# Let X and Y be two events such that P(X/Y) = 1/2 P(Y/X) = 1/3 and P(X∩Y) = 1/6 Which of the following is/are correct?

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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) = $\frac{1}{3}$

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