9$. We are drawing two balls one after the other. We need to find the probability that the first ball is red and the second ball is black under two different conditions: when the first ball is replaced before drawing the second, and when it is not replaced.

Let's define the events:

$R_1$: The first ball drawn is red.

$B_2$: The second ball drawn is black.

We want to find $P(R_1 \cap B_2)$.

Solution - With Replacement:

When the first ball is drawn and then replaced, the composition of the bag remains the same for the second draw. This means the two draws are independent events.

Total number of balls = 9

Number of red balls = 4

Number of black balls = 5

Probability of drawing a red ball on the first draw ($P(R_1)$):

Since the first ball is replaced, the bag still has 9 balls (4 red and 5 black) for the second draw.

Probability of drawing a black ball on the second draw ($P(B_2)$):

For independent events, $P(R_1 \cap B_2) = P(R_1) \times P(B_2)$.

Substitute the probabilities from (1) and (2):

Solution - Without Replacement:

When the first ball is drawn and NOT replaced, the composition of the bag changes for the second draw. This means the two draws are dependent events.

Probability of drawing a red ball on the first draw ($P(R_1)$):

After drawing one red ball and not replacing it, the bag contains $9 - 1 = 8$ balls.

The remaining balls are $4 - 1 = 3$ red balls and 5 black balls.

Probability of drawing a black ball on the second draw, given that the first ball was red ($P(B_2 | R_1)$):

For dependent events, the probability of both events occurring is given by the multiplication rule for dependent events:

Substitute the probabilities from (1) and (3) into equation (4):

Simplify the fraction:

Summary of Results:

Probability of the first ball being red and the second ball being black:

With replacement: $\boldsymbol{\frac{20}{81}}$

Without replacement: $\boldsymbol{\frac{5}{18}}$

Problem Analysis:

We have three boxes with different compositions of red and black balls. A box is chosen randomly, and then a ball is drawn from that box. We are given that the drawn ball is red and we need to find the probability that it came from a specific box (Box B).

This is a conditional probability problem, specifically a case where we need to use Bayes' Theorem.

Let's define the events:

$A$: The chosen box is Box A.

$B$: The chosen box is Box B.

$C$: The chosen box is Box C.

$R$: The drawn ball is red.

We are given that a box is chosen at random. Since there are three boxes, the probability of choosing any specific box is equal:

Now, let's find the conditional probability of drawing a red ball given that a specific box was chosen:

From Box A (2 red, 3 black; Total 5 balls):

From Box B (4 red, 1 black; Total 5 balls):

From Box C (3 red, 4 black; Total 7 balls):

We want to find the probability that the box was Box B given that the ball drawn is red, i.e., $P(B|R)$.

Solution:

Using Bayes' Theorem, the probability of event B given event R is:

To use this formula, we first need to calculate the total probability of drawing a red ball, $P(R)$. We use the Law of Total Probability, considering that the ball must come from either Box A, Box B, or Box C:

Substitute the probabilities from (1), (2), (3), (4), (5), and (6) into equation (8):

Combine the first two terms and simplify the third term:

Simplify the fraction $\frac{6}{15}$:

Find a common denominator (35) and add the fractions:

Now, substitute the values of $P(R|B)$ from (5), $P(B)$ from (2), and $P(R)$ from (9) into Bayes' Theorem (equation 7):

Calculate the numerator:

So, we have:

To divide by a fraction, multiply by its reciprocal:

Simplify by cancelling common factors:

The probability that the red ball came from Box B is $\boldsymbol{\frac{28}{57}}$.

Problem Analysis:

We are selecting a committee of 5 persons from a group containing 8 men and 4 women. The total number of persons in the group is $8 + 4 = 12$. The selection is done without regard to order, so this is a problem involving combinations.

We need to find the probability of two specific compositions of the committee:

i) Exactly 3 men and 2 women.

ii) At least 3 women (which means 3 women and 2 men, OR 4 women and 1 man).

Solution:

Total number of people = 8 men + 4 women = 12.

Committee size = 5.

The total number of ways to form a committee of 5 from 12 people is given by the combination formula $\binom{n}{k} = \frac{n!}{k!(n-k)!}$, where $n$ is the total number of items and $k$ is the number of items to choose.

