Question

Mark the correct alternative in the following question:

$$\begin{gathered} \mathrm{Let}A\mathrm{and}B\mathrm{be}\mathrm{two}\mathrm{events}\mathrm{such}\mathrm{that}\mathrm{P}\left(A\right)=0.6,\mathrm{P}\left(B\right)=0.2,\mathrm{P}\left(A|B\right)=0.5.\mathrm{Then}\mathrm{P}\left(\overline{A}|\overline{B}\right)\mathrm{equals} \\ \left(\mathrm{a}\right)\frac{1}{10}\left(\mathrm{b}\right)\frac{3}{10}\left(\mathrm{c}\right)\frac{3}{8}\left(\mathrm{d}\right)\frac{6}{7} \end{gathered}$$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \mathrm{P}\left(A\right)=0.6,\mathrm{P}\left(B\right)=0.2,\mathrm{P}\left(A|B\right)=0.5 \\ \mathrm{As},\mathrm{P}\left(A|B\right)=0.5 \\ \Rightarrow \frac{\mathrm{P}\left(A\cap B\right)}{\mathrm{P}\left(B\right)}=0.5 \\ \Rightarrow \mathrm{P}\left(A\cap B\right)=0.5\times \mathrm{P}\left(B\right) \\ \Rightarrow \mathrm{P}\left(A\cap B\right)=0.5\times 0.2 \\ \Rightarrow \mathrm{P}\left(A\cap B\right)=0.1 \\ \mathrm{Also},\mathrm{P}\left(\overline{\mathit{B}}\right)=1-\mathrm{P}\left(B\right)=1-0.2=0.8 \\ \mathrm{As},\mathrm{P}\left(A\cup B\right)=\mathrm{P}\left(A\right)+\mathrm{P}\left(B\right)-\mathrm{P}\left(A\cap B\right)=0.6+0.2-0.1=0.7 \\ \mathrm{Now}, \\ \mathrm{P}\left(\overline{A}|\overline{B}\right)=\frac{\mathrm{P}\left(\overline{A}\cap \overline{B}\right)}{\mathrm{P}\left(\overline{B}\right)} \\ =\frac{\mathrm{P}\left(\overline{A\cup B}\right)}{0.8} \\ =\frac{1-\mathrm{P}\left(A\cup B\right)}{0.8} \\ =\frac{1-0.7}{0.8} \\ =\frac{0.3}{0.8} \\ =\frac{3}{8} \end{gathered}$$

Hence, the correct alternative is option (c).

Disclaimer: The answer given in the book is incorrect. The same has been corrected here.