If $P(A) = \frac{3}{10}$, $P(B) = \frac{2}{5}$ and $P(A \cup B) = \frac{3}{5}$, then $P(B \mid A) + P(A \mid B)$ is equal to

$\frac{7}{12}$

The correct answer is Option (4) → $\frac{7}{12}$ ##

Here, $P(A) = \frac{3}{10}$, $P(B) = \frac{2}{5}$ and $P(A \cup B) = \frac{3}{5}$

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

$= \frac{P(A) + P(B) - P(A \cup B)}{P(A)} + \frac{P(A) + P(B) - P(A \cup B)}{P(B)}$

$\left[ ∵P(A \cup B) = P(A) + P(B) - P(A \cap B) \text{ i.e., } P(A \cap B) = P(A) + P(B) - P(A \cup B) \right]$

$= \frac{\frac{3}{10} + \frac{2}{5} - \frac{3}{5}}{\frac{3}{10}} + \frac{\frac{3}{10} + \frac{2}{5} - \frac{3}{5}}{\frac{2}{5}}$

$= \frac{\frac{1}{10}}{\frac{3}{10}} + \frac{\frac{1}{10}}{\frac{2}{5}} = \frac{1}{3} + \frac{1}{4} = \frac{7}{12}$