There are 792 possible committees of 5 that can be formed.

i) Probability of exactly 3 men and 2 women:

To form a committee with exactly 3 men and 2 women, we need to:

- Choose 3 men from the 8 men.

- Choose 2 women from the 4 women.

Number of ways to choose 3 men from 8 = $\binom{8}{3}$.

Number of ways to choose 2 women from 4 = $\binom{4}{2}$.

The number of ways to form the committee with exactly 3 men and 2 women is the product of the number of ways to choose the men and the number of ways to choose the women:

The probability of this event is the number of favorable outcomes divided by the total number of possible outcomes:

Simplify the fraction:

ii) Probability of at least 3 women:

"At least 3 women" in a committee of 5 means the committee can have:

Case 1: Exactly 3 women and 2 men.

Case 2: Exactly 4 women and 1 man.

We calculate the number of ways for each case and sum them up (since these cases are mutually exclusive).

Case 1: Exactly 3 women and 2 men

Number of ways to choose 3 women from 4 = $\binom{4}{3}$.

Number of ways to choose 2 men from 8 = $\binom{8}{2}$.

Case 2: Exactly 4 women and 1 man

Number of ways to choose 4 women from 4 = $\binom{4}{4}$.

Number of ways to choose 1 man from 8 = $\binom{8}{1}$.

The total number of favorable outcomes for "at least 3 women" is the sum of the ways in Case 1 and Case 2:

The probability of this event is the number of favorable outcomes divided by the total number of possible outcomes:

Simplify the fraction:

Summary of Probabilities:

i) Probability that the committee includes exactly 3 men and 2 women = $\boldsymbol{\frac{14}{33}}$

ii) Probability that the committee includes at least 3 women = $\boldsymbol{\frac{5}{33}}$

Problem Analysis:

We are tossing a coin repeatedly with a stopping condition: either a head appears, or we perform a maximum of 4 tosses. We need to list all possible outcomes (the sample space) and then find the probability of a specific event: getting a head on the third toss under these rules.

Solution:

Sample Space:

Let H denote getting a Head and T denote getting a Tail on a single toss.

The experiment proceeds as follows:

Toss 1: We can get H or T.

- If we get H on the first toss, the experiment stops. Outcome: H

- If we get T on the first toss, we proceed to the second toss. Outcome: T

Toss 2 (occurs only if Toss 1 was T): We can get H or T.

- If we get H on the second toss (after a T on the first), the experiment stops. Outcome: TH

- If we get T on the second toss (after a T on the first), we proceed to the third toss. Outcome: TT

Toss 3 (occurs only if Toss 1 was T and Toss 2 was T): We can get H or T.

- If we get H on the third toss (after TT), the experiment stops. Outcome: TTH

- If we get T on the third toss (after TT), we proceed to the fourth toss. Outcome: TTT

Toss 4 (occurs only if Toss 1 was T, Toss 2 was T, and Toss 3 was T): We can get H or T.

- If we get H on the fourth toss (after TTT), the experiment stops (because we got a Head). Outcome: TTTH

- If we get T on the fourth toss (after TTT), the experiment stops because we have reached the maximum number of tosses (4). Outcome: TTTT

Combining all these possible outcomes, the sample space ($S$) is:

The total number of outcomes in the sample space is $n(S) = 5$.

Probability of getting a head on the third toss:

For a head to appear on the third toss, the following conditions must be met:

1. The first toss must be a Tail (otherwise the experiment would have stopped after the first toss).

2. The second toss must be a Tail (otherwise the experiment would have stopped after the second toss).

3. The third toss must be a Head (this is the event we are interested in).

If these three conditions are met, the sequence of outcomes is TTH. According to our stopping rule, the experiment stops after the third toss because a head appeared.

This corresponds to the outcome TTH in our sample space.

Let's calculate the probability of the outcome TTH. We assume the coin is fair, so $P(H) = 1/2$ and $P(T) = 1/2$ for each toss, and each toss is independent of the others.

Since the tosses are independent:

The event "getting a head on the third toss" corresponds to the single outcome {TTH} in the sample space. Therefore, the probability of this event is the probability of this specific outcome.

The sample space is $\boldsymbol{S = \{H, TH, TTH, TTTH, TTTT\}}$.

The probability of getting a head on the third toss is $\boldsymbol{\frac{1}{8}}$.

