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: 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: 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: 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: 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: Defining the Sample Space ($S$)

When a fair coin is tossed three times, each toss has two possible outcomes: Head ($H$) or Tail ($T$). The total number of elementary events in the sample space is $2^3 = 8$.

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

Step 2: Defining Events $E$ and $F$

Event $E$ (At least two heads): This event includes outcomes with 2 or 3 heads.

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

$n(E) = 4$

Event $F$ (At most two heads): This event includes outcomes with 0, 1, or 2 heads.

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

$n(F) = 7$

Step 3: Calculating the Intersection ($E \cap F$)

The intersection $E \cap F$ represents outcomes that satisfy both conditions: at least two heads AND at most two heads. This implies outcomes with exactly two heads.

$E \cap F = \{HHT, HTH, THH\}$

$n(E \cap F) = 3$

Step 4: Applying the Conditional Probability Formula

The conditional probability $P(E|F)$ is defined as [By the definition of Conditional Probability]:

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

Calculating individual probabilities based on the uniform sample space:

• $P(E \cap F) = \frac{n(E \cap F)}{n(S)} = \frac{3}{8}$
• $P(F) = \frac{n(F)}{n(S)} = \frac{7}{8}$

Step 5: Final Computation

$P(E|F) = \frac{3/8}{7/8}$

$P(E|F) = \frac{3}{8} \times \frac{8}{7}$

$P(E|F) = \frac{3}{7}$

Final Solution: $P(E|F) = \frac{3}{7}$

Step 1: Extracting Given Variables

From the problem statement, we identify the following probabilities and relations:

• $2P(A) = \frac{5}{13} \implies P(A) = \frac{5}{26}$
• $P(B) = \frac{5}{13}$
• $P(A|B) = \frac{2}{5}$

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

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

To find the union $P(A \cup B)$, we must first determine the intersection of events $A$ and $B$.

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

$P(A \cap B) = \left( \frac{2}{5} \right) \times \left( \frac{5}{13} \right)$

Canceling the common factor $5$ in the numerator and denominator:

$P(A \cap B) = \frac{2}{13}$

Step 3: Applying the Addition Theorem of Probability

The probability of the union of two events is given by the Addition Theorem:

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

Substitute the derived values:

$P(A \cup B) = \frac{5}{26} + \frac{5}{13} - \frac{2}{13}$

Step 4: Algebraic Simplification

To add these fractions, we identify the least common denominator as $26$.

$P(A \cup B) = \frac{5}{26} + \frac{5 \times 2}{13 \times 2} - \frac{2 \times 2}{13 \times 2}$

$P(A \cup B) = \frac{5}{26} + \frac{10}{26} - \frac{4}{26}$

$P(A \cup B) = \frac{5 + 10 - 4}{26}$

$P(A \cup B) = \frac{11}{26}$

Visual Representation of Set Union

Final Solution: $P(A \cup B) = \frac{11}{26}$

Step 1: Defining the Sample Space

Let $B$ represent a boy and $G$ represent a girl. For a family with two children, the sample space $S$ consists of all ordered pairs $(c_1, c_2)$, where $c_1$ is the first-born and $c_2$ is the second-born child.

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

Given that each birth is independent and equally likely, the probability of each outcome is $P(\omega) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$. [By the Multiplication Rule for independent events].

Step 2: Defining the Events

Let $A$ be the event that both children are girls: $A = \{GG\}$.

Let $F$ be the event that the youngest child is a girl. In the ordered pair $(c_1, c_2)$, the youngest child is the second position ($c_2$): $F = \{BG, GG\}$.

Step 3: Calculating Probabilities

The probability of event $F$ is calculated by the number of favorable outcomes divided by the total sample space:

$P(F) = \frac{n(F)}{n(S)} = \frac{2}{4} = \frac{1}{2}$

The intersection of events $A$ and $F$ is the set of outcomes where both are girls AND the youngest is a girl:

$A \cap F = \{GG\} \cap \{BG, GG\} = \{GG\}$

$P(A \cap F) = \frac{n(A \cap F)}{n(S)} = \frac{1}{4}$

Step 4: Applying Conditional Probability Formula

The conditional probability of $A$ given $F$ is defined as [Per the definition of Conditional Probability]:

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

Substituting the calculated values:

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

$P(A|F) = \frac{1}{4} \times \frac{2}{1} = \frac{1}{2}$

Visual Representation of Sample Space

Final Solution: The conditional probability that both children are girls given that the youngest is a girl is $\frac{1}{2}$.

