Let X and Y be two events such that $P(X)=\frac{1}{3}, P(X/Y)=\frac{1}{2}$ and $P(Y/X)=\frac{2}{5}>$ Then,

(a) $P(X'/Y)=\frac{1}{2}$

(b) $P(X ∩ Y)=\frac{1}{5}$

(c) $P(X ∪ Y)=\frac{2}{5}$

(d) $P(Y)=\frac{4}{15}$

(a), (d)

We have,

$P(Y/X)=\frac{2}{5}⇒ \frac{P(X ∩ Y)}{P(X)}=\frac{2}{5}⇒ P(X ∩ Y) =\frac{2}{5}P(X)=\frac{2}{5}×\frac{1}{3}=\frac{2}{15}$

It is given that

$P(X/Y)=\frac{1}{2}⇒ \frac{P(X ∩ Y)}{P(Y)}=\frac{1}{2}⇒ P(Y)=2P(X ∩ Y) =2×\frac{2}{5}=\frac{4}{15}$

So, option (d) is correct

Thus, we have

$P(X)=\frac{1}{3}, P(Y)=\frac{4}{15}$ and $P(X c=\frac{2}{15}$

$∴ P(X ∪ Y) = P(X) + P(Y) - P(X ∩ Y)$

$=\frac{1}{3}+\frac{4}{15}-\frac{2}{15}=\frac{7}{15}$

Now, $P(X'/Y)= 1- P(X/Y)= 1 - \frac{1}{2}=\frac{1}{2}$

So, option (a) is correct.