$\text{ If two events }A\text{ and }B\text{ are such that }P\left(A^C\right)=0.3,P\left(B\right)=0.4\text{ and }P\left(A\cap B^C\right)=0.5\text{ then }P\left[B/\left(A\cup B^C\right)\right]\text{ is equal to }$

$\begin{array}{l}\mathrm{P}\left({\mathrm{A}}^{\mathrm{C}}\right)=0.3\Rightarrow \mathrm{P}\left(\mathrm{A}\right)=0.7\\ \mathrm{P}\left(\mathrm{B}\right)=0.4\Rightarrow \mathrm{P}\left({\mathrm{B}}^{\mathrm{C}}\right)=0.6\\ \mathrm{P}\left(\mathrm{A}\cap {\mathrm{B}}^{\mathrm{C}}\right)=0.5\Rightarrow \mathrm{P}\left(\mathrm{A}\right)-\mathrm{P}(\mathrm{A}\cap \mathrm{B})=0.5\\ \text{ or, }\mathrm{P}(\mathrm{A}\cap \mathrm{B})=0.7-0.5=0.2\\ \mathrm{P}\left[\mathrm{B}/\left(\mathrm{A}\cap {\mathrm{B}}^{\mathrm{C}}\right)\right]=\frac{\mathrm{P}\left[\mathrm{B}\cap \left(\mathrm{A}\cup {\mathrm{B}}^{\mathrm{C}}\right)\right]}{\mathrm{P}\left(\mathrm{A}\cup {\mathrm{B}}^{\mathrm{C}}\right)}----\left(1\right)\end{array}$

$\text{ Now, }B\cap \left(A\cup B^C\right)=A\cap B$

$\therefore \mathrm{P}\left[\mathrm{B}\cap \left(\mathrm{A}\cup {\mathrm{B}}^{\mathrm{C}}\right)\right]=\mathrm{P}(\mathrm{A}\cap \mathrm{B})=0.2-----\left(2\right)$

$\text{ and }\mathrm{P}\left(\mathrm{A}\cup {\mathrm{B}}^{\mathrm{C}}\right)=\mathrm{P}\left(\mathrm{A}\right)+\mathrm{P}\left({\mathrm{B}}^{\mathrm{C}}\right)-\mathrm{P}\left(\mathrm{A}\cap {\mathrm{B}}^{\mathrm{C}}\right)$

                        $=0.7+0.6-0.5=0.8-----\left(3\right)$

From (1), (2) and (3) 

$\text{ Required probability }=\frac{0.2}{0.8}=\frac{1}{4}$