If $A$ and $B$ are two events such that $P(A) =\frac{1}{2}, P(B) =\frac{1}{3}, P(B|A) =\frac{1}{4}$, then $P(A|B)$ is:

$\frac{3}{8}$

The correct answer is Option (3) → $\frac{3}{8}$

Given:

$P(A) = \frac{1}{2}$,    $P(B) = \frac{1}{3}$,    $P(B|A) = \frac{1}{4}$

Use the identity:

$P(B|A) = \frac{P(A \cap B)}{P(A)} \Rightarrow \frac{1}{4} = \frac{P(A \cap B)}{1/2}$

$\Rightarrow P(A \cap B) = \frac{1}{4} \cdot \frac{1}{2} = \frac{1}{8}$

Now, use:

$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{1/8}{1/3} = \frac{3}{8}$