Question

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

• A. $P(XUY)=\frac{2}{3}$
• 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)

The correct options are
A $P(XUY)=\frac{2}{3}$
B $X$ and $Y$ are independent

$P(X\mid Y)=\frac{P(X\cap Y)}{P\left(Y\right)}=\frac{1}{2}$

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

$\Rightarrow P\left(Y\right)=\frac{1}{3}$

$P(Y\mid X)=\frac{P(Y\cap X)}{P\left(X\right)}=\frac{\frac{1}{6}}{P\left(X\right)}=\frac{1}{3}$

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

$\therefore P(X\cup Y)=P\left(X\right)+P\left(Y\right)-P(X\cap Y)$

$=\frac{1}{2}+\frac{1}{3}-\frac{1}{6}$

$=\frac{2}{3}$

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

$\therefore P(X\cap Y)=P\left(X\right)\cdot P\left(Y\right)\Rightarrow$ X and Y are

independent.

$P(X^c\cap Y)=P\left(Y\right)-P(X\cap Y)$

$=\frac{1}{3}-\frac{1}{6}=\frac{1}{6}$.