Given Variables & Initial Setup

Based on the problem statement, we define the following events:

• Let $A$ be the event that subsystem A fails. Thus, $P(A) = 0.2$.
• Let $B$ be the event that subsystem B fails.
• The probability that B fails alone is given as $P(B \cap A^c) = 0.15$.
• The probability that both A and B fail is given as $P(A \cap B) = 0.15$.

Step 1: Determine the total probability of event B

The total probability of event $B$ failing is the sum of the probability that B fails while A fails, and the probability that B fails while A does not fail.

$P(B) = P(A \cap B) + P(B \cap A^c)$ [By the Law of Total Probability]

$P(B) = 0.15 + 0.15 = 0.30$

Step 2: Applying the Definition of Conditional Probability

We are required to find $P(A \text{ fails} | B \text{ has failed})$, which is denoted as $P(A|B)$.

The formula for conditional probability is:

$P(A|B) = \frac{P(A \cap B)}{P(B)}$ [Per the Axiom of Conditional Probability]

Step 3: Substitution and Calculation

Substitute the values derived in the previous steps:

$P(A|B) = \frac{0.15}{0.30}$

$P(A|B) = \frac{15}{30} = 0.5$

Visual Representation of Event Spaces

Final Solution: $P(A|B) = 0.5$

Step 1: Formal Definition of Given Probabilities

Let $A$ be the event that subsystem $A$ fails, and $B$ be the event that subsystem $B$ fails. Based on the problem statement, we define the following probabilities:

• $P(A) = 0.2$ [Probability that A fails]
• $P(B \cap A^c) = 0.15$ [Probability that B fails alone, where $A^c$ denotes the event that A does not fail]
• $P(A \cap B) = 0.15$ [Probability that both A and B fail]

Step 2: Logical Derivation for "A fails alone"

The event "A fails alone" can be represented as the event that $A$ occurs, but $B$ does not occur. In set-theoretic notation, this is $P(A \cap B^c)$.

By the Law of Total Probability or the properties of set operations, the event $A$ can be expressed as the union of two mutually exclusive events:
$A = (A \cap B) \cup (A \cap B^c)$

Consequently, by the Addition Theorem of Probability for mutually exclusive events:
$P(A) = P(A \cap B) + P(A \cap B^c)$

Step 3: Algebraic Calculation

We rearrange the equation to isolate the target probability $P(A \cap B^c)$:

$P(A \cap B^c) = P(A) - P(A \cap B)$

Substituting the known values into the equation:

$P(A \cap B^c) = 0.2 - 0.15$

$P(A \cap B^c) = 0.05$

Step 4: Venn Diagram Representation

The diagram illustrates the intersection $P(A \cap B) = 0.15$, the "A alone" region $P(A \cap B^c) = 0.05$, and the "B alone" region $P(B \cap A^c) = 0.15$.

Final Solution: $P(\text{A fails alone}) = 0.05$

Step 1: Defining the Sample Space

Let the gender of a child be represented by $B$ (Boy) or $G$ (Girl). For a family with two children, the sample space $S$ consists of all possible ordered pairs $(Child_1, Child_2)$:

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

Each outcome is considered equally likely, where $|S| = 4$. [Per the Classical Definition of Probability]

Step 2: Defining the Events

Let $E$ be the event that both children are females. Thus, $E = \{GG\}$.

Let $F$ be the event that the elder child is a female. The elder child corresponds to the first position in the ordered pair. Thus, $F = \{GG, GB\}$.

Step 3: Applying Conditional Probability Formula

We are required to find the probability of event $E$ given that event $F$ has occurred, denoted as $P(E|F)$.

The formula for conditional probability is:

$P(E|F) = \frac{P(E \cap F)}{P(F)}$ [By the definition of Conditional Probability]

Step 4: Calculating Intersections and Probabilities

First, identify the intersection of events $E$ and $F$:

$E \cap F = \{GG\} \cap \{GG, GB\} = \{GG\}$

Calculate the individual probabilities based on the sample space size:

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

Step 5: Final Computation

Substitute the values into the conditional probability formula:

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

Final Solution: The probability that both children are females, given that the elder child is a female, is $\frac{1}{2}$.

