Let $A$ and $B$ be two events such that $P(A) = \frac{3}{8}, P(B) = \frac{5}{8}$ and $P(A \cup B) = \frac{3}{4}$. Then $P(A \mid B) \cdot P(A' \mid B)$ is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → $\frac{6}{25}$ ##
Here, $P(A) = \frac{3}{8}, P(B) = \frac{5}{8}$ and $P(A \cup B) = \frac{3}{4}$
$∵P(A \cup B) = P(A) + P(B) - P(A \cap B)$
$\Rightarrow P(A \cap B) = \frac{3}{8} + \frac{5}{8} - \frac{3}{4} = \frac{3 + 5 - 6}{8} = \frac{2}{8} = \frac{1}{4}$
$∵P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{1/4}{5/8} = \frac{8}{20} = \frac{2}{5}$
and $P(A' \mid B) = \frac{P(A' \cap B)}{P(B)} = \frac{P(B) - P(A \cap B)}{P(B)}$
$= \frac{\frac{5}{8} - \frac{1}{4}}{\frac{5}{8}} = \frac{\frac{5 - 2}{8}}{\frac{5}{8}} = \frac{3}{5}$
$∴P(A \mid B) \cdot P(A' \mid B) = \frac{2}{5} \times \frac{3}{5} = \frac{6}{25}$