Question

A, B and C are three mutually exclusive and exhaustive events associated with a random experiment. If \(P(B) = 3/2 \, P(A)\) and \(P(C) = 1/2 \, P(B)\), Then \(P(A)\) will be

• A. 3/4
• B. 4/13
• C. 1/13
• D. 1/3

Hint:

Whenever events are declared "mutually exclusive and exhaustive", immediately write their sum as equal to 1. Expressing all variables in terms of a single common term makes algebraic simplification straightforward.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Mutually exclusive events cannot occur at the same time, meaning their intersection is empty:
\[ P(A \cap B) = P(B \cap C) = P(A \cap C) = 0 \] Exhaustive events cover the entire sample space, which means the sum of their individual probabilities must equal exactly 1.
Key Formula or Approach:
For three mutually exclusive and exhaustive events \(A\), \(B\), and \(C\):
\[ P(A) + P(B) + P(C) = 1 \]

Step 2: Detailed Explanation:
Let us express all probabilities in terms of \(P(A)\) using the relationships given in the problem statement:
1. We are given:
\[ P(B) = \frac{3}{2} P(A) \] 2. We are also given:
\[ P(C) = \frac{1}{2} P(B) \] Substitute the expression for \(P(B)\) into this equation:
\[ P(C) = \frac{1}{2} \left( \frac{3}{2} P(A) \right) = \frac{3}{4} P(A) \] 3. Substitute these expressions into the sum-of-probabilities formula:
\[ P(A) + \frac{3}{2} P(A) + \frac{3}{4} P(A) = 1 \] 4. To solve the equation, find a common denominator (which is 4):
\[ P(A) \left( 1 + \frac{6}{4} + \frac{3}{4} \right) = 1 \] \[ P(A) \left( \frac{4}{4} + \frac{6}{4} + \frac{3}{4} \right) = 1 \] \[ P(A) \left( \frac{13}{4} \right) = 1 \] 5. Solve for \(P(A)\):
\[ P(A) = \frac{4}{13} \] Thus, the probability of event \(A\) is \(\frac{4}{13}\).

Step 3: Final Answer:
The correct option is (B).