Step 1: Define the Sample Space

Let $B$ denote a boy and $G$ denote a girl. For a family with two children, the sample space $S$ is the set of all possible ordered pairs $(c_1, c_2)$, where $c_1$ is the first child and $c_2$ is the second child.

$S = \{ (B, B), (B, G), (G, B), (G, G) \}$

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

Step 2: Identify Event Sets

Let $E$ be the event that both children are girls, and $F$ be the event that at least one child is a girl.

• Event $E$ (Both are girls): $E = \{ (G, G) \}$
• Event $F$ (At least one is a girl): $F = \{ (B, G), (G, B), (G, G) \}$

[By the definition of conditional probability, we require the intersection of these sets.]

Step 3: Intersection and Probabilities

Find the intersection $E \cap F$:

$E \cap F = \{ (G, G) \} \cap \{ (B, G), (G, B), (G, G) \} = \{ (G, G) \}$

The number of favorable outcomes are:

• $n(E \cap F) = 1$
• $n(F) = 3$

Since each outcome is equally likely with probability $P(\omega) = \frac{1}{4}$, the probabilities are:

$P(E \cap F) = \frac{n(E \cap F)}{n(S)} = \frac{1}{4}$

$P(F) = \frac{n(F)}{n(S)} = \frac{3}{4}$

Step 4: Applying the Conditional Probability Formula

The conditional probability $P(E|F)$ is defined as:

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

Substituting the calculated values:

$P(E|F) = \frac{1/4}{3/4}$

$P(E|F) = \frac{1}{3}$

Visual Representation of Sample Space

Final Solution: The conditional probability that both children are girls given that at least one is a girl is $\frac{1}{3}$.

Given Variables & Initial Setup

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

The probabilities of the defined events are based on the assumption of a fair die, where each outcome has a probability of $\frac{1}{6}$.

• $E = \{1, 3, 5\}$
• $F = \{2, 3\}$
• $G = \{2, 3, 4, 5\}$

Step 1: Calculating $P((E \cup F) | G)$

By the definition of conditional probability: $P(A|B) = \frac{P(A \cap B)}{P(B)}$ [By the Definition of Conditional Probability].

First, we find $(E \cup F)$: $E \cup F = \{1, 3, 5\} \cup \{2, 3\} = \{1, 2, 3, 5\}$.

Next, we find the intersection $(E \cup F) \cap G$: $(E \cup F) \cap G = \{1, 2, 3, 5\} \cap \{2, 3, 4, 5\} = \{2, 3, 5\}$.

The probability of the intersection is: $P((E \cup F) \cap G) = \frac{n((E \cup F) \cap G)}{n(S)} = \frac{3}{6} = \frac{1}{2}$.

The probability of $G$ is: $P(G) = \frac{n(G)}{n(S)} = \frac{4}{6} = \frac{2}{3}$.

Applying the formula: $P((E \cup F)|G) = \frac{P((E \cup F) \cap G)}{P(G)} = \frac{3/6}{4/6} = \frac{3}{4}$.

Step 2: Calculating $P((E \cap F) | G)$

First, we find $(E \cap F)$: $E \cap F = \{1, 3, 5\} \cap \{2, 3\} = \{3\}$.

Next, we find the intersection $(E \cap F) \cap G$: $(E \cap F) \cap G = \{3\} \cap \{2, 3, 4, 5\} = \{3\}$.

The probability of this intersection is: $P((E \cap F) \cap G) = \frac{n((E \cap F) \cap G)}{n(S)} = \frac{1}{6}$.

Applying the formula: $P((E \cap F)|G) = \frac{P((E \cap F) \cap G)}{P(G)} = \frac{1/6}{4/6} = \frac{1}{4}$.

Visual Representation of Sets

Final Solution: $P((E \cup F)|G) = \frac{3}{4}$ and $P((E \cap F)|G) = \frac{1}{4}$

Step 1: Defining the Sample Space ($S$)

Let the members of the family be represented by $M$ (Mother), $F$ (Father), and $S$ (Son). The total number of ways the three individuals can be arranged in a line is given by the permutations of 3 distinct objects: $3! = 3 \times 2 \times 1 = 6$.

The sample space $S$ consists of the following outcomes:

$S = \{MFS, MSF, FMS, FSM, SMF, SFM\}$

Therefore, $|S| = 6$.

