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 Question:
Mutually exclusive and exhaustive events mean their probabilities are independent and their sum is exactly 1.
Step 2: Key Formula or Approach:
For exhaustive events A, B, and C:
\[ P(A) + P(B) + P(C) = 1 \]
Step 3: Detailed Explanation:
Express \( P(B) \) and \( P(C) \) in terms of \( P(A) \):
Given, \( P(B) = \frac{3}{2} P(A) \).
Given, \( P(C) = \frac{1}{2} P(B) = \frac{1}{2} \left( \frac{3}{2} P(A) \right) = \frac{3}{4} P(A) \).
Substitute these into the sum equation:
\[ P(A) + \frac{3}{2} P(A) + \frac{3}{4} P(A) = 1 \]
Find a common denominator (4):
\[ \frac{4 P(A) + 6 P(A) + 3 P(A)}{4} = 1 \]
\[ \frac{13 P(A)}{4} = 1 \]
\[ P(A) = \frac{4}{13} \]
Step 4: Final Answer:
The value of \( P(A) \) is \( 4/13 \).