Problem Analysis:

We are given three events A, B, and C from a random experiment. We need to express various combinations of these events using standard set notation symbols like union ($\cup$), intersection ($\cap$), and complement (').

Solution:

i) Only A occurs:

This means that event A happens, and simultaneously, event B does not happen (i.e., B' occurs), and event C does not happen (i.e., C' occurs). The word "and" corresponds to the intersection of these events.

In set notation:

ii) At least one of the events occurs:

This means that event A happens, or event B happens, or event C happens, or any combination of A, B, and C happens. The phrase "at least one" corresponds to the union of the events.

In set notation:

iii) Exactly two of the events occur:

This means that one of the following mutually exclusive situations occurs:

• A and B occur, but C does not occur ($A \cap B \cap C'$)
• A and C occur, but B does not occur ($A \cap C \cap B'$)
• B and C occur, but A does not occur ($B \cap C \cap A'$)

The event "exactly two occur" is the union of these three possibilities.

In set notation:

iv) None of the events occur:

This means that event A does not occur (A' occurs), AND event B does not occur (B' occurs), AND event C does not occur (C' occurs). The word "and" corresponds to the intersection of these events.

In set notation:

Alternatively, this event is the complement of "at least one of the events occurs" (from part ii)). Using De Morgan's Law, $(A \cup B \cup C)' = A' \cap B' \cap C'$.

Summary of Events in Set Notation:

i) Only A occurs: $\boldsymbol{A \cap B' \cap C'}$

ii) At least one of the events occurs: $\boldsymbol{A \cup B \cup C}$

iii) Exactly two of the events occur: $\boldsymbol{(A \cap B \cap C') \cup (A \cap C \cap B') \cup (B \cap C \cap A')}$

iv) None of the events occur: $\boldsymbol{A' \cap B' \cap C'}$

Problem Analysis:

We are given information about the proportion of male and female students who are taller than 1.8 meters, as well as the overall proportion of female students in the college. We are asked to find the probability that a randomly selected student, who is known to be taller than 1.8 meters, is male.

This is a conditional probability problem where the condition is that the student is taller than 1.8 meters, and we want to find the probability of the student being male given this condition. This type of problem is solved using Bayes' Theorem.

Let's define the events:

$M$: The selected student is male.

$F$: The selected student is female.

$T$: The selected student is taller than 1.8 meters.

We are given the following probabilities:

Since the students are either male or female, the probability of a student being male is the complement of being female:

We want to find the probability that the student is male given that they are taller than 1.8 meters, which is $P(M|T)$.

Solution:

According to Bayes' Theorem, the probability of event M given event T is:

To use this formula, we first need to calculate the overall probability of a student being taller than 1.8 meters, $P(T)$. We can find this using the Law of Total Probability. Since a student must be either male or female, events M and F form a partition of the student population.

Substitute the given probabilities into equation (2):

This means that $2.2\%$ of the students in the college are taller than 1.8 meters.

Now, substitute the values of $P(T|M)$ (0.04), $P(M)$ (0.40), and $P(T)$ (0.022) from equation (3) into Bayes' Theorem (equation 1):

To simplify, we can multiply the numerator and denominator by 1000:

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

The probability that the student is male given that they are taller than 1.8 meters is $\boldsymbol{\frac{8}{11}}$.

Problem Analysis:

We are throwing a standard six-sided die three times independently. We want to find the probability of getting an odd number on at least one of these throws.

Solution:

When a standard die is thrown, the possible outcomes are $\{1, 2, 3, 4, 5, 6\}$.

The odd numbers are $\{1, 3, 5\}$. The probability of getting an odd number in a single throw is:

The even numbers are $\{2, 4, 6\}$. The probability of getting an even number in a single throw is:

Let $A$ be the event of getting an odd number at least once in three throws.

The complement of event $A$, denoted by $A'$, is the event of NOT getting an odd number at least once. This means getting no odd numbers in any of the three throws.

Getting no odd numbers in three throws is the same as getting an even number on the first throw, AND an even number on the second throw, AND an even number on the third throw.

Since each die throw is an independent event, the probability of the intersection of these events is the product of their individual probabilities:

The probability of event $A$ (getting an odd number at least once) is $1$ minus the probability of its complement $A'$ (getting no odd numbers):

Substitute the value of $P(A')$ from equation (1) into equation (2):

The probability of getting an odd number at least once in three throws is $\boldsymbol{\frac{7}{8}}$.

