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Let $X$ and $Y$ are two events such that $P(X \cup Y=) P X \cap(Y . \quad)$
Statement 1: $P\left(X \cap Y^{\prime}=\dot{P} X^{\prime} \cap(Y=0 \quad)\right.$
Statement 2: $P(X) P Y \in 2) P X \cap Y(\quad)$
Options:
Statement 1: $P\left(X \cap Y^{\prime}=\dot{P} X^{\prime} \cap(Y=0 \quad)\right.$
Statement 2: $P(X) P Y \in 2) P X \cap Y(\quad)$
Solution:
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Verified Answer
The correct answer is:
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation of Statement 1.
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation of Statement 1.
Let $X$ and $Y$ be two events such that
$$
P(X \cup Y)=P(X \cap Y)
$$
We know
$$
\begin{aligned}
& P(X \cup Y)=P(X)+P(Y)-P(X \cap Y) \\
& P(X \cap Y)=P(X)+P(Y)-P(X \cap Y) \\
& \text { (from }(1) \\
& \Rightarrow P(X)+P(Y)=2 P(X \cap Y)
\end{aligned}
$$
Hence, Statement $-2$ is true.
Now, $P\left(X \cap Y^{\prime}\right)=P(X)-P(X \cap Y)$
and $P\left(X^{\prime} \cap Y\right)=P(Y)-P(X \cap Y)$
This implies statement-1 is also true.
$$
P(X \cup Y)=P(X \cap Y)
$$
We know
$$
\begin{aligned}
& P(X \cup Y)=P(X)+P(Y)-P(X \cap Y) \\
& P(X \cap Y)=P(X)+P(Y)-P(X \cap Y) \\
& \text { (from }(1) \\
& \Rightarrow P(X)+P(Y)=2 P(X \cap Y)
\end{aligned}
$$
Hence, Statement $-2$ is true.
Now, $P\left(X \cap Y^{\prime}\right)=P(X)-P(X \cap Y)$
and $P\left(X^{\prime} \cap Y\right)=P(Y)-P(X \cap Y)$
This implies statement-1 is also true.
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