If $A$ and $B$ are two events such that $P(A) = \frac{1}{2}$, $P(B) = \frac{1}{3}$ and $P(A \mid B) = \frac{1}{4}$, then $P(A' \cap B')$ is equal to

$\frac{1}{4}$

The correct answer is Option (3) → $\frac{1}{4}$ ##

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

$∵P(A \mid B) = \frac{P(A \cap B)}{P(B)}$

$\Rightarrow P(A \cap B) = P(A \mid B) \cdot P(B) = \frac{1}{4} \times \frac{1}{3} = \frac{1}{12}$

Now, $P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B)$

$= 1 - [P(A) + P(B) - P(A \cap B)]$

$= 1 - \left[ \frac{1}{2} + \frac{1}{3} - \frac{1}{12} \right] = 1 - \left[ \frac{6 + 4 - 1}{12} \right]$

$= 1 - \frac{9}{12} = \frac{3}{12} = \frac{1}{4}$