Given:

A and B are two events associated with a random experiment.

To Prove:

Proof:

We are given that $P(A) \neq 0$. The definition of the conditional probability of event B given event A is:

We are given that $P(B|A) = 1$. Substitute this value into the formula:

Since $P(A) \neq 0$, we can multiply both sides of equation (1) by $P(A)$:

Equation (2) tells us that the probability of event A is equal to the probability of the intersection of events A and B.

By the definition of set intersection, the set of outcomes in $A \cap B$ is always a subset of the set of outcomes in A. That is, $A \cap B \subseteq A$.

For any two events (sets of outcomes) E and F, if $E \subseteq F$, then it necessarily follows that $P(E) \leq P(F)$. Thus, from $A \cap B \subseteq A$, we have $P(A \cap B) \leq P(A)$.

However, from equation (2), we found that $P(A \cap B) = P(A)$. This equality implies that there are no outcomes in A that are not also in $A \cap B$.

In other words, if an outcome $\omega$ is in the event A ($\omega \in A$), then for $P(A) = P(A \cap B)$ to be true, that outcome $\omega$ must also be in the event $A \cap B$ ($\omega \in A \cap B$).

By the definition of intersection, if $\omega \in A \cap B$, then $\omega \in A$ AND $\omega \in B$.

So, if we start with an outcome $\omega \in A$, it must follow that $\omega \in B$.

This is precisely the definition of a subset: if every element of set A is also an element of set B, then A is a subset of B ($A \subseteq B$).

Therefore, we have proven that if $P(A) \neq 0$ and $P(B|A) = 1$, then $A \subseteq B$.

Hence proved.

Problem Analysis:

We are performing a fixed number of independent trials (tossing a biased coin 4 times). For each trial, there are only two possible outcomes: getting a head (success) or getting a tail (failure). The probability of success is constant for each trial ($0.6$). We want to find the probability of getting a specific number of successes (exactly 2 heads). This is a problem that can be solved using the binomial probability distribution.

Solution:

Let $X$ be the number of heads obtained in 4 tosses.

This is a binomial experiment with the following parameters:

• Number of trials, $n = 4$.
• Probability of success (getting a Head), $p = P(H) = 0.6$.
• Probability of failure (getting a Tail), $q = P(T) = 1 - p = 1 - 0.6 = 0.4$.

We want to find the probability of getting exactly 2 heads, so we are looking for $P(X=2)$.

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

where $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ is the binomial coefficient, representing the number of ways to choose $k$ successes in $n$ trials.

In this case, $n=4$, $k=2$, $p=0.6$, and $q=0.4$. Substitute these values into the formula:

First, calculate the binomial coefficient $\binom{4}{2}$:

Next, calculate the powers of $p$ and $q$:

Now, substitute these values back into equation (1):

Perform the multiplication:

The probability of getting exactly 2 heads in 4 tosses of this biased coin is $\boldsymbol{0.3456}$.

Problem Analysis:

We are selecting a fixed number of items (10) from a manufacturing process where each item is either defective or not defective. The probability of an item being defective is constant (5%) and independent for each item. This scenario fits the model of a binomial probability distribution.

Solution:

Let $X$ be the number of defective items in the sample of 10.

This is a binomial experiment with the following parameters:

• Number of trials (items selected), $n = 10$.
• Probability of success (an item is defective), $p = 5\% = 0.05$.
• Probability of failure (an item is not defective), $q = 1 - p = 1 - 0.05 = 0.95$.

The probability of getting exactly $k$ defective items in $n$ trials is given by the binomial probability formula:

Probability that none of the items are defective:

This means we want to find the probability that the number of defective items is exactly 0 ($k=0$).

Substitute $n=10$, $k=0$, $p=0.05$, and $q=0.95$ into equation (1):

Calculate the terms:

So, the probability is:

Using a calculator, $(0.95)^{10} \approx 0.5987369$.

Probability that at least one item is defective:

"At least one item is defective" means that the number of defective items $X$ is greater than or equal to 1 ($X \geq 1$).

The complement of the event "at least one item is defective" ($X \geq 1$) is the event "none of the items are defective" ($X=0$).

