Question

Let $E_1$ and $E_2$ be two independent events of a random experiment such that
$P(E_1) = \frac{1}{2}, \quad P(E_1 \cup E_2) = \frac{2}{3}$.
Then match the items of List-I with the items of List-II:

The correct match is:

• A. \( A \to \text{iii}, B \to \text{iv}, C \to \text{i}, D \to \text{v} \)
• B. \( A \to \text{iii}, B \to \text{i}, C \to \text{v}, D \to \text{ii} \)
• C. \( A \to \text{i}, B \to \text{v}, C \to \text{iii}, D \to \text{iv} \)
• D. \( A \to \text{v}, B \to \text{i}, C \to \text{iii}, D \to \text{ii} \)

Hint:

For independent events, use the formula \( P(A \cap B) = P(A) P(B) \) and conditional probability definitions to compute probabilities efficiently.

Answer 1

The Correct Option is B

Step 1: Find \( P(E_2) \)
Using the formula for the union of two independent events: \[ P(E_1 \cup E_2) = P(E_1) + P(E_2) - P(E_1 \cap E_2). \] Substituting given values: \[ \frac{2}{3} = \frac{1}{2} + P(E_2) - P(E_1) P(E_2). \] Since \( E_1 \) and \( E_2 \) are independent: \[ P(E_1 \cap E_2) = P(E_1) P(E_2). \] \[ \frac{2}{3} = \frac{1}{2} + P(E_2) - \frac{1}{2} P(E_2). \] Solving for \( P(E_2) \): \[ P(E_2) = \frac{1}{3}. \] Step 2: Compute conditional probabilities
\[ P(E_1/E_2) = \frac{P(E_1 \cap E_2)}{P(E_2)} = \frac{\frac{1}{2} \times \frac{1}{3}}{\frac{1}{3}} = \frac{1}{2}. \] \[ P(E_2/E_1) = \frac{P(E_1 \cap E_2)}{P(E_1)} = \frac{\frac{1}{2} \times \frac{1}{3}}{\frac{1}{2}} = \frac{2}{3}. \] \[ P(E_1 \cap E_2) = P(E_1) P(E_2) = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6}. \] Step 3: Conclusion
\[ \boxed{A \to iii, B \to i, C \to v, D \to ii.} \]