Answer: (b)
P(X/Y) = \(\frac{1}{2}\) P (Y/X) = \(\frac{1}{3}\) P(X ∩ Y) = \(\frac{1}{6}\)
\(\therefore P(\frac{X}{Y}) =\frac{P(X\,\cap\,Y)}{P(Y)}\)
\(\implies\frac{1}{2} =\frac{1/6}{P(Y)}\implies P(Y)=\frac{1}{3}\) ... (i)
Now \(P(\frac{Y}{X}) =\frac{P(Y\,\cap\,X)}{P(X)} =\frac{P(X\,\cap\,Y)}{P(X)}\)
\(\implies \frac{1}{3}=\frac{1}{6}\times\frac{1}{P(X)}\implies P(X)= \frac{1}{2}\) ... (ii)
\(\therefore P(X\,\cup\,Y)= P(X)+P(Y) -P(X\,\cap\,Y) =\frac{1}{2}+\frac{1}{3}-\frac{1}{6} = \frac{2}{3}\) ... (iii)
\({P(X\,\cap\,Y)}=\frac{1}{6}\) and P(X). P(Y) = \(\frac{1}{2}.\frac{1}{3}=\frac{1}{6}\)
⇒\({P(X\,\cap\,Y)}\) = P(X).P(Y)
⇒X and Y are independent events
Now \({P(X^c\,\cap\,Y)}\) =P(Y) - \({P(X\,\cap\,Y)}\) = \(\frac{1}{3}-\frac{1}{6}=\frac{1}{6}\)