Using the complement rule, the probability of getting at least one defective item is:

Substitute the value of $P(X=0)$ from equation (2) into equation (3):

Using the approximate value of $(0.95)^{10}$:

The probability that none of the items are defective is $\boldsymbol{(0.95)^{10}}$ (approximately $\boldsymbol{0.5987}$).

The probability that at least one item is defective is $\boldsymbol{1 - (0.95)^{10}}$ (approximately $\boldsymbol{0.4013}$).

Problem Analysis:

We are given the proportion of items produced by two machines (A and B) and the probability of an item being defective depending on which machine produced it. We are asked to find the probability that a randomly selected item, which is known to be defective, was produced by Machine B.

This is a conditional probability problem, specifically a case where we need to use Bayes' Theorem.

Let's define the events:

$A$: The selected item was produced by Machine A.

$B$: The selected item was produced by Machine B.

$D$: The selected item is defective.

We are given the following probabilities:

We want to find the probability that the item was produced by Machine B given that it is defective, which is $P(B|D)$.

Solution:

According to Bayes' Theorem, the probability of event B given event D is:

To use this formula, we first need to calculate the overall probability of an item being defective, $P(D)$. We can find this using the Law of Total Probability. Since an item must be produced by either Machine A or Machine B (and not both), events A and B form a partition of the sample space of production sources.

Substitute the given probabilities into equation (2):

This means that $2.2\%$ of the total items produced are defective.

Now, substitute the values of $P(D|B)$ (0.05), $P(B)$ (0.30), and $P(D)$ (0.022) from equation (3) into Bayes' Theorem (equation 1):

To simplify, we can multiply the numerator and denominator by 1000:

This fraction cannot be simplified further.

The probability that the defective item was produced by Machine B is $\boldsymbol{\frac{15}{22}}$.

Given:

A and B are two events associated with a random experiment on a sample space $S$.

To Prove:

Proof:

We can express the union of events A and B, $A \cup B$, as the union of three mutually exclusive (disjoint) events:

1. The outcomes that are in A but not in B: $A \cap B'$

2. The outcomes that are in B but not in A: $B \cap A'$

3. The outcomes that are in both A and B: $A \cap B$

So, we can write the union $A \cup B$ as:

Since the events $(A \cap B')$, $(B \cap A')$, and $(A \cap B)$ are mutually exclusive (disjoint), the probability of their union is the sum of their individual probabilities, by the third axiom of probability:

Now consider event A. Event A can be expressed as the union of two mutually exclusive events:

1. Outcomes in A that are also in B: $A \cap B$

2. Outcomes in A that are not in B: $A \cap B'$

So, we can write event A as:

Since $(A \cap B)$ and $(A \cap B')$ are mutually exclusive, the probability of A is the sum of their probabilities:

From equation (2), we can express $P(A \cap B')$ as:

Similarly, consider event B. Event B can be expressed as the union of two mutually exclusive events:

1. Outcomes in B that are also in A: $B \cap A$ (which is the same as $A \cap B$)

2. Outcomes in B that are not in A: $B \cap A'$

So, we can write event B as:

Since $(A \cap B)$ and $(B \cap A')$ are mutually exclusive, the probability of B is the sum of their probabilities:

From equation (4), we can express $P(B \cap A')$ as:

Now, substitute the expressions for $P(A \cap B')$ from equation (3) and $P(B \cap A')$ from equation (5) into equation (1):

Simplify the expression:

This is the required result.

Hence proved.

Problem Analysis:

We are drawing a set of 3 cards simultaneously from a standard deck of 52 cards. Since the order in which the cards are drawn does not matter (as they are drawn simultaneously), this is a problem involving combinations.

We need to find the probability of three specific outcomes for the set of 3 cards drawn.

Solution:

Total number of cards in the deck = 52.

Number of cards drawn = 3.

The total number of ways to choose 3 cards from a deck of 52 is given by $\binom{52}{3}$.

Calculate $\binom{52}{3}$:

i) Probability of getting three Kings:

There are 4 Kings in a deck of 52 cards.

The number of ways to choose 3 Kings from the 4 Kings is given by $\binom{4}{3}$.

The probability of getting three Kings is the number of ways to get three Kings divided by the total number of ways to choose 3 cards:

Simplify the fraction:

ii) Probability of getting two Queens and one King:

There are 4 Queens and 4 Kings in a deck.

The number of ways to choose 2 Queens from the 4 Queens is $\binom{4}{2}$.