Step 2: Defining Events $E$ and $F$

Event $E$ (Son on one end):

The outcomes where the son ($S$) is at the first or third position are:

$E = \{MFS, MSF, FMS, SFM\}$

$|E| = 4$

Event $F$ (Father in the middle):

The outcomes where the father ($F$) is in the second position are:

$F = \{MFS, SFM\}$

$|F| = 2$

Step 3: Calculating the Intersection ($E \cap F$)

The intersection represents the set of outcomes where both conditions are satisfied (Son on an end AND Father in the middle):

$E \cap F = \{MFS, SFM\}$

$|E \cap F| = 2$

Step 4: Computing Probabilities

Using the classical definition of probability, $P(A) = \frac{n(A)}{n(S)}$, where $n$ is the number of favorable outcomes and $n(S)$ is the total number of equally likely outcomes [Per Kolmogorov's Axioms]:

$P(F) = \frac{|F|}{|S|} = \frac{2}{6} = \frac{1}{3}$

$P(E \cap F) = \frac{|E \cap F|}{|S|} = \frac{2}{6} = \frac{1}{3}$

Step 5: Applying the Conditional Probability Formula

The conditional probability $P(E|F)$ is defined as the probability of event $E$ occurring, given that event $F$ has already occurred. The formula is:

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

[By Definition of Conditional Probability for $P(F) > 0$]

Substituting the values derived in Step 4:

$P(E|F) = \frac{1/3}{1/3} = 1$

Visual Representation of Sample Space Arrangement

Final Solution: $P(E|F) = 1$

Step 1: Establishing the Sample Space and Event Sets

For a fair six-sided die, the sample space $S$ is defined as $S = \{1, 2, 3, 4, 5, 6\}$. The total number of outcomes is $n(S) = 6$.

The given events are:

• $E = \{1, 3, 5\} \implies n(E) = 3$
• $F = \{2, 3\} \implies n(F) = 2$
• $G = \{2, 3, 4, 5\} \implies n(G) = 4$

Step 2: Calculating Individual Probabilities

Using the classical definition of probability, $P(A) = \frac{n(A)}{n(S)}$ [Per the Axioms of Probability]:

• $P(E) = \frac{3}{6} = \frac{1}{2}$
• $P(G) = \frac{4}{6} = \frac{2}{3}$

Step 3: Finding the Intersection of Events

To find the conditional probability $P(E|G)$ and $P(G|E)$, we require the intersection set $E \cap G$ [Per the definition of Conditional Probability]:

$E \cap G = \{1, 3, 5\} \cap \{2, 3, 4, 5\} = \{3, 5\}$

$n(E \cap G) = 2$

$P(E \cap G) = \frac{n(E \cap G)}{n(S)} = \frac{2}{6} = \frac{1}{3}$

Step 4: Applying the Conditional Probability Formula

The formula for conditional probability is given by $P(A|B) = \frac{P(A \cap B)}{P(B)}$, where $P(B) > 0$.

Part (ii-a): Calculation of $P(E|G)$

Part (ii-b): Calculation of $P(G|E)$

Visual Representation of Sample Space Intersections

Final Solution: $P(E|G) = \frac{1}{2}$ and $P(G|E) = \frac{2}{3}$

Step 1: Defining the Sample Space ($S$)

When two distinguishable dice (a black die and a red die) are rolled, the sample space $S$ consists of all possible ordered pairs $(b, r)$, where $b$ is the result of the black die and $r$ is the result of the red die. Since each die has 6 faces:

$n(S) = 6 \times 6 = 36$ [By the Fundamental Counting Principle].

Step 2: Defining the Events

Let $E$ be the event of obtaining a sum greater than 9.

Let $F$ be the event that the black die results in a 5.

Step 3: Calculating Event $F$ and $E \cap F$

Event $F$ (Black die shows 5):

$F = \{(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)\}$

$n(F) = 6$

Event $E$ (Sum > 9):

The possible pairs for a sum > 9 are $(4,6), (5,5), (5,6), (6,4), (6,5), (6,6)$.

Intersection $E \cap F$ (Sum > 9 AND Black die is 5):

$E \cap F = \{(5, 5), (5, 6)\}$

$n(E \cap F) = 2$

Step 4: Applying the Conditional Probability Formula

The conditional probability of $E$ given $F$ is defined as:

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

Since the sample space is uniform, we can use the ratio of the number of outcomes:

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

Step 5: Final Calculation

$P(E|F) = \frac{2}{6}$

$P(E|F) = \frac{1}{3}$

Final Solution: The conditional probability is $\frac{1}{3}$.

Step 1: Identifying Given Parameters

Based on the provided information, we have the following probabilities for events $A$ and $B$ within a sample space $S$:

• Probability of event $B$, denoted as $P(B) = 0.5$.
• Probability of the intersection of events $A$ and $B$, denoted as $P(A \cap B) = 0.32$.

