Three events $A, B$ and $C$ have probabilities $\frac{2}{5}, \frac{1}{3}$ and $\frac{1}{2}$, respectively. Given that $P(A \cap C) = \frac{1}{5}$ and $P(B \cap C) = \frac{1}{4}$, find the values of $P(C | B)$ and $P(A' \cap C')$.

$\frac{3}{4}$ and $\frac{3}{10}$

The correct answer is Option (1) → $\frac{3}{4}$ and $\frac{3}{10}$ ##

Here, $P(A) = \frac{2}{5}, P(B) = \frac{1}{3}, P(C) = \frac{1}{2}, P(A \cap C) = \frac{1}{5}$ and $P(B \cap C) = \frac{1}{4}$

$∴P(C | B) = \frac{P(B \cap C)}{P(B)} = \frac{1/4}{1/3} = \frac{3}{4}$

and $P(A' \cap C') = P(A \cup C)' = 1 - P(A \cup C) = 1 - [P(A) + P(C) - P(A \cap C)]$

$= 1 - \left[ \frac{2}{5} + \frac{1}{2} - \frac{1}{5} \right] = 1 - \left[ \frac{4 + 5 - 2}{10} \right] = 1 - \frac{7}{10} = \frac{3}{10}$