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
Answer & explanation
Correct answer: option 3
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 $\frac{3}{5}$.