If A and B are two events such that  $P\left(\overline{A}\right)=0.3,P\left(B\right)=0.4$ and  $P\left(A\cap \overline{B}\right)=0.5$ then  $P(B/A\cup \overline{B)}$ = 

$\text{Given}P\left(\overline{A}\right)=0.3,\Rightarrow P\left(A\right)=0.7\text{and}\mathrm{P}\left(\mathrm{B}\right)=0.4$
$\Rightarrow P\left(\overline{B}\right)=0.6$
$\begin{array}{rl} & \text{Now,}P\left(\frac{B}{A\cup \overline{B}}\right)=\frac{P(B\cap (A\cup \overline{B}\left)\right)}{P(A\cup \overline{B})}\\ & P(B\cap (A\cup \overline{B}\left)\right)=P\left(\right(B\cap A)\cup (B\cap \overline{B}\left)\right)\end{array}$
$\begin{array}{rl} & =P(A\cap B)\\ & =P\left(A\right)-P(A\cap \overline{B})\\ & =0.2\end{array}$
$\begin{array}{rl} & P(A\cup \overline{B})=P\left(A\right)+P\left(\overline{B}\right)-P(A\cap \overline{B})\\ & =0.4+0.6-0.5=0.8\\ & \therefore P\left(\frac{B}{A\cup \overline{B}}\right)=\frac{0.2}{0.8}=\frac{1}{4}=0.25\end{array}$