Question

If X and Y are two events such that $P\left(X|Y\right)=\frac{1}{2},P\left(Y|X\right)=\frac{1}{3}andP(X\cap Y)=\frac{1}{6}.$ Then, 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)

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

(i) Conditional probability. i.e. $P\left(A|B\right)=\frac{P(A\cap B)}{P\left(B\right)}$
(ii) $P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$
(iii) Independent event, then $P(A\cap B)=P\left(A\right).P\left(B\right)$
Here, $P\left(X|Y\right)=\frac{1}{2},P\left(Y|X\right)=\frac{1}{3}$
and $P(X\cap Y)=\frac{1}{6}$
$\therefore P\left(X|Y\right)=\frac{P(X\cap Y)}{P\left(Y\right)}\Rightarrow \frac{1}{2}=\frac{\frac{1}{6}}{P\left(Y\right)}\Rightarrow P\left(Y\right)=\frac{1}{3}...\left(i\right)$
$P\left(Y|X\right)=\frac{1}{3}\Rightarrow \frac{P(X\cap Y)}{P\left(X\right)}=\frac{1}{3}$
$\Rightarrow \frac{1}{6}=\frac{1}{3}P\left(X\right)$
$\therefore P\left(X\right)=\frac{1}{2}$ ...(ii)
$P(X\cup Y)=P\left(X\right)+P\left(Y\right)-P(X\cap Y)$
$=\frac{1}{2}+\frac{1}{3}-\frac{1}{6}=\frac{2}{3}$ ...(iii)
$P(X\cap Y)=\frac{1}{6}andP\left(X\right).P\left(Y\right)=\frac{1}{2}.\frac{1}{3}=\frac{1}{6}\Rightarrow P(X\cap Y)=P\left(X\right).P\left(Y\right)$
i.e. independent events
$\therefore P(X^C\cap Y)=P\left(Y\right)-P(X\cap Y)=\frac{1}{3}-\frac{1}{6}=\frac{1}{6}$