Step 1: Analyzing the Given Conditional Probability

We are provided with the condition $P(B|A) = 1$, where $P(A) \neq 0$. The definition of conditional probability for any two events $A$ and $B$ in a sample space $S$ is given by:

$P(B|A) = \frac{P(A \cap B)}{P(A)}$ [Definition of Conditional Probability]

Step 2: Substituting and Solving for Set Relations

Substituting the given value $P(B|A) = 1$ into the definition:

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

$P(A) = P(A \cap B)$ [Algebraic manipulation]

Step 3: Logical Inference on Set Membership

The identity $P(A \cap B) = P(A)$ implies that the probability of the intersection of $A$ and $B$ is equal to the probability of $A$ itself. In the context of set theory and probability measures:

• Since $A \cap B \subseteq A$, the equality $P(A \cap B) = P(A)$ indicates that all elements of $A$ are contained within the intersection of $A$ and $B$.
• This leads to the condition $A \subseteq B$.

Step 4: Geometric Representation of $A \subseteq B$

Step 5: Theoretical Conclusion

Given $P(B|A) = 1$ and $P(A) \neq 0$, it follows that:

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

By set-theoretic axioms, this indicates that the event $A$ is a subset of event $B$ ($A \subset B$). Specifically, if the occurrence of $A$ guarantees the occurrence of $B$, $A$ must be entirely contained within the region defined by $B$.

Final Solution: A \subset B

Step 1: Analyzing the Conditional Probability Expression

The given condition is $P(A|B) > P(A)$. Per the definition of conditional probability for any two events $A$ and $B$ where $P(B) > 0$, the probability of $A$ given $B$ is expressed as:

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

[By the Multiplicative Rule of Probability]

Step 2: Substitution and Algebraic Manipulation

Substituting the definition into the given inequality:

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

Assuming $P(B) > 0$, we can multiply both sides by $P(B)$ without reversing the inequality:

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

Step 3: Evaluating the Implications for Event B

To determine the relation between $P(B|A)$ and $P(B)$, we use the symmetry of the intersection formula, $P(A \cap B) = P(B \cap A)$. Applying the conditional probability definition for $B$ given $A$ (assuming $P(A) > 0$):

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

From our inequality $P(A \cap B) > P(A) \cdot P(B)$, divide both sides by $P(A)$:

$\frac{P(A \cap B)}{P(A)} > \frac{P(A) \cdot P(B)}{P(A)}$

$P(B|A) > P(B)$

Step 4: Visualizing Event Dependency

Step 5: Theoretical Conclusion

The condition $P(A|B) > P(A)$ implies that the occurrence of event $B$ increases the probability of the occurrence of event $A$. This is the definition of positive correlation between two events in probability theory. Mathematically, this is equivalent to $P(B|A) > P(B)$.

Final Solution: The correct relation is $P(B|A) > P(B)$.

Step 1: Defining the Sample Space

Let the sample space $S$ represent all possible gender combinations for two children, where $B$ denotes a boy and $G$ denotes a girl. The birth order matters:

The total number of outcomes is $n(S) = 4$.

Step 2: Defining Events

Let $E$ be the event that both children are males, and $F$ be the event that at least one child is male.

• $E = \{BB\}$
• $F = \{BB, BG, GB\}$

Per the definition of events, the intersection of $E$ and $F$ is:

Step 3: Applying Conditional Probability

We are required to find $P(E|F)$, the probability that both children are males given that at least one is a male. [By the Definition of Conditional Probability]:

Calculating the individual probabilities:

• $n(E \cap F) = 1 \implies P(E \cap F) = \frac{1}{4}$
• $n(F) = 3 \implies P(F) = \frac{3}{4}$

Step 4: Algebraic Computation

Substituting the values into the formula:

Visual Representation of Sample Space

Final Solution: The probability that both children are males, given that at least one is male, is $\frac{1}{3}$.

Step 1: Formal Representation of the Given Condition

The problem provides the following equation involving the probabilities of two events, $A$ and $B$, within a sample space $S$:

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

[Given: The probability of the union of two events is defined by the Addition Theorem of Probability: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$].

