Let E and F be two events such that $P(E) =\frac{1}{3}, P(F) =\frac{1}{4}$ and $P(E∩F) =\frac{1}{5}$. Then the value of $P(F|E)$ is equal to

$\frac{3}{5}$

The correct answer is Option (3) → $\frac{3}{5}$ **

$P(E)=\frac{1}{3},\; P(F)=\frac{1}{4},\; P(E\cap F)=\frac{1}{5}$

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

$=\frac{\frac{1}{5}}{\frac{1}{3}}=\frac{1}{5}\cdot 3=\frac{3}{5}$

The value of $P(F\mid E)$ is $\frac{3}{5}$.