Question

Let $X$ and $Y$ are two events such that $P\left(X\cup Y=\right)PX\cap \left(Y\cdot \right)$
Statement 1: $P\left(X\cap Y^{\prime }\right)=P{X}^{\prime }\cap \left(Y=0\right)$
Statement 2: $P\left(X\right)+PY\left(=2\right)PX\cap Y\left(\right)$

• A. Statement 1 is false, Statement 2 is true
• B. Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation of Statement 1.
• C. Statement 1 is true, Statement 2 is false
• D. Statement 1 is true, Statement 2 is true; Statement 2 is a correct explanation of Statement 1

Answer 1

Answer: (B)

Let $X$ and $y$ be two events such that
$P\left(X\cup Y\right)=P\left(X\cap Y\right)...\left(1\right)$
We know
$P\left(X\cup Y\right)=P\left(X\right)+P\left(Y\right)-P\left(X\cap Y\right)$
$P\left(X\cap Y\right)=P\left(X\right)+P\left(Y\right)-P\left(X\cap Y\right)$
$\left(from\left(1\right)\right)<br/>\Rightarrow P\left(X\right)+P\left(Y\right)=2P\left(X\cap Y\right)$
Hence, Statement - 2 is true.
Now, $P\left(X\cap Y^{\prime }\right)=P\left(X\right)-P\left(Y\right)-P\left(X\cap Y\right)$
and $P\left(X^{\prime }\cap Y\right)=P\left(Y\right)-P\left(X\cap Y\right)$
This implies statement-1 is also true.