Step 2: Algebraic Simplification

By substituting the definition of the union of two events into the given equation, we derive the following logical equivalence:

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

[By the Axiom of Probability, for any two sets $A$ and $B$, $A \subseteq (A \cup B)$].

Step 3: Logical Deduction of Set Relations

If $P(A \cup B) = P(A)$, this implies that the event $(A \cup B)$ is effectively equivalent to event $A$ in terms of probability measure. Since $A \subseteq (A \cup B)$ is always true, the condition $P(A \cup B) = P(A)$ implies that the set $B$ must be a subset of $A$.

Specifically:

• If $B \subseteq A$, then $A \cup B = A$.
• Consequently, $P(A \cup B) = P(A)$.

[Per the Monotonicity Property of Probability: If $B \subseteq A$, then $P(B) \leq P(A)$ and $P(A \cap B) = P(B)$].

Step 4: Verification via the Inclusion-Exclusion Principle

Revisiting the original equation: $P(A) + P(B) - P(A \cap B) = P(A)$.

Subtract $P(A)$ from both sides:

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

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

This result is the necessary and sufficient condition for $B \subseteq A$. If the probability of the intersection of two events is equal to the probability of one of the events, that event must be contained within the other.

Step 5: Synthesis of Results

The given equation simplifies to $P(B) = P(A \cap B)$, which holds if and only if $B$ is a subset of $A$ ($B \subseteq A$).

Final Solution: B \subseteq A

Step 1: Definition of Events and Probabilities

Let $M$ be the event that the selected person is male, and $W$ be the event that the selected person is female. Let $G$ be the event that the selected person has grey hair.

Given that there are equal numbers of males and females, the prior probabilities are:

• $P(M) = 0.5$
• $P(W) = 0.5$

The conditional probabilities of having grey hair given the gender are:

• $P(G|M) = 5\% = 0.05$
• $P(G|W) = 0.25\% = 0.0025$

Step 2: Total Probability of Having Grey Hair

Per the Law of Total Probability, the probability that a randomly selected person has grey hair is given by:

$P(G) = P(M) \cdot P(G|M) + P(W) \cdot P(G|W)$

$P(G) = (0.5 \times 0.05) + (0.5 \times 0.0025)$

$P(G) = 0.025 + 0.00125$

$P(G) = 0.02625$

Step 3: Applying Bayes' Theorem

We aim to find the probability that the person is male given they have grey hair, denoted as $P(M|G)$. According to Bayes' Theorem:

$P(M|G) = \frac{P(M) \cdot P(G|M)}{P(G)}$

Substituting the calculated values:

$P(M|G) = \frac{0.5 \times 0.05}{0.02625}$

$P(M|G) = \frac{0.025}{0.02625}$

Step 4: Algebraic Simplification

To simplify the fraction, multiply both the numerator and denominator by 100,000:

$P(M|G) = \frac{2500}{2625}$

Divide both terms by their greatest common divisor ($125$):

$P(M|G) = \frac{2500 \div 125}{2625 \div 125} = \frac{20}{21}$

Visual Representation of Probability Space

Final Solution: The probability that the person with grey hair is male is $\frac{20}{21} \approx 0.9524$ or $95.24\%$.

Step 1: Definition of Events

Let $E$ be the event that the patient suffers a heart attack. Let $A_1$ and $A_2$ be the events of choosing the meditation/yoga course and the drug prescription, respectively.

• $P(A_1) = 0.5$ (Given: patient chooses one of two options with equal probability)
• $P(A_2) = 0.5$

The initial probability of a heart attack is $0.40$.