The number of ways to choose 1 King from the 4 Kings is $\binom{4}{1}$.

The number of ways to get two Queens AND one King is the product of the number of ways to choose the Queens and the number of ways to choose the Kings (by the multiplication principle):

The probability of getting two Queens and one King is the number of favorable outcomes divided by the total number of ways to choose 3 cards:

Simplify the fraction:

iii) Probability of getting three cards of the same suit:

There are 4 suits (Spades, Hearts, Diamonds, Clubs), and each suit has 13 cards.

Getting three cards of the same suit means getting 3 cards that are all Spades, OR all Hearts, OR all Diamonds, OR all Clubs.

These are four mutually exclusive events.

The number of ways to choose 3 cards from a single suit (e.g., 3 Spades from 13 Spades) is $\binom{13}{3}$.

Since there are 4 such suits, the total number of ways to get three cards of the same suit is $4 \times \binom{13}{3}$.

The probability of getting three cards of the same suit is the number of favorable outcomes divided by the total number of ways to choose 3 cards:

Simplify the fraction:

$P(\text{three cards of same suit}) = \frac{\cancel{286}^{22}}{\cancel{5525}_{425}} = \frac{22}{425}$

Summary of Probabilities:

i) Probability of getting three Kings = $\boldsymbol{\frac{1}{5525}}$

ii) Probability of getting two Queens and one King = $\boldsymbol{\frac{6}{5525}}$

iii) Probability of getting three cards of the same suit = $\boldsymbol{\frac{22}{425}}$

Problem Analysis:

We are given information about the proportion of bulbs produced by three different plants (I, II, and III) and the percentage of defective bulbs from each plant. A bulb is selected randomly from the total production and is found to be defective. We need to determine the probability that this defective bulb came from Plant III.

This is a conditional probability problem where we know the outcome (the bulb is defective) and want to find the probability of a specific prior cause (it was produced by Plant III). This calls for the use of Bayes' Theorem.

Let's define the events:

$P_1$: The selected bulb was produced by Plant I.

$P_2$: The selected bulb was produced by Plant II.

$P_3$: The selected bulb was produced by Plant III.

$D$: The selected bulb is defective.

We are given the following probabilities:

We want to find the probability that the bulb was produced by Plant III given that it is defective, which is $P(P_3|D)$.

Solution:

According to Bayes' Theorem, the probability of event $P_3$ given event $D$ is:

To use this formula, we first need to calculate the overall probability of a randomly selected bulb being defective, $P(D)$. We can find this using the Law of Total Probability. Since every bulb is produced by exactly one of the three plants, events $P_1, P_2, P_3$ form a partition of the sample space of production sources.

Substitute the given probabilities into equation (2):

This means that $2.7\%$ of the total bulbs produced are defective.

Now, substitute the values of $P(D|P_3)$ (0.04), $P(P_3)$ (0.20), and $P(D)$ (0.027) from equation (3) into Bayes' Theorem (equation 1):

To simplify, we can multiply the numerator and denominator by 1000:

This fraction cannot be simplified further.

The probability that the defective bulb was produced by Plant III is $\boldsymbol{\frac{8}{27}}$.

Given:

A and B are two independent events associated with a random experiment.

By the definition of independence, $P(A \cap B) = P(A)P(B)$.

To Prove:

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

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

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

Proof:

We know that for any event E, $P(E') = 1 - P(E)$.

Thus, $P(A') = 1 - P(A)$ and $P(B') = 1 - P(B)$.

i) Proof that $A$ and $B'$ are independent:

Event A can be written as the union of two mutually exclusive events: the outcomes in A that are also in B ($A \cap B$), and the outcomes in A that are not in B ($A \cap B'$).

Since $(A \cap B)$ and $(A \cap B')$ are disjoint, the probability of A is the sum of their probabilities:

Rearranging this equation, we get:

Since A and B are independent, $P(A \cap B) = P(A)P(B)$. Substitute this into the equation:

Factor out $P(A)$:

Since $1 - P(B) = P(B')$, we have:

This satisfies the condition for independence between A and B'.

ii) Proof that $A'$ and $B$ are independent:

Event B can be written as the union of two mutually exclusive events: the outcomes in B that are also in A ($A \cap B$), and the outcomes in B that are not in A ($A' \cap B$).

