Question

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

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