if A ,B and C are three mutually exclusive and exhaustive events . Find P(A) , if ` P (B ) = 3/2 P(A ) and P(C ) = 1/2 P(B) `

Options

A. `8/13`
B. `5/13`
C. `4/13`
D. `9/13`

Answer

To solve the problem, we need to find the probability of event A (P(A)) given that events A, B, and C are mutually exclusive and exhaustive. We also know the relationships between the probabilities of these events.

### Step-by-Step Solution:

1. **Define the Probability of A**:
Let's assume the probability of event A is \( P(A) = x \).

2. **Express the Probability of B**:
According to the problem, \( P(B) = \frac{3}{2} P(A) \). Substituting \( P(A) \):
\[ P(B) = \frac{3}{2} x \]

3. **Express the Probability of C**:
The problem states that \( P(C) = \frac{1}{2} P(B) \). Substituting \( P(B) \):
\[ P(C) = \frac{1}{2} \left(\frac{3}{2} x\right) = \frac{3}{4} x \]

4. **Set Up the Equation for Exhaustive Events**:
Since A, B, and C are mutually exclusive and exhaustive, the sum of their probabilities must equal 1:
\[ P(A) + P(B) + P(C) = 1 \]
Substituting the expressions we derived:
\[ x + \frac{3}{2} x + \frac{3}{4} x = 1 \]

5. **Combine the Terms**:
To combine the terms, we need a common denominator. The least common multiple of 1, 2, and 4 is 4. Thus, we rewrite each term:
\[ x = \frac{4}{4} x, \quad \frac{3}{2} x = \frac{6}{4} x, \quad \frac{3}{4} x = \frac{3}{4} x \]
Now, substituting back:
\[ \frac{4}{4} x + \frac{6}{4} x + \frac{3}{4} x = 1 \]
Combining the fractions:
\[ \frac{4 + 6 + 3}{4} x = 1 \]
\[ \frac{13}{4} x = 1 \]

6. **Solve for x**:
To find \( x \), multiply both sides by \( \frac{4}{13} \):
\[ x = \frac{4}{13} \]

7. **Conclusion**:
Therefore, the probability of event A is:
\[ P(A) = \frac{4}{13} \]