Question

Let $X$ and $Y$ be two events such that $P\left(X|Y\right)=\frac{1}{2},P\left(Y|X\right)=\frac{1}{3}$ and $P(X\cap Y)=\frac{1}{6}$ . Which of the following is (are) correct?

• A. $P(X\cup Y)=\frac{2}{3}$
• B. $X$ and $Y$ are independent
• C. $X$ and $Y$ are not independent
• D. $P(X^c\cap Y)=\frac{1}{3}$

Answer 1

Answer: (A), (B)

$X$ and $Y$ are independent

$P\left(X\right)\cdot P\left(Y|X\right)=P\left(Y\right)\cdot P\left(X|Y\right)=P(X\cap Y)$

$\Rightarrow P\left(X\right)=\frac{1/6}{1/3}=\frac{1}{2}$

$P\left(Y\right)=\frac{1/6}{1/2}=\frac{1}{3}$

$\therefore P(X\cup Y)=\frac{1}{2}+\frac{1}{3}-\frac{1}{6}=\frac{2}{3}$

$P(X\cap Y)=\frac{1}{6}$

and $P\left(X\right)\cdot P\left(Y\right)=\frac{1}{2}\times \frac{1}{3}=\frac{1}{6}$

$\Rightarrow X\text{and}Y$ are independent events.

$P(X^c\cap Y)=P\left(X^c\right)\cdot P\left(Y\right)=\frac{1}{2}\times \frac{1}{3}=\frac{1}{6}$