Question

Choose the correct alternative in the following question:

$$\begin{gathered} \mathrm{If}\mathrm{P}\left(A\right)=\frac{3}{10},\mathrm{P}\left(B\right)=\frac{2}{5}\mathrm{and}\mathrm{P}\left(A\cup B\right)=\frac{3}{5},\mathrm{then}\mathrm{P}\left(A|B\right)+\mathrm{P}\left(B|A\right)\mathrm{equals} \\ \left(\mathrm{a}\right)\frac{1}{4}\left(\mathrm{b}\right)\frac{7}{12}\left(\mathrm{c}\right)\frac{5}{12}\left(\mathrm{d}\right)\frac{1}{3} \end{gathered}$$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \mathrm{P}\left(A\right)=\frac{3}{10},\mathrm{P}\left(B\right)=\frac{2}{5}\mathrm{and}\mathrm{P}\left(A\cup B\right)=\frac{3}{5} \\ \mathrm{As},\mathrm{P}\left(A\cup B\right)=\frac{3}{5} \\ \Rightarrow \mathrm{P}\left(A\right)+\mathrm{P}\left(B\right)-\mathrm{P}\left(A\cap B\right)=\frac{3}{5} \\ \Rightarrow \frac{3}{10}+\frac{2}{5}-\mathrm{P}\left(A\cap B\right)=\frac{3}{5} \\ \Rightarrow \frac{3+4}{10}-\mathrm{P}\left(A\cap B\right)=\frac{3}{5} \\ \Rightarrow \mathrm{P}\left(A\cap B\right)=\frac{7}{10}-\frac{3}{5} \\ \Rightarrow \mathrm{P}\left(A\cap B\right)=\frac{7-6}{10} \\ \Rightarrow \mathrm{P}\left(A\cap B\right)=\frac{1}{10} \\ \mathrm{Now}, \\ \mathrm{P}\left(A|B\right)+\mathrm{P}\left(B|A\right)=\frac{\mathrm{P}\left(A\cap B\right)}{\mathrm{P}\left(B\right)}+\frac{\mathrm{P}\left(A\cap B\right)}{\mathrm{P}\left(A\right)} \\ =\frac{\left({\displaystyle \frac{1}{10}}\right)}{\left({\displaystyle \frac{2}{5}}\right)}+\frac{\left({\displaystyle \frac{1}{10}}\right)}{\left({\displaystyle \frac{3}{10}}\right)} \\ =\frac{5}{10\times 2}+\frac{10}{10\times 3} \\ =\frac{1}{4}+\frac{1}{3} \\ =\frac{3+4}{12} \\ =\frac{7}{12} \end{gathered}$$

Hence, the correct alternative is option (b).

Disclaimer: The option (b) given in the book is incorrect as the probability of any event is always less than 1. The same has been corrected here.