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

Solution:

We have,

  $P(Y/X)=\frac{2}{5}\Rightarrow \frac{P(X\cap Y)}{P\left(X\right)}=\frac{2}{5}\Rightarrow P(X\cap Y)=\frac{2}{5}P\left(X\right)=\frac{2}{5}\times \frac{1}{3}=\frac{2}{15}$

It is given that

$P(X/Y)=\frac{1}{2}\Rightarrow \frac{P(X\cap Y)}{P\left(Y\right)}=\frac{1}{2}\Rightarrow P\left(Y\right)=2P(X\cap Y)=2\times \frac{2}{15}=\frac{4}{15}$

So, option $\left(d\right)$ is correct. 

Thus, we have

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

$\begin{array}{r}\therefore P(X\cup Y)=P\left(X\right)+P\left(Y\right)-P(X\cap Y)\\ =\frac{1}{3}+\frac{4}{15}-\frac{2}{15}=\frac{7}{15}\end{array}$

So,