If A and B are two events such that `P(A) = 1/4 , P(A//B) = 1/2 and P(B//A)= 2/3 ` then `P(B)` is equal to

Options

A. `1/3`
B. `1/6`
C. `1/2`
D. `2/3`

Answer

To find \( P(B) \) given the probabilities \( P(A) = \frac{1}{4} \), \( P(A|B) = \frac{1}{2} \), and \( P(B|A) = \frac{2}{3} \), we can use the definitions of conditional probability and the formula for the intersection of events.

### Step-by-step Solution:

1. **Use the definition of conditional probability**:
\[ P(A|B) = \frac{P(A \cap B)}{P(B)} \]
From the problem, we know \( P(A|B) = \frac{1}{2} \). Therefore, we can express this as:
\[ \frac{1}{2} = \frac{P(A \cap B)}{P(B)} \quad \text{(1)} \]

2. **Use the definition of conditional probability again**:
\[ P(B|A) = \frac{P(A \cap B)}{P(A)} \]
From the problem, we know \( P(B|A) = \frac{2}{3} \) and \( P(A) = \frac{1}{4} \). Thus, we can express this as:
\[ \frac{2}{3} = \frac{P(A \cap B)}{\frac{1}{4}} \quad \text{(2)} \]

3. **Solve equation (2) for \( P(A \cap B) \)**:
Rearranging equation (2):
\[ P(A \cap B) = \frac{2}{3} \times \frac{1}{4} = \frac{2}{12} = \frac{1}{6} \]

4. **Substitute \( P(A \cap B) \) back into equation (1)**:
Now substitute \( P(A \cap B) = \frac{1}{6} \) into equation (1):
\[ \frac{1}{2} = \frac{\frac{1}{6}}{P(B)} \]

5. **Cross-multiply to solve for \( P(B) \)**:
Cross-multiplying gives:
\[ P(B) \cdot \frac{1}{2} = \frac{1}{6} \]
Therefore,
\[ P(B) = \frac{1}{6} \cdot 2 = \frac{2}{6} = \frac{1}{3} \]

Thus, the probability \( P(B) \) is equal to \( \frac{1}{3} \).

### Final Answer:
\[ P(B) = \frac{1}{3} \]