If A,B,C are three mutually exclusive and exhaustive events of an experiment such that `3P(A)=2P(B)=P(C)`, then 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 of event A, given that A, B, and C are mutually exclusive and exhaustive events, and that \(3P(A) = 2P(B) = P(C)\).

### Step-by-Step Solution:

1. **Understanding the Relationships**:
We know that:
- \(P(A) + P(B) + P(C) = 1\) (since A, B, and C are exhaustive).
- \(3P(A) = 2P(B) = P(C)\) (given relationship).

2. **Let’s Introduce a Variable**:
Let \(P(A) = x\). Then, according to the given relationships:
- \(P(B) = \frac{3}{2}x\) (from \(3P(A) = 2P(B)\))
- \(P(C) = 3x\) (from \(3P(A) = P(C)\))

3. **Substituting into the Exhaustive Condition**:
Now we substitute \(P(A)\), \(P(B)\), and \(P(C)\) into the equation \(P(A) + P(B) + P(C) = 1\):
\[ x + \frac{3}{2}x + 3x = 1 \]

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

5. **Solving for x**:
Now, we can solve for \(x\):
\[ x = \frac{2}{11} \]

6. **Finding \(P(A)\)**:
Since \(P(A) = x\), we have:
\[ P(A) = \frac{2}{11} \]

### Final Answer:
Thus, \(P(A) = \frac{2}{11}\).