Key Concepts and Formulas

• Conditional Probability: The probability of event $A$ occurring given that event $B$ has already occurred is defined as: $P(A|B) = \frac{P(A \cap B)}{P(B)}, \quad \text{provided } P(B) > 0$ Similarly, $P(B|A) = \frac{P(A \cap B)}{P(A)}$, provided $P(A) > 0$.
• Probability of Union of Events: For any two events $A$ and $B$: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$
• Complementary Events: The probability of the complement of an event $A$, denoted $A'$, is given by: $P(A') = 1 - P(A)$
• Probability of Intersection with Complement: The probability of event $A$ occurring and event $B$ not occurring is: $P(A \cap B') = P(A) - P(A \cap B)$ Similarly, $P(A' \cap B) = P(B) - P(A \cap B)$.
• De Morgan's Laws for Probabilities: The probability of neither A nor B occurring is: $P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B)$
• Independent Events: Two events $A$ and $B$ are independent if and only if $P(A \cap B) = P(A) \cdot P(B)$.

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Step-by-Step Solution

We are given the following information:

• $P(E_1|E_2) = \frac{1}{2}$
• $P(E_2|E_1) = \frac{3}{4}$
• $P(E_1 \cap E_2) = \frac{1}{8}$

Our strategy is to first calculate the individual probabilities $P(E_1)$ and $P(E_2)$ using the given conditional probabilities and the probability of their intersection. Once these are determined, we will evaluate each option.

Step 1: Determine the Individual Probabilities $P(E_1)$ and $P(E_2)$

We use the definition of conditional probability to find $P(E_1)$ and $P(E_2)$.

Calculating $P(E_2)$: We know $P(E_1|E_2) = \frac{P(E_1 \cap E_2)}{P(E_2)}$. Substitute the given values: $\frac{1}{2} = \frac{1/8}{P(E_2)}$ To solve for $P(E_2)$, we rearrange the equation: $P(E_2) = \frac{1/8}{1/2} = \frac{1}{8} \times 2 = \frac{2}{8} = \frac{1}{4}$ So, $P(E_2) = \frac{1}{4}$.

Calculating $P(E_1)$: Similarly, we know $P(E_2|E_1) = \frac{P(E_1 \cap E_2)}{P(E_1)}$. Substitute the given values: $\frac{3}{4} = \frac{1/8}{P(E_1)}$ To solve for $P(E_1)$, we rearrange the equation: $P(E_1) = \frac{1/8}{3/4} = \frac{1}{8} \times \frac{4}{3} = \frac{4}{24} = \frac{1}{6}$ So, $P(E_1) = \frac{1}{6}$.

Summary of Probabilities: We have found the following probabilities:

• $P(E_1) = \frac{1}{6}$
• $P(E_2) = \frac{1}{4}$
• $P(E_1 \cap E_2) = \frac{1}{8}$ (given)

Step 2: Evaluate Each Given Option

Now we will systematically check each option using the probabilities calculated above.

Option (A): Check if $P(E_1 \cap E_2) = P(E_1) \cdot P(E_2)$ This option tests for the independence of events $E_1$ and $E_2$.

• Left Hand Side (LHS): $P(E_1 \cap E_2) = \frac{1}{8}$ (given).
• Right Hand Side (RHS): $P(E_1) \cdot P(E_2) = \frac{1}{6} \cdot \frac{1}{4} = \frac{1}{24}$. Comparing LHS and RHS: $\frac{1}{8} \neq \frac{1}{24}$. Therefore, option (A) is incorrect. Events $E_1$ and $E_2$ are not independent.

Option (B): Check if $P(E'_1 \cap E'_2) = P(E'_1) \cdot P(E_2)$ First, calculate the probabilities of the complementary events:

• $P(E'_1) = 1 - P(E_1) = 1 - \frac{1}{6} = \frac{5}{6}$.

• $P(E'_2) = 1 - P(E_2) = 1 - \frac{1}{4} = \frac{3}{4}$.

