If $P(A) = \frac{2}{5}$, $P(B) = \frac{3}{10}$ and $P(A \cap B) = \frac{1}{5}$, then $P(A' \mid B') \cdot P(B' \mid A')$ is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $\frac{25}{42}$ ##
Here, $P(A) = \frac{2}{5}$, $P(B) = \frac{3}{10}$ and $P(A \cap B) = \frac{1}{5}$
$P(A' \mid B') = \frac{P(A' \cap B')}{P(B')} = \frac{P((A \cup B)')}{P(B')} = \frac{1 - P(A \cup B)}{1 - P(B)}$
$= \frac{1 - [P(A) + P(B) - P(A \cap B)]}{1 - P(B)}$
$= \frac{1 - \left( \frac{2}{5} + \frac{3}{10} - \frac{1}{5} \right)}{1 - \frac{3}{10}}$
$= \frac{1 - \left( \frac{4 + 3 - 2}{10} \right)}{\frac{7}{10}} = \frac{1 - \frac{1}{2}}{\frac{7}{10}} = \frac{5}{7}$
and $P(B' \mid A') = \frac{P(B' \cap A')}{P(A')} = \frac{1 - P(A \cup B)}{1 - P(A)} = \frac{1 - [P(A) + P(B) - P(A \cap B)]}{1 - P(A)}$
$= \frac{1 - \frac{1}{2}}{1 - \frac{2}{5}} = \frac{1/2}{3/5} = \frac{5}{6} \quad \left[ ∵P(A) + P(B) - P(A \cup B) = \frac{1}{2} \right]$
$∴P(A' \mid B') \cdot P(B' \mid A') = \frac{5}{7} \times \frac{5}{6} = \frac{25}{42}$