Since $(A \cap B)$ and $(A' \cap B)$ are disjoint, the probability of B is the sum of their probabilities:

Rearranging this equation, we get:

Since A and B are independent, $P(A \cap B) = P(A)P(B)$. Substitute this into the equation:

Factor out $P(B)$:

Since $1 - P(A) = P(A')$, we have:

This satisfies the condition for independence between A' and B.

iii) Proof that $A'$ and $B'$ are independent:

Using De Morgan's Law, the intersection of the complements $A' \cap B'$ is equal to the complement of the union $(A \cup B)'$.

The probability of the complement of an event is 1 minus the probability of the event:

Using the Addition Rule for Probabilities, $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. Substitute this into the equation:

Since A and B are independent, $P(A \cap B) = P(A)P(B)$. Substitute this into the equation:

Factor the expression by grouping terms:

Factor out the common term $(1 - P(A))$:

Since $1 - P(A) = P(A')$ and $1 - P(B) = P(B')$, we have:

This satisfies the condition for independence between A' and B'.

Thus, we have proven that if events A and B are independent, then A and B' are independent, A' and B are independent, and A' and B' are independent.

Hence proved.

Problem Analysis:

We are drawing a set of 2 cards simultaneously from a standard deck of 52 cards without replacement. Since the order does not matter when drawing simultaneously, this is a problem involving combinations. The phrase "without replacement" is implicit in the combination calculation when selecting multiple items at once.

We need to find the probability of two specific outcomes for the set of 2 cards drawn.

Solution:

Total number of cards in the deck = 52.

Number of cards drawn = 2.

The total number of ways to choose 2 cards from a deck of 52 is given by $\binom{52}{2}$.

Calculate $\binom{52}{2}$:

i) Probability that both are of the same suit:

There are 4 suits (Spades, Hearts, Diamonds, Clubs), and each suit has 13 cards.

For the two cards to be of the same suit, we must choose one of the 4 suits AND then choose 2 cards from the 13 cards within that chosen suit.

Number of ways to choose a suit = $\binom{4}{1} = 4$.

Number of ways to choose 2 cards from that suit = $\binom{13}{2}$.

The number of ways to get two cards of the same suit is the product of the number of ways to choose a suit and the number of ways to choose 2 cards from that suit:

The probability of getting two cards of the same suit is the number of favorable outcomes divided by the total number of possible outcomes:

Simplify the fraction by dividing both numerator and denominator by common factors (e.g., 6):

ii) Probability that one is a King and the other is an Ace:

We need to choose 1 King from the 4 Kings in the deck AND 1 Ace from the 4 Aces in the deck.

Number of ways to choose 1 King from 4 Kings = $\binom{4}{1}$.

Number of ways to choose 1 Ace from 4 Aces = $\binom{4}{1}$.

The number of ways to get one King AND one Ace is the product of the number of ways to choose a King and the number of ways to choose an Ace:

The probability of getting one King and one Ace is the number of favorable outcomes divided by the total number of ways to choose 2 cards:

Simplify the fraction by dividing both numerator and denominator by 2:

Summary of Probabilities:

i) Probability that both are of the same suit = $\boldsymbol{\frac{4}{17}}$

ii) Probability that one is a King and the other is an Ace = $\boldsymbol{\frac{8}{663}}$

Problem Analysis:

We have a group of 100 people. We are given the number of people who like Coffee, the number who like Tea, and the number who like both. We need to find two probabilities for a randomly selected person:

1. The probability that the person likes at least one of the drinks (Coffee or Tea or both).

2. The probability that the person likes Tea, given that they like Coffee.

Let's define the events:

$C$: The selected person likes Coffee.

$T$: The selected person likes Tea.

The total number of people is 100. We can convert the given numbers into probabilities by dividing by the total number of people.

The number of people who like both Coffee and Tea is given. This corresponds to the intersection of events C and T.

Solution:

i) Probability that the person likes at least one of the drinks:

"At least one of the drinks" means the person likes Coffee OR Tea (or both). This is the probability of the union of events C and T, $P(C \cup T)$.

We use the Addition Rule for Probabilities:

Substitute the given probabilities into equation (1):

This means $50\%$ of the people like at least one of the drinks.

ii) Probability that they also like Tea, given that they like Coffee:

This is a conditional probability. We are given that the person likes Coffee (event C has occurred), and we want to find the probability that they also like Tea (event T). This is denoted as $P(T|C)$.

