Let X and Y be events such that P(X cup Y) = | vedclass

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(D) Given $P(X \cup Y) = P(X \cap Y)$.We know that $P(X \cup Y) = P(X) + P(Y) - P(X \cap Y)$.Substituting the given condition: $P(X \cap Y) = P(X) + P

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Statement-$1$ is false,Statement-$2$ is true.

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(D) Given $P(X \cup Y) = P(X \cap Y)$.We know that $P(X \cup Y) = P(X) + P(Y) - P(X \cap Y)$.Substituting the given condition: $P(X \cap Y) = P(X) + P(Y) - P(X \cap Y)$,which implies $P(X) + P(Y) = 2P(X \cap Y)$. Thus,Statement-$2$ is true.Also,$P(X \cup Y) = P(X \cap Y)$ implies that the events $X$ and $Y$ must be identical $(X = Y)$,because $P(X \cup Y) \ge P(X)$ and $P(X \cup Y) \ge P(Y)$,while $P(X \cap Y) \le P(X)$ and $P(X \cap Y) \le P(Y)$.If $X = Y$,then $P(X \cup Y) = P(X)$ and $P(X \cap Y) = P(X)$.So $P(X) = 2P(X)$,which implies $P(X) = 0$. Consequently,$P(Y) = 0$ and $P(X \cap Y) = 0$.Now,$P(X' \cap Y') = P((X \cup Y)') = 1 - P(X \cup Y) = 1 - 0 = 1$.Since $P(X' \cap Y') = 1 
eq 0$,Statement-$1$ is false.

Let $X$ and $Y$ be events such that $P(X \cup Y) = P(X \cap Y).$
Statement-$1$: $P(X \cap Y) = P(X' \cap Y') = 0$
Statement-$2$: $P(X) + P(Y) = 2P(X \cap Y).$

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Let $X$ and $Y$ be events such that $P(X \cup Y) = P(X \cap Y).$Statement-$1$: $P(X \cap Y) = P(X' \cap Y') = 0$Statement-$2$: $P(X) + P(Y) = 2P(X \cap Y).$