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: Given Variables & Theoretical Foundation

We are provided with the probabilities of two independent events, $A$ and $B$:

• $P(A) = \frac{3}{5}$
• $P(B) = \frac{1}{5}$

The core theoretical principle governing this problem is the Multiplication Theorem for Independent Events. Two events $A$ and $B$ are defined as independent if and only if the occurrence of one does not affect the probability of the occurrence of the other. Mathematically, this is expressed as:

$P(A \cap B) = P(A) \times P(B)$ [By definition of Independent Events]

Step 2: Mathematical Execution

Substitute the given values into the multiplication identity:

$P(A \cap B) = \left( \frac{3}{5} \right) \times \left( \frac{1}{5} \right)$

Step 3: Arithmetic Computation

Multiplying the numerators and the denominators separately:

$P(A \cap B) = \frac{3 \times 1}{5 \times 5}$

$P(A \cap B) = \frac{3}{25}$

Step 4: Geometric Representation of Independent Probabilities

The intersection $P(A \cap B)$ represents the joint probability of both events occurring simultaneously, which, for independent events, is the area of the overlapping region in the sample space defined by the product of their individual probabilities.

Final Solution: $P(A \cap B) = \frac{3}{25}$

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: Theoretical Analysis of the Subset Condition

Given the condition $A \subset B$, by the definition of a subset in probability theory, every sample point in event $A$ is also contained within event $B$. Consequently, the intersection of $A$ and $B$ is simply $A$:

$A \cap B = A$ [Definition of Subset]

Therefore, the probability of the intersection is:

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

Step 2: Evaluating Conditional Probability

The conditional probability of $A$ given $B$ is defined by the formula [Per the Axioms of Probability]:

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

Substituting the result from Step 1 into this formula:

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

Step 3: Logical Deduction of the Inequality

Since $A \subset B$, it follows that $P(A) \leq P(B)$ [Monotonicity Property of Probability]. Given $P(B) \neq 0$, we analyze the possible values for $P(A|B)$:

• Since $P(A) \leq P(B)$, then $\frac{P(A)}{P(B)} \leq 1$.
• Since $P(A) \geq 0$ and $P(B) > 0$, then $\frac{P(A)}{P(B)} \geq 0$.

Step 4: Comparative Analysis of Options

In standard textbook problems of this type, the objective is to identify the relationship between $P(A|B)$ and $P(A)$.

Final Solution: The correct relationship is $P(A|B) \geq P(A)$, as $P(A|B) = \frac{P(A)}{P(B)}$ and $P(B) \leq 1$.

Step 1: Definition of Conditional Probability

The conditional probability of an event occurring, given that another event has already occurred, is defined by the formula [Per the Axioms of Probability Theory]:

$P(A|B) = \frac{P(A \cap B)}{P(B)}$, where $P(B) > 0$

$P(B|A) = \frac{P(B \cap A)}{P(A)}$, where $P(A) > 0$

Step 2: Equating the Given Conditions

According to the problem statement, we are given that $P(A|B) = P(B|A)$. Substituting the definitions from Step 1 into this equality:

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

Step 3: Algebraic Simplification

Since the intersection of sets is commutative, we know that $P(A \cap B) = P(B \cap A)$ [Commutative Law of Set Theory]. We can rewrite the equation as:

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

Step 4: Logical Analysis of Cases

Case 1: If $P(A \cap B) \neq 0$

We can divide both sides of the equation by $P(A \cap B)$. This yields:

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

Taking the reciprocal of both sides leads to:

$P(A) = P(B)$

Case 2: If $P(A \cap B) = 0$

If the intersection is a null event, the equation becomes $0 = 0$, which is trivially true regardless of $P(A)$ or $P(B)$, provided the conditional probabilities exist. However, in the context of standard probability problems of this type, the result $P(A) = P(B)$ is the primary deductive conclusion.

Visual Representation of Event Space

Final Solution: $P(A) = P(B)$ (Given that $P(A \cap B) \neq 0$)

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: Defining Events and Initial Probabilities

Let the events be defined as follows:

• $E_1$: The event of selecting Bag I.
• $E_2$: The event of selecting Bag II.
• $A$: The event of drawing a red ball.

