Let $X$ and $Y$ be two events such that $P(X \cup Y) = P(X) \cap P(Y)$. Statement 1: $P\left(X \cap Y^{\prime}\right) = P\left(X^{\prime} \cap Y\right)$. Statement 2: $P(X) P(Y) = 2 P(X \cap Y)$.

Question

Aicharya · Accepted Answer

Let $X$ and $Y$ be two events such that $ P(X \cup Y)=P(X \cap Y) $ We know $ $\begin{aligned} & P(X \cup Y)=P(X)+P(Y)-P(X \cap Y) \ & P(X \cap Y)=P(X)+P(Y)-P(X \cap Y) \ & 	ext{(from }(1) \ & \Rightarrow P(X)+P(Y)=2 P(X \cap Y) \end{aligned}$ $ Hence, Statement $-2$ is true. Now, $P\left(X \cap Y^{\prime}ight)=P(X)-P(X \cap Y)$ and $P\left(X^{\prime} \cap Yight)=P(Y)-P(X \cap Y)$ This implies statement $-1$ is also true

Let $X$ and $Y$ be two events such that $P(X \cup Y) = P(X) \cap P(Y)$. Statement 1: $P\left(X \cap…

Solution