Question

Mark the correct alternative in the following question:

$$\begin{gathered} \mathrm{If}\mathrm{P}\left(B\right)=\frac{3}{5},\mathrm{P}\left(A|B\right)=\frac{1}{2}\mathrm{and}\mathrm{P}\left(A\cup B\right)=\frac{4}{5},\mathrm{then}\mathrm{P}\left(B|\overline{A}\right)= \\ \left(\mathrm{a}\right)\frac{1}{5}\left(\mathrm{b}\right)\frac{3}{10}\left(\mathrm{c}\right)\frac{1}{2}\left(\mathrm{d}\right)\frac{3}{5} \end{gathered}$$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \mathrm{P}\left(B\right)=\frac{3}{5},\mathrm{P}\left(A|B\right)=\frac{1}{2}\mathrm{and}\mathrm{P}\left(A\cup B\right)=\frac{4}{5} \\ \mathrm{As},\mathrm{P}\left(A|B\right)=\frac{1}{2} \\ \Rightarrow \frac{\mathrm{P}\left(A\cap B\right)}{\mathrm{P}\left(B\right)}=\frac{1}{2} \\ \Rightarrow \mathrm{P}\left(A\cap B\right)=\frac{1}{2}\times \mathrm{P}\left(B\right) \\ \Rightarrow \mathrm{P}\left(A\cap B\right)=\frac{1}{2}\times \frac{3}{5} \\ \Rightarrow \mathrm{P}\left(A\cap B\right)=\frac{3}{10} \\ \mathrm{And},\mathrm{P}\left(B\cap \overline{A}\right)=\mathrm{P}\left(B\right)-\mathrm{P}\left(A\cap B\right) \\ =\frac{3}{5}-\frac{3}{10} \\ =\frac{3}{10} \\ \mathrm{Also},\mathrm{P}\left(A\cup B\right)=\mathrm{P}\left(A\right)+\mathrm{P}\left(B\right)-\mathrm{P}\left(A\cap B\right) \\ \Rightarrow \frac{4}{5}=\mathrm{P}\left(A\right)+\frac{3}{5}-\frac{3}{10} \\ \Rightarrow \frac{4}{5}=\mathrm{P}\left(A\right)+\frac{3}{10} \\ \Rightarrow \mathrm{P}\left(A\right)=\frac{4}{5}-\frac{3}{10} \\ \Rightarrow \mathrm{P}\left(A\right)=\frac{5}{10} \\ \Rightarrow \mathrm{P}\left(A\right)=\frac{1}{2} \\ \mathrm{So},\mathrm{P}\left(\overline{A}\right)=1-\mathrm{P}\left(A\right)=1-\frac{1}{2}=\frac{1}{2} \\ \mathrm{Now}, \\ \mathrm{P}\left(B|\overline{A}\right)=\frac{\mathrm{P}\left(B\cap \overline{A}\right)}{P\left(\overline{A}\right)} \\ =\frac{\left({\displaystyle \frac{3}{10}}\right)}{\left({\displaystyle \frac{1}{2}}\right)} \\ =\frac{3\times 2}{10} \\ =\frac{3}{5} \end{gathered}$$

Hence, the correct alternative is option (d).