Step 2: Calculating Conditional Probabilities

We determine the reduced probability of a heart attack given the chosen course:

• If the patient chooses meditation ($A_1$), the risk is reduced by $30\%$: $P(E|A_1) = 0.40 \times (1 - 0.30) = 0.40 \times 0.70 = 0.28$ [By application of percentage reduction]
• If the patient chooses the drug ($A_2$), the risk is reduced by $25\%$: $P(E|A_2) = 0.40 \times (1 - 0.25) = 0.40 \times 0.75 = 0.30$ [By application of percentage reduction]

Step 3: Application of Bayes' Theorem

We are required to find $P(A_1|E)$, the probability that the patient followed meditation/yoga given they suffered a heart attack. [Per Bayes' Theorem: $P(A_i|E) = \frac{P(A_i)P(E|A_i)}{\sum P(A_j)P(E|A_j)}$]

Calculating the total probability $P(E)$:

Step 4: Final Probability Computation

Visual Representation of Probability Distribution

Final Solution: The probability that the patient followed a course of meditation and yoga is $\frac{14}{29} \approx 0.4827$.

Step 1: Defining Events and Prior Probabilities

Let $E$ be the event that the marble drawn is red. Let $A, B, C,$ and $D$ be the events of selecting boxes A, B, C, and D, respectively. Since there are four boxes and one is selected at random:

$P(A) = P(B) = P(C) = P(D) = \frac{1}{4}$ [By the principle of equal likelihood]

Step 2: Conditional Probabilities of Drawing a Red Marble

Based on the provided distribution (A: 1 Red, 6 White; B: 2 Red, 2 White; C: 3 Red, 1 White; D: 4 Red, 0 White), we calculate the probability of drawing a red marble given a specific box selection:

• $P(E|A) = \frac{1}{1+6} = \frac{1}{7}$
• $P(E|B) = \frac{2}{2+2} = \frac{2}{4} = \frac{1}{2}$
• $P(E|C) = \frac{3}{3+1} = \frac{3}{4}$
• $P(E|D) = \frac{4}{4+0} = \frac{4}{4} = 1$

Step 3: Calculating Total Probability $P(E)$

Per the Law of Total Probability: $P(E) = P(A)P(E|A) + P(B)P(E|B) + P(C)P(E|C) + P(D)P(E|D)$

$P(E) = \left( \frac{1}{4} \times \frac{1}{7} \right) + \left( \frac{1}{4} \times \frac{1}{2} \right) + \left( \frac{1}{4} \times \frac{3}{4} \right) + \left( \frac{1}{4} \times 1 \right)$

$P(E) = \frac{1}{4} \left( \frac{1}{7} + \frac{1}{2} + \frac{3}{4} + 1 \right) = \frac{1}{4} \left( \frac{8 + 28 + 42 + 56}{56} \right) = \frac{1}{4} \left( \frac{134}{56} \right) = \frac{67}{112}$

Step 4: Applying Bayes' Theorem

To find the probability that the marble was drawn from a specific box given it is red, we use: $P(X|E) = \frac{P(X)P(E|X)}{P(E)}$ where $X \in \{A, B, C, D\}$.

Final Solution: The probabilities that the red marble was drawn from boxes A, B, C, and D are $\frac{4}{67}, \frac{14}{67}, \frac{21}{67},$ and $\frac{28}{67}$ respectively.

Step 1: Identifying the Conditional Probability Formula

The conditional probability of an event $B$ occurring given that event $A$ has already occurred is defined by the multiplication rule of probability:

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

[Per the definition of conditional probability for any two events $A$ and $B$ where $P(A) \neq 0$]

Step 2: Analyzing the Given Condition

The problem states that $A \cap B = \phi$, where $\phi$ denotes the empty set (or null event). In probability theory, the probability of an impossible or empty event is defined as:

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

[By the Axiom of Probability, the measure of the empty set in the sample space $S$ is zero]

Step 3: Calculating $P(B|A)$

Substitute the value obtained in Step 2 into the formula established in Step 1:

$P(B|A) = \frac{0}{P(A)}$

Since the problem provides the constraint $P(A) \neq 0$, the expression is well-defined and evaluates as follows:

$P(B|A) = 0$

Visual Representation of Mutually Exclusive Events

Final Solution: $P(B|A) = 0$

Step 1: Definition of Events and Initial Probabilities

Let the events be defined as follows:

• $E_1$: The ball transferred from Bag I to Bag II is red.
• $E_2$: The ball transferred from Bag I to Bag II is black.
• $A$: The ball drawn from Bag II is red.