Since the bag is selected at random, the prior probabilities are:

• $P(E_1) = \frac{1}{2}$ [Per the definition of random selection]
• $P(E_2) = \frac{1}{2}$ [Per the definition of random selection]

Step 2: Calculating Conditional Probabilities

Based on the contents of the bags:

• Bag I contains $4$ red and $4$ black balls (Total = $8$ balls).
• Bag II contains $2$ red and $6$ black balls (Total = $8$ balls).

The conditional probabilities of drawing a red ball given the selected bag are:

• $P(A|E_1) = \frac{4}{8} = \frac{1}{2}$ [By the classical definition of probability]
• $P(A|E_2) = \frac{2}{8} = \frac{1}{4}$ [By the classical definition of probability]

Step 3: Calculating Total Probability of Drawing a Red Ball

Using the Law of Total Probability [Per the Law of Total Probability]:

Step 4: Applying Bayes' Theorem

We seek the posterior probability $P(E_1|A)$, the probability that the ball came from Bag I given it is red [Per Bayes' Theorem]:

Visual Representation of Probability Tree

Final Solution: The probability that the ball is drawn from the first bag is $\frac{2}{3}$.

Step 1: Identifying Given Parameters

To determine the conditional probability $P(A|B)$, we utilize the provided empirical values for the events $A$ and $B$ within a probability space:

• $P(A) = 0.8$
• $P(B) = 0.5$
• $P(B|A) = 0.4$

Step 2: Theoretical Foundation

The definition of conditional probability for any two events $A$ and $B$, where $P(A) > 0$ and $P(B) > 0$, is given by [Per the Definition of Conditional Probability]:

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

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

Step 3: Calculating the Intersection $P(A \cap B)$

Using the definition $P(B|A) = \frac{P(A \cap B)}{P(A)}$, we rearrange the terms to isolate the intersection probability [By Algebraic Manipulation]:

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

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

$P(A \cap B) = 0.32$

Step 4: Deriving $P(A|B)$

Substituting the intersection value $P(A \cap B) = 0.32$ and the given $P(B) = 0.5$ into the formula for $P(A|B)$:

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

$P(A|B) = \frac{32}{50}$

$P(A|B) = 0.64$

Visual Representation of Probability Space

Final Solution: P(A|B) = 0.64

Step 1: Establishing the Conditional Probability Formula

To determine the conditional probability $P(A|B)$, which represents the probability of event $A$ occurring given that event $B$ has already occurred, we employ the fundamental definition of conditional probability:
$P(A|B) = \frac{P(A \cap B)}{P(B)}$ [Per the definition of Conditional Probability, provided $P(B) > 0$].

Step 2: Calculating the Intersection $P(A \cap B)$

We are given the probabilities $P(A)$, $P(B)$, and the union $P(A \cup B)$. We utilize the Addition Theorem of Probability to find the intersection:
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$ [Per the Inclusion-Exclusion Principle].

Substituting the given values into the equation:
$\frac{7}{11} = \frac{6}{11} + \frac{5}{11} - P(A \cap B)$

Isolating $P(A \cap B)$:
$P(A \cap B) = \frac{6}{11} + \frac{5}{11} - \frac{7}{11}$
$P(A \cap B) = \frac{11 - 7}{11}$
$P(A \cap B) = \frac{4}{11}$

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

Now, applying the values to the conditional probability formula established in Step 1:
$P(A|B) = \frac{\frac{4}{11}}{\frac{5}{11}}$

Simplifying the complex fraction by multiplying the numerator and denominator by $11$:
$P(A|B) = \frac{4}{11} \times \frac{11}{5}$
$P(A|B) = \frac{4}{5}$

Final Solution: $P(A|B) = \frac{4}{5}$

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: Establishing the Theoretical Framework

To determine the conditional probability $P(B|A)$, we utilize the definition of conditional probability, which states that for two events $A$ and $B$ within a sample space, where $P(A) > 0$:
$P(B|A) = \frac{P(A \cap B)}{P(A)}$ [By Definition of Conditional Probability]

