Question

If A, B, and C are mutually exclusive and exhaustive events of a random experiment such that P(B) = (3/2) P(A) and P(C) = (1/2) P(B), then P(A ∪ C) equals:

• A. \( \frac{10}{13} \)
• B. \( \frac{3}{13} \)
• C. \( \frac{6}{13} \)
• D. \( \frac{7}{13} \)

Hint:

For mutually exclusive events, use the rule \( P(A \cup B) = P(A) + P(B) \) and substitute the given relationships between the probabilities.

Answer 1

The Correct Option is D

We are given that \( A \), \( B \), and \( C \) are mutually exclusive and exhaustive events, which means: \[ P(A) + P(B) + P(C) = 1 \] We are also given: \[ P(B) = \frac{3}{2} P(A) \quad \text{and} \quad P(C) = \frac{1}{2} P(B) \] Step 1: Express all probabilities in terms of \( P(A) \) From the given relations: \[ P(B) = \frac{3}{2} P(A) \] \[ P(C) = \frac{1}{2} P(B) = \frac{1}{2} \times \frac{3}{2} P(A) = \frac{3}{4} P(A) \] Step 2: Solve for \( P(A) \) Substitute these expressions into the equation \( P(A) + P(B) + P(C) = 1 \): \[ P(A) + \frac{3}{2} P(A) + \frac{3}{4} P(A) = 1 \] To simplify, multiply through by 4 to eliminate the fractions: \[ 4P(A) + 6P(A) + 3P(A) = 4 \] \[ 13P(A) = 4 \] \[ P(A) = \frac{4}{13} \] Step 3: Calculate \( P(A \cup C) \) Since \( A \) and \( C \) are mutually exclusive: \[ P(A \cup C) = P(A) + P(C) = \frac{4}{13} + \frac{3}{4} \times \frac{4}{13} = \frac{4}{13} + \frac{3}{13} = \frac{7}{13} \] Thus, the correct answer is \( \frac{7}{13} \).