Bag I contains 3 red balls and 4 black balls (Total = 7 balls).
Bag II contains 4 red balls and 5 black balls (Total = 9 balls).

Probabilities of transferring balls from Bag I:

• $P(E_1) = \frac{3}{7}$
• $P(E_2) = \frac{4}{7}$

Step 2: Conditional Probabilities of Drawing a Red Ball

If $E_1$ occurs, Bag II receives a red ball. The composition of Bag II becomes (4+1) red and 5 black (Total = 10 balls).
$P(A|E_1) = \frac{5}{10} = \frac{1}{2}$

If $E_2$ occurs, Bag II receives a black ball. The composition of Bag II becomes 4 red and (5+1) black (Total = 10 balls).
$P(A|E_2) = \frac{4}{10} = \frac{2}{5}$

Step 3: Application of Bayes' Theorem

We are required to find the probability that the transferred ball was black, given that the drawn ball is red, i.e., $P(E_2|A)$.

Per Bayes' Theorem [Theorems of Total Probability and Bayes' Rule]:

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

Step 4: Algebraic Computation

Substitute the established values into the formula:

$P(E_2|A) = \frac{\frac{4}{7} \cdot \frac{4}{10}}{\left(\frac{3}{7} \cdot \frac{5}{10}\right) + \left(\frac{4}{7} \cdot \frac{4}{10}\right)}$

$P(E_2|A) = \frac{\frac{16}{70}}{\frac{15}{70} + \frac{16}{70}}$

$P(E_2|A) = \frac{\frac{16}{70}}{\frac{31}{70}}$

$P(E_2|A) = \frac{16}{31}$

Final Solution: The probability that the transferred ball is black is $\frac{16}{31}$.

Step 1: Analyzing the Timeframe of a Leap Year

A standard year consists of $365$ days, whereas a leap year consists of $366$ days [Per the Gregorian calendar definition].

We convert the total number of days into weeks:

$366 \text{ days} = 52 \text{ weeks} + 2 \text{ days}$

This implies that every leap year is guaranteed to have $52$ Tuesdays, as $52 \times 7 = 364$ days cover $52$ full cycles of the week.

Step 2: Determining the Sample Space of Remaining Days

The remaining $2$ days (the $365^{th}$ and $366^{th}$ days) can be any two consecutive days of the week. We denote the sample space $S$ as follows:

• $S = \{(\text{Sunday, Monday}), (\text{Monday, Tuesday}), (\text{Tuesday, Wednesday}), (\text{Wednesday, Thursday}), (\text{Thursday, Friday}), (\text{Friday, Saturday}), (\text{Saturday, Sunday})\}$

The total number of outcomes is $n(S) = 7$.

Step 3: Defining the Favorable Event

For the year to contain exactly $53$ Tuesdays, one of the two remaining days must be a Tuesday. Let $E$ be the event of selecting such a pair:

• $E = \{(\text{Monday, Tuesday}), (\text{Tuesday, Wednesday})\}$

The number of favorable outcomes is $n(E) = 2$.

Step 4: Calculation of Probability

Using the Classical Definition of Probability [Probability $P(E) = \frac{n(E)}{n(S)}$]:

$P(E) = \frac{2}{7}$

Visual Representation of Sample Space

Final Solution: The probability that a leap year contains 53 Tuesdays is $\frac{2}{7}$.

Step 1: Identifying the Distribution Model

The problem involves a fixed number of independent Bernoulli trials, where each trial (selecting a person) has two possible outcomes: right-handed (success) or not right-handed (failure). Since the probability of success is constant, we apply the Binomial Distribution formula:

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

Where:

• $n = 10$ (Total number of people in the sample)
• $p = 0.9$ (Probability that a person is right-handed)
• $q = 1 - p = 0.1$ (Probability that a person is not right-handed)
• $X$ is the random variable representing the number of right-handed people.

