If A, B and C are three mutually exclusive and exhaustive events of an experiment such that
3P(A) = 2P(B) = P(C), then find the value of P(A).

Answer

To solve the problem, we need to find the value of \( P(A) \) given the relationships between the probabilities of three mutually exclusive and exhaustive events \( A \), \( B \), and \( C \).

### Step-by-step Solution:

1. **Understanding the Relationships**:
We are given that:
\[ 3P(A) = 2P(B) = P(C) \]
Let’s denote \( P(A) = x \), \( P(B) = y \), and \( P(C) = z \). From the given relationships, we can express \( y \) and \( z \) in terms of \( x \):
\[ P(B) = \frac{3}{2}P(A) = \frac{3}{2}x \]
\[ P(C) = 3P(A) = 3x \]

2. **Using the Property of Exhaustive Events**:
Since \( A \), \( B \), and \( C \) are mutually exclusive and exhaustive, their probabilities must sum to 1:
\[ P(A) + P(B) + P(C) = 1 \]
Substituting the expressions for \( P(B) \) and \( P(C) \):
\[ x + \frac{3}{2}x + 3x = 1 \]

3. **Combining the Terms**:
To combine the terms, we need a common denominator. The common denominator for \( 1 \), \( \frac{3}{2} \), and \( 3 \) is \( 2 \):
\[ x + \frac{3}{2}x + \frac{6}{2}x = 1 \]
This simplifies to:
\[ \frac{2}{2}x + \frac{3}{2}x + \frac{6}{2}x = 1 \]
\[ \frac{2 + 3 + 6}{2}x = 1 \]
\[ \frac{11}{2}x = 1 \]

4. **Solving for \( x \)**:
To find \( x \), we multiply both sides by \( \frac{2}{11} \):
\[ x = \frac{2}{11} \]

5. **Conclusion**:
Thus, the value of \( P(A) \) is:
\[ P(A) = \frac{2}{11} \]

### Final Answer:
\[ P(A) = \frac{2}{11} \]