The formula for conditional probability is:

Note that $P(T \cap C)$ is the same as $P(C \cap T)$, which is the probability of liking both.

Substitute the given probabilities $P(C \cap T) = 0.20$ and $P(C) = 0.40$ into the formula:

This means that among the people who like Coffee, $50\%$ of them also like Tea.

Summary of Probabilities:

Probability that the person likes at least one of the drinks = $\boldsymbol{0.50}$ or $\boldsymbol{50\%}$ or $\boldsymbol{\frac{1}{2}}$.

Probability that they also like Tea, given that they like Coffee = $\boldsymbol{0.50}$ or $\boldsymbol{50\%}$ or $\boldsymbol{\frac{1}{2}}$.

Problem Analysis:

We have two individuals, A and B, each with a certain probability of speaking the truth when stating a fact. They state the same fact, and we want to find the probability that they contradict each other. Contradiction happens if one speaks the truth and the other lies.

We assume that A speaking the truth is independent of B speaking the truth.

Let's define the events:

$T_A$: A speaks the truth.

$L_A$: A lies (does not speak the truth).

$T_B$: B speaks the truth.

$L_B$: B lies (does not speak the truth).

We are given the probabilities that A and B speak the truth:

The probabilities that they lie are the complements of speaking the truth:

They contradict each other when stating the same fact if one states the fact is true while the other states it is false. This can happen in two mutually exclusive ways:

Case 1: A speaks the truth AND B lies ($T_A \cap L_B$).

Case 2: A lies AND B speaks the truth ($L_A \cap T_B$).

Solution:

Since A and B speaking the truth (or lying) are assumed to be independent events, we can use the multiplication rule for independent events.

Probability of Case 1 (A speaks truth and B lies):

Probability of Case 2 (A lies and B speaks truth):

The event that they contradict each other is the union of Case 1 and Case 2. Since these cases are mutually exclusive, the probability of contradiction is the sum of their probabilities:

To express this probability as a percentage, multiply by 100:

They are likely to contradict each other in $\boldsymbol{35\%}$ of the cases.

Problem Analysis:

We are given information about the accuracy of a diagnostic test for a disease within a population. Specifically, we know the probability of a positive test result given whether a person has the disease or not, and the overall prevalence of the disease in the population. We are asked to find the probability that a person actually has the disease, given that their test result is positive.

This is a conditional probability problem, where the condition is the test result, and we want the probability of the underlying state (having the disease). This is a typical scenario for applying Bayes' Theorem.

Let's define the events:

$D$: The selected person has the disease.

$D'$: The selected person does not have the disease.

$P$: The test result is positive.

We are given the following probabilities:

Since a person either has the disease or does not, the probability of not having the disease is the complement of having it:

We want to find the probability that a person has the disease given that they tested positive, which is $P(D|P)$.

Solution:

According to Bayes' Theorem, the probability of event D given event P is:

To use this formula, we first need to calculate the overall probability of a randomly selected person testing positive, $P(P)$. We can find this using the Law of Total Probability. The event of testing positive can occur in two mutually exclusive ways: the person has the disease and tests positive ($D \cap P$), or the person does not have the disease and tests positive ($D' \cap P$).

Using the definition of conditional probability, $P(A \cap B) = P(A|B) P(B)$, we can write:

Substitute the known probabilities into equation (2):

Perform the multiplications:

Sum the values:

This is the overall probability of a randomly selected person testing positive.

Now, substitute the values of $P(P|D)$ (0.99), $P(D)$ (0.001), and $P(P)$ (0.005985) from equation (3) into Bayes' Theorem (equation 1):

To simplify the fraction, multiply the numerator and denominator by 100,000 (to remove decimals):

Simplify the fraction $\frac{990}{5985}$. Both are divisible by 5:

Both are divisible by 9 (sum of digits $1+9+8=18$ and $1+1+9+7=18$):

The fraction $\frac{22}{133}$ cannot be simplified further as the prime factors of 22 are 2 and 11, and $133 = 7 \times 19$.

The probability that a randomly selected person who tests positive actually has the disease is $\boldsymbol{\frac{22}{133}}$. This is approximately $\boldsymbol{0.1654}$ or $\boldsymbol{16.54\%}$.