Question

Mark the correct alternative in the following question:

$$\begin{gathered} \mathrm{Let}A\mathrm{and}B\mathrm{are}\mathrm{two}\mathrm{events}\mathrm{such}\mathrm{that}\mathrm{P}\left(A\right)=\frac{3}{8},\mathrm{P}\left(B\right)=\frac{5}{8}\mathrm{and}\mathrm{P}\left(A\cup B\right)=\frac{3}{4}.\mathrm{Then}\mathrm{P}\left(A|B\right)\times \mathrm{P}\left(\overline{A}\cap B\right)\mathrm{is}\mathrm{equals}\mathrm{to} \\ \left(\mathrm{a}\right)\frac{2}{5}\left(\mathrm{b}\right)\frac{3}{8}\left(\mathrm{c}\right)\frac{3}{20}\left(\mathrm{d}\right)\frac{6}{25} \end{gathered}$$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \mathrm{P}\left(A\right)=\frac{3}{8},\mathrm{P}\left(B\right)=\frac{5}{8}\mathrm{and}\mathrm{P}\left(A\cup B\right)=\frac{3}{4} \\ \mathrm{As},\mathrm{P}\left(A\cup B\right)=\frac{3}{4} \\ \Rightarrow \mathrm{P}\left(A\right)+\mathrm{P}\left(B\right)-\mathrm{P}\left(A\cap B\right)=\frac{3}{4} \\ \Rightarrow \frac{3}{8}+\frac{5}{8}-\mathrm{P}\left(A\cap B\right)=\frac{3}{4} \\ \Rightarrow \frac{8}{8}-\mathrm{P}\left(A\cap B\right)=\frac{3}{4} \\ \Rightarrow 1-\mathrm{P}\left(A\cap B\right)=\frac{3}{4} \\ \Rightarrow \mathrm{P}\left(A\cap B\right)=1-\frac{3}{4} \\ \Rightarrow \mathrm{P}\left(A\cap B\right)=\frac{1}{4} \\ \mathrm{Also}, \\ \mathrm{P}\left(\overline{A}\cap B\right)=\mathrm{P}\left(B\right)-\mathrm{P}\left(A\cap B\right) \\ =\frac{5}{8}-\frac{1}{4} \\ =\frac{5-2}{8} \\ =\frac{3}{8} \\ \mathrm{Now}, \\ \mathrm{P}\left(A|B\right)\times \mathrm{P}\left(\overline{A}|B\right)=\frac{\mathrm{P}\left(A\cap B\right)}{\mathrm{P}\left(B\right)}\times \frac{\mathrm{P}\left(\overline{A}\cap B\right)}{\mathrm{P}\left(B\right)} \\ =\frac{\left({\displaystyle \frac{1}{4}}\right)}{\left({\displaystyle \frac{5}{8}}\right)}\times \frac{\left({\displaystyle \frac{3}{8}}\right)}{\left({\displaystyle \frac{5}{8}}\right)} \\ =\frac{8}{4\times 5}\times \frac{3\times 8}{5\times 8} \\ =\frac{2}{5}\times \frac{3}{5} \\ =\frac{6}{25} \end{gathered}$$

Hence, the correct alternative is option (d).