Key Concepts and Formulas

1. Conditional Probability: The probability of an event $X$ occurring given that event $Y$ has already occurred is defined as: $P(X \mid Y) = \frac{P(X \cap Y)}{P(Y)}, \quad \text{provided } P(Y) > 0$ This formula is crucial for finding unknown probabilities of individual events or their intersection when conditional probabilities are provided.

2. Probability of Union (Addition Rule): The probability that at least one of two events, $A$ or $B$, occurs is given by: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ This accounts for the overlap between events $A$ and $B$.

3. Complement Rule and De Morgan's Laws:

   • Complement Rule: The probability that an event $A$ does not occur (denoted $A'$) is $P(A') = 1 - P(A)$.
   • De Morgan's Law for Intersection of Complements: The probability that neither $A$ nor $B$ occurs is $P(A' \cap B') = P((A \cup B)')$. This means "not A AND not B" is equivalent to "not (A OR B)".
   • Identity for $P(A' \cup B)$: The probability of "A not occurring OR B occurring" can be conveniently expressed as $P(A' \cup B) = 1 - P(A \cap B')$, where $P(A \cap B')$ is the probability of $A$ occurring but $B$ not occurring, which can be calculated as $P(A) - P(A \cap B)$.

Step-by-Step Solution

We are given the following probabilities for events A and B:

• $P(B \mid A)=\frac{2}{5}$
• $P(A \mid B)=\frac{1}{7}$
• $P(A \cap B)=\frac{1}{9}$

We need to verify the truthfulness of two statements: (S1) $P\left(A^{\prime} \cup B\right)=\frac{5}{6}$ (S2) $P\left(A^{\prime} \cap B^{\prime}\right)=\frac{1}{18}$

Step 1: Calculate the individual probabilities $P(A)$ and $P(B)$.

• Why this step? To evaluate statements (S1) and (S2), which involve unions and complements, we first need to determine the base probabilities $P(A)$ and $P(B)$. The given conditional probabilities and the probability of the intersection allow us to derive these.

1. Calculate $P(A)$: We use the conditional probability formula $P(B \mid A) = \frac{P(A \cap B)}{P(A)}$. Substituting the given values: $\frac{2}{5} = \frac{\frac{1}{9}}{P(A)}$ To solve for $P(A)$, we rearrange the equation: $P(A) = \frac{\frac{1}{9}}{\frac{2}{5}} = \frac{1}{9} \times \frac{5}{2} = \frac{5}{18}$

2. Calculate $P(B)$: Similarly, we use the conditional probability formula $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$. Substituting the given values: $\frac{1}{7} = \frac{\frac{1}{9}}{P(B)}$ To solve for $P(B)$, we rearrange the equation: $P(B) = \frac{\frac{1}{9}}{\frac{1}{7}} = \frac{1}{9} \times \frac{7}{1} = \frac{7}{9}$

• Intermediate Result: We have determined $P(A) = \frac{5}{18}$ and $P(B) = \frac{7}{9}$. These values are essential for the subsequent calculations.

Step 2: Evaluate Statement (S1) $P\left(A^{\prime} \cup B\right)=\frac{5}{6}$.

• Why this step? We need to verify if the probability of the event "A does not occur OR B occurs" matches the value stated in (S1). We will use a standard probability identity to calculate this.

1. Apply the identity for $P(A' \cup B)$: The event $A' \cup B$ (A does not occur OR B occurs) is the complement of the event $A \cap B'$ (A occurs AND B does not occur). Therefore, we can write: $P(A' \cup B) = 1 - P(A \cap B')$

2. Calculate $P(A \cap B')$: The probability of $A$ occurring but $B$ not occurring is $P(A \cap B') = P(A) - P(A \cap B)$. This represents the portion of event $A$ that does not overlap with event $B$. Using the values from Step 1 and the given information: $P(A \cap B') = \frac{5}{18} - \frac{1}{9}$ To subtract these fractions, we find a common denominator, which is 18: $P(A \cap B') = \frac{5}{18} - \frac{1 \times 2}{9 \times 2} = \frac{5}{18} - \frac{2}{18} = \frac{3}{18} = \frac{1}{6}$

3. Calculate $P(A' \cup B)$: Now, substitute $P(A \cap B')$ back into the identity from step 2.1: $P(A' \cup B) = 1 - \frac{1}{6} = \frac{6}{6} - \frac{1}{6} = \frac{5}{6}$

4. Compare with (S1): Our calculated value of $P(A' \cup B) = \frac{5}{6}$ matches the value stated in (S1). Therefore, Statement (S1) is true.

Step 3: Evaluate Statement (S2) $P\left(A^{\prime} \cap B^{\prime}\right)=\frac{1}{18}$.

• Why this step? We need to verify if the probability of the event "neither A nor B occurs" matches the value stated in (S2). De Morgan's Law provides an efficient way to calculate this.

1. Apply De Morgan's Law: According to De Morgan's Law, the probability that neither $A$ nor $B$ occurs is equal to the probability of the complement of their union: $P(A' \cap B') = P((A \cup B)')$

2. Calculate $P(A \cup B)$: First, we find the probability of the union of $A$ and $B$ using the Addition Rule: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. Using the values from Step 1 and the given information: $P(A \cup B) = \frac{5}{18} + \frac{7}{9} - \frac{1}{9}$ To perform the addition and subtraction, we use a common denominator of 18: $P(A \cup B) = \frac{5}{18} + \frac{7 \times 2}{9 \times 2} - \frac{1 \times 2}{9 \times 2}$ $P(A \cup B) = \frac{5}{18} + \frac{14}{18} - \frac{2}{18}$ $P(A \cup B) = \frac{5 + 14 - 2}{18} = \frac{19 - 2}{18} = \frac{17}{18}$

3. Calculate $P(A' \cap B')$: Now, using the complement rule with the result from step 3.2: $P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B)$ $P(A' \cap B') = 1 - \frac{17}{18} = \frac{18}{18} - \frac{17}{18} = \frac{1}{18}$

4. Compare with (S2): Our calculated value of $P(A' \cap B') = \frac{1}{18}$ matches the value stated in (S2). Therefore, Statement (S2) is true.

Common Mistakes & Tips

• Fractional Arithmetic: A common source of errors is incorrect addition, subtraction, or division of fractions. Always ensure common denominators are found correctly.
• Misapplication of Identities: Be precise when using probability identities. For example, $P(A' \cup B)$ is not the same as $P(A' \cap B')$, and their formulas differ. Visualizing with Venn diagrams can help clarify these relationships.
• Step-by-Step Calculation: Break down the problem into smaller, manageable steps. Calculate intermediate probabilities ($P(A)$, $P(B)$, $P(A \cup B)$) accurately before using them in subsequent formulas.

Summary

This problem involved applying various fundamental probability formulas to verify two given statements. We began by calculating the individual probabilities $P(A)$ and $P(B)$ using the provided conditional probabilities and the intersection probability. Subsequently, we evaluated Statement (S1) by expressing $P(A' \cup B)$ as $1 - P(A \cap B')$ and calculating $P(A \cap B')$. For Statement (S2), we used De Morgan's Law to relate $P(A' \cap B')$ to $P((A \cup B)')$ and then applied the Addition Rule and Complement Rule. Both statements were found to be true based on our calculations.

The final answer is $\boxed{A}$ which corresponds to option (A).