If A and B be two events such that P(A) = 0.6, P(B) = 0.2 and P(A/B) = 0.5, then P( A'/ B') is equal to

Given $\mathrm{P}(\mathrm{A}/\mathrm{B})=0.5\Rightarrow \frac{\mathrm{P}(\mathrm{A}\cap \mathrm{B})}{\mathrm{P}\left(\mathrm{B}\right)}=0.5$

$\begin{array}{l}\Rightarrow \mathrm{P}(\mathrm{A}\cap \mathrm{B})=\left(0.5\right)\times \mathrm{P}\left(\mathrm{B}\right)=0.5\times 0.2=0.1\\ \Rightarrow \mathrm{P}(\mathrm{A}\cup \mathrm{B})=\mathrm{P}\left(\mathrm{A}\right)+\mathrm{P}\left(\mathrm{B}\right)-\mathrm{P}(\mathrm{A}\cap \mathrm{B})\\ =0.6+0.2-0.1=0.7\end{array}$

Hence $\mathrm{P}\left({\mathrm{A}}^{\mathrm{\prime }}/{\mathrm{B}}^{\mathrm{\prime }}\right)=\frac{\mathrm{P}\left({\mathrm{A}}^{\mathrm{\prime }}\cap {\mathrm{B}}^{\mathrm{\prime }}\right)}{\mathrm{P}\left({\mathrm{B}}^{\mathrm{\prime }}\right)}=\frac{\mathrm{P}\left((\mathrm{A}\cup \mathrm{B}{)}^{\mathrm{\prime }}\right)}{1-\mathrm{P}\left(\mathrm{B}\right)}$

$=\frac{1-\mathrm{P}(\mathrm{A}\cup \mathrm{B})}{1-\mathrm{P}\left(\mathrm{B}\right)}=\frac{1-0.7}{1-0.2}=\frac{3}{8}$