Question

If $ A $, $ B $, and $ C $ are mutually exclusive and exhaustive events of a random experiment such that $ P(B) = \frac{3}{2} P(A) $ and $ P(C) = \frac{1}{2} P(B) $, then $ P(A \cup C) $ equals to

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

Hint:

When events are mutually exclusive and exhaustive, you can calculate the total probability of their union by adding the probabilities of each event.

Answer 1

The Correct Option is D

We are given that \( A \), \( B \), and \( C \) are mutually exclusive and exhaustive events. This means: \[ P(A \cup B \cup C) = 1 \] The probability of the union of mutually exclusive events is simply the sum of their individual probabilities: \[ P(A \cup B \cup C) = P(A) + P(B) + P(C) \] We also know: \[ P(B) = \frac{3}{2} P(A) \] \[ P(C) = \frac{1}{2} P(B) \] Substitute \( P(B) \) into the equation for \( P(C) \): \[ P(C) = \frac{1}{2} \times \frac{3}{2} P(A) = \frac{3}{4} P(A) \] Now substitute \( P(B) = \frac{3}{2} P(A) \) and \( P(C) = \frac{3}{4} P(A) \) into the total probability equation: \[ 1 = P(A) + \frac{3}{2} P(A) + \frac{3}{4} P(A) \] Factor out \( P(A) \): \[ 1 = P(A) \left( 1 + \frac{3}{2} + \frac{3}{4} \right) \] Simplify the expression inside the parentheses: \[ 1 = P(A) \left( \frac{4}{4} + \frac{6}{4} + \frac{3}{4} \right) = P(A) \times \frac{13}{4} \] Solve for \( P(A) \): \[ P(A) = \frac{4}{13} \] Now, to find \( P(A \cup C) \), we use the formula for the union of mutually exclusive events: \[ P(A \cup C) = P(A) + P(C) \] Substitute the value of \( P(A) \) and \( P(C) \): \[ P(A \cup C) = \frac{4}{13} + \frac{3}{4} \times \frac{4}{13} \] Simplify: \[ P(A \cup C) = \frac{4}{13} + \frac{12}{52} = \frac{4}{13} + \frac{3}{13} = \frac{7}{13} \] Thus, the probability \( P(A \cup C) = \frac{7}{13} \).