Step 2: Defining the Theoretical Framework

To find the conditional probability of event $A$ given that event $B$ has already occurred, denoted as $P(A|B)$, we utilize the axiomatic definition of conditional probability [Per Kolmogorov's Axioms for Probability]:

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

This formula is valid under the condition that $P(B) > 0$. Since $P(B) = 0.5 \neq 0$, the calculation is well-defined.

Step 3: Execution of Calculation

Substitute the given numerical values into the formula:

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

To simplify the fraction, multiply both the numerator and the denominator by $100$ to eliminate decimals:

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

Dividing both by $2$ to reach the simplest fraction form:

$P(A|B) = \frac{16}{25}$

Converting the fraction to a decimal representation:

$P(A|B) = 0.64$

Step 4: Visual Representation of the Intersection

The diagram illustrates the intersection $P(A \cap B)$ relative to the total set $B$. Given that the condition restricts the sample space to $B$, the probability $P(A|B)$ represents the ratio of the overlap to the total area of circle $B$.

Final Solution: $P(A|B) = 0.64$

Step 1: Identifying the Conditional Probability Formula

The conditional probability of an event $A$, given that event $B$ has occurred, is defined by the formula:
$P(A|B) = \frac{P(A \cap B)}{P(B)}$
where $P(B) > 0$. [Per the Kolmogorov Axioms of Probability theory regarding conditional space].

Step 2: Evaluating the Given Variables

• $P(A) = \frac{1}{2}$
• $P(B) = 0$

Step 3: Analyzing Mathematical Constraints

The definition of conditional probability $P(A|B)$ inherently assumes the existence of the event $B$ as a non-empty set within the sample space $S$. Specifically, the denominator $P(B)$ must be strictly greater than zero for the ratio to be well-defined in the field of real numbers $\mathbb{R}$.

In this scenario, since $P(B) = 0$, the expression for $P(A|B)$ becomes:
$P(A|B) = \frac{P(A \cap B)}{0}$
Division by zero is undefined in standard arithmetic and set-theoretic probability measures. Consequently, $P(A|B)$ is not defined when $P(B) = 0$.

Visual Representation of the Null Event Space

Logical Conclusion

Since the condition for conditional probability requires the probability of the conditioning event to be non-zero, and the given $P(B) = 0$, the value is mathematically indeterminate.

Final Solution: The probability $P(A|B)$ is undefined.

Step 1: Defining the Sample Space

When two dice (one black, one red) are rolled, each die has $6$ independent outcomes. The total number of outcomes in the sample space $S$ is: $n(S) = 6 \times 6 = 36$.

Let the outcome be represented as $(b, r)$, where $b$ is the result of the black die and $r$ is the result of the red die, $b, r \in \{1, 2, 3, 4, 5, 6\}$.

Step 2: Defining the Events

Let $A$ be the event of obtaining a sum of $8$. $A = \{(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)\}$ $n(A) = 5$.

Let $F$ be the event that the red die resulted in a number less than $4$. $F = \{(b, r) : r < 4, r \in \{1, 2, 3\}\}$ Since $b$ can be any value from $1$ to $6$ and $r \in \{1, 2, 3\}$, the set $F$ is: $F = \{(1,1), (2,1), (3,1), (4,1), (5,1), (6,1), (1,2), (2,2), (3,2), (4,2), (5,2), (6,2), (1,3), (2,3), (3,3), (4,3), (5,3), (6,3)\}$ $n(F) = 6 \times 3 = 18$.

Step 3: Calculating the Intersection $A \cap F$

We identify elements present in both $A$ and $F$. These are the outcomes where the sum is $8$ AND the red die is less than $4$: $A \cap F = \{(5, 3), (6, 2)\}$ $n(A \cap F) = 2$.

Step 4: Applying the Conditional Probability Formula

[Per the definition of Conditional Probability: $P(A|F) = \frac{P(A \cap F)}{P(F)}$]

The formula can be simplified to: $P(A|F) = \frac{n(A \cap F)}{n(F)}$

Substituting the values derived: $P(A|F) = \frac{2}{18}$

Step 5: Simplification

Dividing the numerator and the denominator by their greatest common divisor ($2$): $P(A|F) = \frac{1}{9}$

Visual Representation of Sample Space subset $F$

Final Solution: The conditional probability is $\frac{1}{9}$.

