Question

Choose the correct alternative in the following question:

$$\begin{gathered} \mathrm{If}\mathrm{P}\left(A\right)=\frac{2}{5},\mathrm{P}\left(B\right)=\frac{3}{10}\mathrm{and}\mathrm{P}\left(A\cap B\right)=\frac{1}{5},\mathrm{then},\mathrm{P}\left(\overline{A}|\overline{B}\right)\mathrm{P}\left(\overline{B}|\overline{A}\right)\mathrm{is}\mathrm{equal}\mathrm{to} \\ \left(\mathrm{a}\right)\frac{5}{6}\left(\mathrm{b}\right)\frac{5}{7}\left(\mathrm{c}\right)\frac{25}{42}\left(\mathrm{d}\right)1 \end{gathered}$$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \mathrm{P}\left(A\right)=\frac{2}{5},\mathrm{P}\left(B\right)=\frac{3}{10}\mathrm{and}\mathrm{P}\left(A\cap B\right)=\frac{1}{5} \\ \mathrm{Also},\mathrm{P}\left(\overline{A}\right)=1-\mathrm{P}\left(A\right)=1-\frac{2}{5}=\frac{5-2}{5}=\frac{3}{5}\mathrm{and} \\ \mathrm{P}\left(\overline{B}\right)=1-\mathrm{P}\left(B\right)=1-\frac{3}{10}=\frac{10-3}{10}=\frac{7}{10} \\ \mathrm{As},\mathrm{P}\left(A\cup B\right)=\mathrm{P}\left(A\right)+\mathrm{P}\left(B\right)-\mathrm{P}\left(A\cap B\right) \\ =\frac{2}{5}+\frac{3}{10}-\frac{1}{5} \\ =\frac{4+3-2}{10} \\ =\frac{5}{10} \\ =\frac{1}{2} \\ \mathrm{Also},\mathrm{P}\left(\overline{A}\cap \overline{B}\right)=\mathrm{P}\left(\overline{A\cup B}\right) \\ =1-\mathrm{P}\left(A\cup B\right) \\ =1-\frac{1}{2} \\ =\frac{1}{2} \\ \mathrm{Now}, \\ \mathrm{P}\left(\overline{A}|\overline{B}\right)\times \mathrm{P}\left(\overline{B}|\overline{A}\right)=\frac{\mathrm{P}\left(\overline{A}\cap \overline{B}\right)}{\mathrm{P}\left(\overline{B}\right)}\times \frac{\mathrm{P}\left(\overline{A}\cap \overline{B}\right)}{\mathrm{P}\left(\overline{A}\right)} \\ =\frac{\left({\displaystyle \frac{1}{2}}\right)}{\left({\displaystyle \frac{7}{10}}\right)}\times \frac{\left({\displaystyle \frac{1}{2}}\right)}{\left({\displaystyle \frac{3}{5}}\right)} \\ =\frac{10}{2\times 7}\times \frac{5}{3\times 2} \\ =\frac{5}{7}\times \frac{5}{6} \\ =\frac{25}{42} \end{gathered}$$

Hence, the correct alternative is option (c).