If A,B,C are three mutually exclusive and exhaustive events of an experiment such 3P(A) = 2I = P(C ) , the P(A) is equal to .

Options

A. `1/11`
B. `2/11`
C. `5/11`
D. `6/11`

Answer

To solve the problem, we need to find the probability \( P(A) \) 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 this common value as \( p \). Thus, we can express the probabilities of events \( A, B, \) and \( C \) in terms of \( p \):
\[ P(A) = \frac{p}{3}, \quad P(B) = \frac{p}{2}, \quad P(C) = p \]

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

3. **Finding a Common Denominator**:
To solve the equation, we need a common denominator. The least common multiple of 3, 2, and 1 is 6. Rewriting each term with a denominator of 6:
\[ \frac{2p}{6} + \frac{3p}{6} + \frac{6p}{6} = 1 \]

4. **Combining the Terms**:
Now, combine the fractions:
\[ \frac{2p + 3p + 6p}{6} = 1 \]
This simplifies to:
\[ \frac{11p}{6} = 1 \]

5. **Solving for \( p \)**:
To find \( p \), multiply both sides by 6:
\[ 11p = 6 \]
Now, divide by 11:
\[ p = \frac{6}{11} \]

6. **Finding \( P(A) \)**:
Now that we have \( p \), we can find \( P(A) \):
\[ P(A) = \frac{p}{3} = \frac{6/11}{3} = \frac{6}{33} = \frac{2}{11} \]

### Final Answer:
Thus, the probability \( P(A) \) is:
\[ \boxed{\frac{2}{11}} \]