Let X and Y be two events such that P(X/Y) = \(\frac{1}{2}\) , P(Y/X)

1.	Let X and Y be two events such that P(X/Y) = \(\frac{1}{2}\) , P(Y/X) =\(\frac{1}{3}\) and P(X∩Y) =\(\frac{1}{6}\)Which of the following is/are correct? (a) P(X \(\cup\) Y) = 2/3 (b) X and Y are independent (c) X and Y are not independent (d) P(XC∩Y) = \(\frac{1}{3}\)
Answer» 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) - \({P(X\,\cap\,Y)}\) = \(\frac{1}{3}-\frac{1}{6}=\frac{1}{6}\)

Let X and Y be two events such that P(X/Y) = \(\frac{1}{2}\) , P(Y/X) =\(\frac{1}{3}\) and P(X∩Y) =\(\frac{1}{6}\)Which of the following is/are correct? (a) P(X \(\cup\) Y) = 2/3 (b) X and Y are independent (c) X and Y are not independent (d) P(XC∩Y) = \(\frac{1}{3}\)

Answer»

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) - \({P(X\,\cap\,Y)}\) = \(\frac{1}{3}-\frac{1}{6}=\frac{1}{6}\)