Step 1: Given Variables and Theoretical Foundation

To determine the probability of the intersection of two events $A$ and $B$, we utilize the definition of conditional probability. The given parameters are:

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

The conditional probability of event $B$ given that event $A$ has occurred is defined as: [By the Definition of Conditional Probability]

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

Step 2: Rearrangement of the Formula

To isolate the target variable $P(A \cap B)$, we multiply both sides of the equation by $P(A)$:

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

Step 3: Numerical Substitution and Calculation

Substitute the provided values into the derived formula:

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

$P(A \cap B) = 0.32$

Visual Representation of Sample Space

The intersection $P(A \cap B)$ represents the probability that both events $A$ and $B$ occur simultaneously. Based on the calculation derived from the conditional probability definition, the product of the probability of $A$ and the conditional probability of $B$ given $A$ yields the joint probability.

Final Solution: $P(A \cap B) = 0.32$

Step 1: Defining the Sample Space of the Experiment

Let $D$ represent the outcome of the die throw and $C$ represent the outcome of the coin toss. The experiment follows a branching logic based on the outcome of the first die ($D_1$):

• If $D_1 \in \{3, 6\}$, throw the die again ($D_2$). The possible outcomes are $(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)$.
• If $D_1 \in \{1, 2, 4, 5\}$, toss a coin ($C$). The possible outcomes are $(1, H), (1, T), (2, H), (2, T), (4, H), (4, T), (5, H), (5, T)$.

Total sample space $S$ consists of $12 + 8 = 20$ equally likely paths of the initial choice, but we must weight them by probability. Each $D_1$ has $P = 1/6$.

Step 2: Defining Events $A$ and $B$

Let $A$ be the event 'the coin shows a tail' and $B$ be the event 'at least one die shows a 3'.

The event $B$ occurs if $D_1 = 3$ (regardless of $D_2$), or if $D_1 \in \{1, 2, 4, 5, 6\}$ and the second event results in a 3. Specifically:

• From first branch ($D_1 \in \{3, 6\}$):
  • If $D_1 = 3$, $B$ occurs for any $D_2 \in \{1, 2, 3, 4, 5, 6\}$. Probability $= (1/6) \times (6/6) = 6/36$.
  • If $D_1 = 6$, $B$ occurs only if $D_2 = 3$. Probability $= (1/6) \times (1/6) = 1/36$.
• From second branch ($D_1 \in \{1, 2, 4, 5\}$):
  • $B$ cannot occur because the second experiment is a coin toss.

Total $P(B) = 6/36 + 1/36 = 7/36$. [By the Addition Theorem for mutually exclusive events].

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

Event $A$ is 'the coin shows a tail'. This only occurs when $D_1 \in \{1, 2, 4, 5\}$.

Event $B$ is 'at least one die shows a 3'. This only occurs when $D_1 \in \{3, 6\}$.

Since the conditions for $A$ and $B$ are mutually exclusive regarding the first die roll ($D_1$), there is no sample point that satisfies both $A$ and $B$ simultaneously.

$A \cap B = \emptyset \implies P(A \cap B) = 0$.

Step 4: Applying the Conditional Probability Formula

The conditional probability $P(A|B)$ is defined as:

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

Substituting the values derived:

$P(A|B) = \frac{0}{7/36} = 0$.

Visual Representation of Sample Space Logic

Final Solution: The conditional probability $P(A|B) = 0$.

Given Variables & Initial Setup

The sample space $S$ for rolling a fair die is defined as: $S = \{1, 2, 3, 4, 5, 6\}$.

The total number of outcomes in the sample space is $n(S) = 6$.

The events are defined as:

• $E = \{1, 3, 5\}$
• $F = \{2, 3\}$
• $G = \{2, 3, 4, 5\}$

The probability of any event $A$ is given by $P(A) = \frac{n(A)}{n(S)}$, assuming a fair die [Definition of Classical Probability].

Step 1: Calculating Individual Probabilities

Based on the sets provided:

• $n(E) = 3 \implies P(E) = \frac{3}{6} = \frac{1}{2}$
• $n(F) = 2 \implies P(F) = \frac{2}{6} = \frac{1}{3}$

Step 2: Calculating Intersection of Events

To compute conditional probability $P(E|F) = \frac{P(E \cap F)}{P(F)}$ and $P(F|E) = \frac{P(F \cap E)}{P(E)}$, we first identify the intersection $E \cap F$ [Definition of Conditional Probability].

$E \cap F = \{1, 3, 5\} \cap \{2, 3\} = \{3\}$

$n(E \cap F) = 1 \implies P(E \cap F) = \frac{1}{6}$

Step 3: Calculating $P(E|F)$

Substituting the values into the conditional probability formula:

Step 4: Calculating $P(F|E)$

Substituting the values into the conditional probability formula:

Visual Representation of Sample Space and Events

Final Solution: $P(E|F) = \frac{1}{2}$ and $P(F|E) = \frac{1}{3}$