$X$ and $Y$ are two events such that $P(X|Y) = 0.2$ and $P(Y) = 0.5$. Find the value of $P(X' \cap Y)$.

$0.4$

The correct answer is Option (3) → $0.4$ ##

Given, $P(X|Y) = 0.2, P(Y) = 0.5$

Find $P(X' \cap Y)$

Writes that $P(X' \cap Y) = P(X' | Y) \times P(Y)$.

Uses the property of conditional probability and simplifies the above equation as:

$P(X' \cap Y) = [1 - P(X|Y)] \times P(Y)$

Substitutes the given values in the above equation to find $P(X' \cap Y)$ as

$P(X' \cap Y) = 0.8 \times 0.5 = 0.4$