Two events $E$ and $F$ are independent. If $P(E) = 0.3$ and $P(E \cup F) = 0.5$, then $P(E \mid F) - P(F \mid E)$ is equal to

$\frac{1}{70}$

The correct answer is Option (3) → $\frac{1}{70}$ ##

Here, $P(E) = 0.3$ and $P(E \cup F) = 0.5$

Let $P(F) = x$

$∵P(E \cup F) = P(E) + P(F) - P(E \cap F)$

$= P(E) + P(F) - P(E) \cdot P(F) \quad [∵P(E \cap F) = P(E) \cdot P(F)]$

$\Rightarrow 0.5 = 0.3 + x - 0.3x$

$\Rightarrow x = \frac{0.5 - 0.3}{0.7} = \frac{2}{7} = P(F)$ ...(i)

$∴P(E \mid F) - P(F \mid E) = \frac{P(E \cap F)}{P(F)} - \frac{P(F \cap E)}{P(E)}$

$= \frac{P(E \cap F) \cdot P(E) - P(F \cap E) \cdot P(F)}{P(E) \cdot P(F)}$

$= \frac{P(E \cap F) [P(E) - P(F)]}{P(E \cap F)} = P(E) - P(F)$

$= \frac{3}{10} - \frac{2}{7} = \frac{21 - 20}{70} = \frac{1}{70} \quad [\text{from Eq. (i)}]$