Question

If \( A, B, C \) are mutually exclusive and exhaustive events of a sample space \( S \) such that \( P(B) = \frac{3}{2}P(A) \) and \( P(C) = \frac{1}{2}P(B) \), then \( P(A) = \) ______. 

• A. \( \frac{4}{13} \)
• B. \( \frac{3}{13} \)
• C. \( \frac{5}{13} \)
• D. \( \frac{1}{3} \)

Hint:

When dealing with mutually exclusive and exhaustive events, use the fact that the sum of their probabilities is equal to 1 to set up an equation and solve for the unknown probability.

Answer 1

The Correct Option is B

Step 1: Understanding the problem.
We are given that \( A, B, C \) are mutually exclusive and exhaustive events of a sample space \( S \). This means:
\[ P(A) + P(B) + P(C) = 1. \]
We are also given the following relationships between the probabilities:
\[ P(B) = \frac{3}{2} P(A) \quad \text{and} \quad P(C) = \frac{1}{2} P(B). \]

Step 2: Expressing probabilities in terms of \( P(A) \).
From the given relationships, we can express \( P(B) \) and \( P(C) \) in terms of \( P(A) \):
\[ P(B) = \frac{3}{2} P(A), \]
\[ P(C) = \frac{1}{2} \times \frac{3}{2} P(A) = \frac{3}{4} P(A). \]

Step 3: Substituting into the sum of probabilities.
Substitute these expressions for \( P(B) \) and \( P(C) \) into the equation \( P(A) + P(B) + P(C) = 1 \):
\[ P(A) + \frac{3}{2} P(A) + \frac{3}{4} P(A) = 1. \]

Step 4: Simplifying the equation.
Combine the terms on the left-hand side:
\[ P(A) \left( 1 + \frac{3}{2} + \frac{3}{4} \right) = 1. \]
Finding the common denominator:
\[ P(A) \left( \frac{4}{4} + \frac{6}{4} + \frac{3}{4} \right) = 1, \] \[ P(A) \times \frac{13}{4} = 1. \]

Step 5: Solving for \( P(A) \).
Solve for \( P(A) \):
\[ P(A) = \frac{4}{13}. \]
Final Answer:
Thus, the correct answer is:
\[ \boxed{\frac{3}{13}}. \]