NIMCET 2017 Question Paper with solutions

Let ${\mathit{E}}_1$ and ${\mathit{E}}_2$ be the events that man speaks truth and lie, respectively. Also, let A be the event that 1 appear on dice.
Now, according to the question, we know that
$\mathit{P}\left({\mathit{E}}_1\right)$ = $\frac{2}{3}$ , $\mathit{P}\left({\mathit{E}}_2\right)$ = $\frac{1}{3}$
$\mathit{P}\left(\frac{\mathit{A}}{{\mathit{E}}_1}\right)$ = $\frac{1}{6}$ , $\mathit{P}\left(\frac{\mathit{A}}{{\mathit{E}}_2}\right)$ = $\frac{5}{6}$
∴ Required probability
$\mathit{P}\left(\frac{{\mathit{E}}_2}{\mathit{A}}\right)$ = $\frac{\mathit{P}\left({\mathit{E}}_1\right)\mathit{P}\left(\frac{\mathit{A}}{{\mathit{E}}_1}\right)}{\mathit{P}\left({\mathit{E}}_1\right)\mathit{P}\left(\frac{\mathit{A}}{{\mathit{E}}_2}\right)+\mathit{P}\left({\mathit{E}}_2\right)\mathit{P}\left(\frac{\mathit{A}}{{\mathit{E}}_2}\right)}$
= $\frac{\frac{2}{3}\times \frac{1}{6}}{\frac{2}{3}\times \frac{1}{6}+\frac{1}{3}\times \frac{5}{6}}$ = $\frac{2}{2+5}$ = $\frac{2}{7}$