Step 2: Calculating the Intersection $P(A \cap B)$

We are given the values for $P(A)$, $P(B)$, and $P(A \cup B)$. We utilize the Addition Theorem of Probability to isolate $P(A \cap B)$:
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$ [Addition Theorem of Probability]

Rearranging the equation to solve for $P(A \cap B)$:
$P(A \cap B) = P(A) + P(B) - P(A \cup B)$

Substituting the given values:
$P(A \cap B) = \frac{6}{11} + \frac{5}{11} - \frac{7}{11}$
$P(A \cap B) = \frac{6 + 5 - 7}{11} = \frac{4}{11}$

Step 3: Computing the Conditional Probability $P(B|A)$

With the intersection $P(A \cap B) = \frac{4}{11}$ and the given $P(A) = \frac{6}{11}$, we apply the formula established in Step 1:
$P(B|A) = \frac{\frac{4}{11}}{\frac{6}{11}}$

Performing the algebraic simplification:
$P(B|A) = \frac{4}{11} \times \frac{11}{6}$
$P(B|A) = \frac{4}{6}$
$P(B|A) = \frac{2}{3}$

Visual Representation of Set Theory

Final Solution: $P(B|A) = \frac{2}{3}$

Step 1: Define Events and Initial Probabilities

Let $E_1$ be the event that the first ball drawn is Red, and $E_2$ be the event that the first ball drawn is Black. The urn initially contains 5 Red ($R$) and 5 Black ($B$) balls, totaling 10 balls.

• $P(E_1) = \frac{5}{10} = \frac{1}{2}$
• $P(E_2) = \frac{5}{10} = \frac{1}{2}$

Step 2: Define Conditional Probabilities for the Second Draw

Let $A$ be the event that the second ball drawn is Red. We must consider the state of the urn after the first ball is replaced and two additional balls of the same color are added.

Step 3: Apply the Law of Total Probability

To find the total probability of event $A$, we use the Law of Total Probability: $P(A) = P(E_1) \cdot P(A|E_1) + P(E_2) \cdot P(A|E_2)$

Substituting the values derived above:

$P(A) = \left( \frac{1}{2} \times \frac{7}{12} \right) + \left( \frac{1}{2} \times \frac{5}{12} \right)$

Step 4: Algebraic Simplification

$P(A) = \frac{7}{24} + \frac{5}{24}$

$P(A) = \frac{12}{24}$

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

Final Solution: The probability that the second ball is red is $\frac{1}{2}$.

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

Given Variables & Initial Setup

The problem provides two independent events $A$ and $B$ with the following probabilities:

• $P(A) = 0.3$
• $P(B) = 0.6$

Since $A$ and $B$ are independent events, the following axiom holds: $P(A \cap B) = P(A) \cdot P(B)$ [By the definition of independence in Probability Theory].

Step 1: Determine the Probability of the Complement Event

To find the probability of "not $B$" (denoted as $B'$ or $B^c$), we utilize the Complement Rule: $P(B') = 1 - P(B)$ [Per the Law of Complements].

$P(B') = 1 - 0.6 = 0.4$

Step 2: Analytical Expression for Independent Complements

If $A$ and $B$ are independent, then $A$ and $B'$ are also independent [Theorem: If $A$ and $B$ are independent, then $A$ and $B^c$ are independent].

Therefore, the probability of the intersection $P(A \cap B')$ is calculated as:

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

Step 3: Numerical Computation

Substituting the given and derived values into the equation:

$P(A \cap B') = 0.3 \times 0.4$

$P(A \cap B') = 0.12$

Visual Representation of Independent Events in Sample Space

Final Solution: $P(A \text{ and not } B) = 0.12$

Step 1: Given Variables & Initial Setup

Based on the provided information, we define the following probabilities:

• $P(A) = 0.8$
• $P(B) = 0.5$
• $P(B|A) = 0.4$

We are required to find $P(A \cup B)$. [Per the Addition Theorem of Probability, $P(A \cup B) = P(A) + P(B) - P(A \cap B)$].

Step 2: Calculating the Intersection $P(A \cap B)$

To compute the union, we must first determine the intersection $P(A \cap B)$. We utilize the definition of conditional probability: [Per the Definition of Conditional Probability, $P(B|A) = \frac{P(A \cap B)}{P(A)}$].

