Question

A, B, and C are three mutually exclusive and exhaustive events associated with a random experiment. Also, \(P(B) = \frac{3{2} P(A)\) and \(P(C) = \frac{1}{2} P(B)\), then \(P(A)\) is:}

• A. \(4/13\)
• B. \(7/13\)
• C. \(2/13\)
• D. \(5/13\)

Hint:

Exam Tip:
For mutually exclusive and exhaustive events: \[ P(A) + P(B) + P(C) = 1 \] Use the given relationships to express all probabilities in terms of one variable.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
A, B, and C are mutually exclusive and exhaustive.
This means: \[ P(A) + P(B) + P(C) = 1 \]

Step 2: Key Formula or Approach:
Given: \[ P(B) = \frac{3}{2} P(A), \quad P(C) = \frac{1}{2} P(B) \]

Step 3: Detailed Explanation:
Substitute \(P(B)\) in terms of \(P(A)\): \[ P(B) = \frac{3}{2} P(A) \] \[ P(C) = \frac{1}{2} P(B) = \frac{1}{2} \cdot \frac{3}{2} P(A) = \frac{3}{4} P(A) \] Now, \[ P(A) + \frac{3}{2} P(A) + \frac{3}{4} P(A) = 1 \] \[ P(A) \left( 1 + \frac{3}{2} + \frac{3}{4} \right) = 1 \] \[ P(A) \left( \frac{4}{4} + \frac{6}{4} + \frac{3}{4} \right) = 1 \Rightarrow P(A) \cdot \frac{13}{4} = 1 \] \[ P(A) = \frac{4}{13} \]

Step 4: Final Answer:
Therefore, option (A) is correct.