If A and B are two independent events such that $\mathrm{P}\left(\mathrm{A}\right)=1/2\text{ and }\mathrm{P}\left(\mathrm{B}\right)=1/5$, then 

Since A and B are independent events, we have

$\begin{array}{l}\mathrm{P}(\mathrm{A}\cap \mathrm{B})=\mathrm{P}\left(\mathrm{A}\right)\mathrm{P}\left(\mathrm{B}\right)=\frac{1}{2}\times \frac{1}{5}=\frac{1}{10}\\ \mathrm{P}(\mathrm{A}/\mathrm{B})=\mathrm{P}\left(\mathrm{A}\right)=\frac{1}{2}\end{array}$

Now,

$\begin{array}{r}\mathrm{P}(\mathrm{A}\cup \mathrm{B})=\mathrm{P}\left(\mathrm{A}\right)+\mathrm{P}\left(\mathrm{B}\right)-\mathrm{P}(\mathrm{A}\cap \mathrm{B})\\ =\frac{1}{2}+\frac{1}{5}-\frac{1}{10}=\frac{3}{5}\end{array}$

Now,

$\begin{array}{l}\mathrm{P}(\mathrm{A}/\mathrm{A}\cup \mathrm{B})=\frac{\mathrm{P}[\mathrm{A}\cap (\mathrm{A}\cup \mathrm{B}\left)\right]}{\mathrm{P}(\mathrm{A}\cup \mathrm{B})}=\frac{\mathrm{P}\left(\mathrm{A}\right)}{\mathrm{P}(\mathrm{A}\cup \mathrm{B})}=\frac{1/2}{3/5}=\frac{5}{6}\\ \mathrm{P}\left[\frac{\mathrm{A}\cap \mathrm{B}}{\left(\overline{\mathrm{A}\cap \mathrm{B}}\right)}\right]=\mathrm{P}\left(\frac{\mathrm{A}\cap \mathrm{B}}{\left(\overline{\mathrm{A}\cap \mathrm{B}}\right)}\right)=0\end{array}$