Rearranging the formula to isolate the intersection:

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

Substituting the given values:

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

$P(A \cap B) = 0.32$

Step 3: Calculating the Union $P(A \cup B)$

Now, substitute the known values into the Addition Theorem formula:

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

$P(A \cup B) = 0.8 + 0.5 - 0.32$

$P(A \cup B) = 1.3 - 0.32$

$P(A \cup B) = 0.98$

Visual Representation (Venn Diagram)

Final Solution: $P(A \cup B) = 0.98$

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: Identifying Variables and Probabilities

The total number of balls in the box is given by the sum of black and red balls:

$N = 10 \text{ (black)} + 8 \text{ (red)} = 18 \text{ total balls}$

Let $B$ be the event of drawing a black ball and $R$ be the event of drawing a red ball. Since the balls are drawn with replacement, the trials are independent, and the probabilities remain constant for each draw.

Probability of drawing a black ball $P(B) = \frac{10}{18} = \frac{5}{9}$

Probability of drawing a red ball $P(R) = \frac{8}{18} = \frac{4}{9}$

Step 2: Defining the Favorable Outcomes

The event "one black and one red" can occur in two mutually exclusive ways when drawing two balls:

• Case 1: The first ball is Black and the second ball is Red $(B, R)$.
• Case 2: The first ball is Red and the second ball is Black $(R, B)$.

Step 3: Calculating Individual Probabilities

Since the draws are independent [By the Multiplication Rule for Independent Events, $P(A \cap B) = P(A) \cdot P(B)$]:

For Case 1 ($B$ then $R$):

$P(B \cap R) = P(B) \times P(R) = \frac{5}{9} \times \frac{4}{9} = \frac{20}{81}$

For Case 2 ($R$ then $B$):

$P(R \cap B) = P(R) \times P(B) = \frac{4}{9} \times \frac{5}{9} = \frac{20}{81}$

Step 4: Combining Probabilities

Because these cases are mutually exclusive [By the Addition Rule for Mutually Exclusive Events, $P(A \cup B) = P(A) + P(B)$]:

$P(\text{one black, one red}) = P(B \cap R) + P(R \cap B)$

$P(\text{one black, one red}) = \frac{20}{81} + \frac{20}{81} = \frac{40}{81}$

Visual Representation of Outcome Space

Final Solution: The probability that one ball is black and the other is red is $\frac{40}{81}$.

Step 1: Establishing the Sample Space (S)

When a fair coin is tossed three times, each toss has two possible outcomes: Head ($H$) or Tail ($T$). The sample space $S$ consists of $2^3 = 8$ equally likely outcomes:

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

The total number of outcomes, $n(S) = 8$.

Step 2: Defining Events E and F

Event $E$ is defined as "head on the third toss." We identify outcomes in $S$ where the third position is $H$:

$E = \{HHH, HTH, THH, TTH\}$

$n(E) = 4$

Event $F$ is defined as "heads on the first two tosses." We identify outcomes in $S$ where the first two positions are $H$:

$F = \{HHH, HHT\}$

$n(F) = 2$

Step 3: Calculating the Intersection (E ∩ F)

The intersection $E \cap F$ contains outcomes that satisfy both conditions (head on the third toss AND heads on the first two tosses):

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

$n(E \cap F) = 1$

Step 4: Applying the Conditional Probability Formula

The probability of event $E$ occurring given that event $F$ has occurred is defined by [Per the definition of Conditional Probability]:

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

Calculating the individual probabilities based on the classical definition $P(A) = \frac{n(A)}{n(S)}$:

• $P(E \cap F) = \frac{n(E \cap F)}{n(S)} = \frac{1}{8}$
• $P(F) = \frac{n(F)}{n(S)} = \frac{2}{8}$

Step 5: Final Computation

Substituting the values into the formula:

$P(E|F) = \frac{1/8}{2/8} = \frac{1}{8} \times \frac{8}{2} = \frac{1}{2}$

Visual Representation of Sample Space subset

Final Solution: $P(E|F) = \frac{1}{2}$

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