Step 2: Defining the Target Probability

We are required to find the probability that at most 6 people are right-handed. This is expressed as:

$P(X \leq 6) = \sum_{k=0}^{6} \binom{10}{k} (0.9)^k (0.1)^{10-k}$

Due to the law of total probability, it is computationally more efficient to calculate the complement and subtract from 1:

$P(X \leq 6) = 1 - [P(X=7) + P(X=8) + P(X=9) + P(X=10)]$

Step 3: Calculating Individual Probabilities

We compute each term $P(X=k) = \binom{10}{k} (0.9)^k (0.1)^{10-k}$:

Step 4: Final Summation and Subtraction

Summing the probabilities for $k > 6$:

$\sum_{k=7}^{10} P(X=k) = 0.0573956 + 0.1937102 + 0.3874205 + 0.3486784 = 0.9872047$

Subtracting from total probability [Per the Complement Rule]:

$P(X \leq 6) = 1 - 0.9872047 = 0.0127953$

Visual Representation of Distribution

Final Solution: The probability that at most 6 people are right-handed is approximately 0.0128.

Step 1: Establishing the Sample Space

A second-order determinant is represented as:

$D = \begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix} = a_{11}a_{22} - a_{12}a_{21}$

Each entry $a_{ij}$ can take a value from the set $\{0, 1\}$. Since there are 4 positions and each position has 2 possible values, the total number of possible determinants is $2^4 = 16$. [Per the Fundamental Counting Principle].

Step 2: Defining the Condition for a Positive Determinant

We require the value of the determinant $D > 0$. Given the expression $D = a_{11}a_{22} - a_{12}a_{21}$, the condition $D > 0$ implies:

$a_{11}a_{22} - a_{12}a_{21} > 0 \implies a_{11}a_{22} > a_{12}a_{21}$

Step 3: Systematic Case Analysis

Since the entries are restricted to $\{0, 1\}$, the products $a_{11}a_{22}$ and $a_{12}a_{21}$ can only take values $0$ or $1$. For the inequality $a_{11}a_{22} > a_{12}a_{21}$ to hold, the only possibility is:

• $a_{11}a_{22} = 1$
• $a_{12}a_{21} = 0$

Step 4: Counting Favorable Outcomes

We analyze the two conditions independently:

Condition 1: $a_{11}a_{22} = 1$
This occurs only if $a_{11} = 1$ AND $a_{22} = 1$. There is $1$ way to satisfy this.

Condition 2: $a_{12}a_{21} = 0$
This occurs if $a_{12} = 0, a_{21} = 0$; OR $a_{12} = 0, a_{21} = 1$; OR $a_{12} = 1, a_{21} = 0$. There are $3$ ways to satisfy this.

By the Multiplication Rule for independent events, the total number of favorable outcomes is $1 \times 3 = 3$.

Step 5: Calculating Probability

The probability $P(E)$ is defined as the ratio of favorable outcomes to the total number of elements in the sample space. [By the Classical Definition of Probability].

$P(D > 0) = \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}} = \frac{3}{16}$

Final Solution: The probability that the value of the determinant is positive is $\frac{3}{16}$.

Step 1: Formal Definition of Conditional Probability

The conditional probability of an event $B$, given that event $A$ has occurred, is defined as:

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

[Per the definition of Conditional Probability provided in the axiomatic framework of probability theory, where $P(A) > 0$.]

Step 2: Analysis of the Given Condition

The problem states that $A$ is a subset of $B$ ($A \subseteq B$). By the set-theoretic definition of a subset, if every element of set $A$ is also an element of set $B$, then the intersection of $A$ and $B$ is equal to set $A$ itself.

$A \cap B = A$

[By the property of set intersection: if $A \subseteq B$, then $A \cap B = A$.]

Step 3: Derivation of the Probability Measure

Applying the result from Step 2 to the probability of the intersection:

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

Step 4: Substitution into the Conditional Probability Formula

Substitute $P(A \cap B) = P(A)$ into the expression derived in Step 1:

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

Step 5: Simplification

Since it is given that $P(A) \neq 0$, the ratio $\frac{P(A)}{P(A)}$ simplifies to unity:

$P(B|A) = 1$

Visual Representation of Subset Inclusion

The Venn diagram illustrates that all outcomes within region $A$ are strictly contained within region $B$. Consequently, if event $A$ is known to have occurred, it is certain that event $B$ has also occurred.

Final Solution: $P(B|A) = 1$