Let X and Y be two 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,we get $P(X \cap Y) = P(

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Statement $1$ is true,Statement $2$ is true; Statement $2$ is a correct explanation of Statement $1$.

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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,we get $P(X \cap Y) = P(X) + P(Y) - P(X \cap Y)$,which implies $P(X) + P(Y) = 2P(X \cap Y)$. Thus,Statement $2$ is true.Now,$P(X \cap Y') = P(X) - P(X \cap Y)$ and $P(X' \cap Y) = P(Y) - P(X \cap Y)$.Since $P(X \cup Y) = P(X \cap Y)$,it implies $P(X \cup Y) - P(X \cap Y) = 0$,which is $P(X \Delta Y) = 0$,where $X \Delta Y$ is the symmetric difference.This means $P(X \cap Y') = 0$ and $P(X' \cap Y) = 0$. Thus,Statement $1$ is true.Since $P(X \cap Y') = P(X) - P(X \cap Y) = 0$ and $P(X' \cap Y) = P(Y) - P(X \cap Y) = 0$,we have $P(X) = P(X \cap Y)$ and $P(Y) = P(X \cap Y)$,which leads to $P(X) + P(Y) = 2P(X \cap Y)$. Therefore,Statement $2$ is the correct explanation of Statement $1$.

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