• LHS: $P(E'_1 \cap E'_2)$ Using De Morgan's Law, $P(E'_1 \cap E'_2) = P((E_1 \cup E_2)')$. This is equal to $1 - P(E_1 \cup E_2)$. First, calculate $P(E_1 \cup E_2)$: $P(E_1 \cup E_2) = P(E_1) + P(E_2) - P(E_1 \cap E_2)$ $P(E_1 \cup E_2) = \frac{1}{6} + \frac{1}{4} - \frac{1}{8}$ Find a common denominator (24): $P(E_1 \cup E_2) = \frac{4}{24} + \frac{6}{24} - \frac{3}{24} = \frac{4+6-3}{24} = \frac{7}{24}$. Now, calculate $P(E'_1 \cap E'_2)$: $P(E'_1 \cap E'_2) = 1 - P(E_1 \cup E_2) = 1 - \frac{7}{24} = \frac{17}{24}$.

• RHS: $P(E'_1) \cdot P(E_2)$ $P(E'_1) \cdot P(E_2) = \frac{5}{6} \cdot \frac{1}{4} = \frac{5}{24}$. Comparing LHS and RHS: $\frac{17}{24} \neq \frac{5}{24}$. Therefore, option (B) is incorrect.

Option (C): Check if $P(E_1 \cap E'_2) = P(E_1) \cdot P(E_2)$

• LHS: $P(E_1 \cap E'_2)$ This represents the probability that $E_1$ occurs but $E_2$ does not. $P(E_1 \cap E'_2) = P(E_1) - P(E_1 \cap E_2)$ $P(E_1 \cap E'_2) = \frac{1}{6} - \frac{1}{8}$ Find a common denominator (24): $P(E_1 \cap E'_2) = \frac{4}{24} - \frac{3}{24} = \frac{1}{24}$.

• RHS: $P(E_1) \cdot P(E_2)$ From Option (A), we calculated $P(E_1) \cdot P(E_2) = \frac{1}{6} \cdot \frac{1}{4} = \frac{1}{24}$. Comparing LHS and RHS: $\frac{1}{24} = \frac{1}{24}$. Therefore, option (C) is correct.

Option (D): Check if $P(E'_1 \cap E_2) = P(E_1) \cdot P(E_2)$

• LHS: $P(E'_1 \cap E_2)$ This represents the probability that $E_2$ occurs but $E_1$ does not. $P(E'_1 \cap E_2) = P(E_2) - P(E_1 \cap E_2)$ $P(E'_1 \cap E_2) = \frac{1}{4} - \frac{1}{8}$ Find a common denominator (8): $P(E'_1 \cap E_2) = \frac{2}{8} - \frac{1}{8} = \frac{1}{8}$.

• RHS: $P(E_1) \cdot P(E_2)$ From Option (A), we calculated $P(E_1) \cdot P(E_2) = \frac{1}{6} \cdot \frac{1}{4} = \frac{1}{24}$. Comparing LHS and RHS: $\frac{1}{8} \neq \frac{1}{24}$. Therefore, option (D) is incorrect.

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Common Mistakes & Tips

• Algebraic Errors with Fractions: Be extremely careful when adding, subtracting, or multiplying fractions. A small calculation mistake can propagate and lead to incorrect results for all subsequent steps. Always simplify fractions to their lowest terms.
• Misinterpreting Conditional Probability: Remember that $P(A|B)$ is not the same as $P(B|A)$ or $P(A \cap B)$. Each has a distinct definition and formula.
• Forgetting De Morgan's Laws: For probabilities involving complements of unions or intersections (e.g., $P(E'_1 \cap E'_2)$), De Morgan's laws are essential for proper calculation.
• Assuming Independence: Do not assume events are independent unless explicitly stated or proven. Always check the condition $P(A \cap B) = P(A) \cdot P(B)$.

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Summary

This problem required us to apply fundamental probability concepts to determine individual event probabilities and then evaluate given statements. We first utilized the definitions of conditional probability with the provided intersection probability to find $P(E_1) = \frac{1}{6}$ and $P(E_2) = \frac{1}{4}$. With these values, we systematically checked each option. Option (C) was found to be the only true statement, as $P(E_1 \cap E'_2) = P(E_1) - P(E_1 \cap E_2) = \frac{1}{6} - \frac{1}{8} = \frac{1}{24}$, which is equal to $P(E_1) \cdot P(E_2) = \frac{1}{6} \cdot \frac{1}{4} = \frac{1}{24}$.

The final answer is $\boxed{C}$.