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

Options

A. `1/12`
B. `3/4`
C. `1/4`
D. `3/16`

Answer

To solve the problem, we need to find \( P(A \cap B) \) using the given probabilities.

### Step-by-Step Solution:

1. **Write down the given probabilities:**
- \( P(A) = \frac{1}{2} \)
- \( P(B) = \frac{1}{3} \)
- \( P(A|B) = \frac{1}{4} \)

2. **Use the formula for conditional probability:**
The conditional probability \( P(A|B) \) is defined as:
\[ P(A|B) = \frac{P(A \cap B)}{P(B)} \]
Rearranging this formula gives us:
\[ P(A \cap B) = P(A|B) \times P(B) \]

3. **Substitute the known values into the equation:**
We know \( P(A|B) = \frac{1}{4} \) and \( P(B) = \frac{1}{3} \). Therefore:
\[ P(A \cap B) = \frac{1}{4} \times \frac{1}{3} \]

4. **Calculate the product:**
\[ P(A \cap B) = \frac{1 \times 1}{4 \times 3} = \frac{1}{12} \]

5. **Final answer:**
Thus, the probability \( P(A \cap B) \) is:
\[ P(A \cap B) = \frac{1}{12} \]

### Summary:
The probability of the intersection of events A and B, \( P(A \cap B) \), is \( \frac{1}{12} \).