Question

A speaks the truth 8 times out of 10 times. A die is tossed. He reports that it was 5. What is the probability that it was actually 5?

Answer 1

Let A denote the event that man reports that 5 occurs and E the event that 5 actually turns up.

∴ P(E) = $\frac{1}{6}$ and $\mathrm{P}\left(\overline{E}\right)=1-\frac{1}{6}=\frac{5}{6}$

Also, $\mathrm{P}\left(\frac{A}{E}\right)$ = Probability that man reports that 5 occurs given that 5 actually turns up = Probability of man speaking the truth = $\frac{8}{10}=\frac{4}{5}$

$\mathrm{P}\left(\frac{A}{\overline{E}}\right)$ = Probability that man reports that 5 occurs given that 5 doesnot turns up = Probability of man not speaking the truth = $1-\frac{4}{5}=\frac{1}{5}$

∴ Required probability = $\mathrm{P}\left(\frac{E}{A}\right)=\frac{\mathrm{P}\left(E\right)\mathrm{P}\left({\displaystyle \frac{\mathit{A}}{\mathit{E}}}\right)}{\mathrm{P}\left(E\right)\mathrm{P}\left({\displaystyle \frac{A}{E}}\right)+\mathrm{P}\left(\overline{\mathit{E}}\right)\mathrm{P}\left({\displaystyle \frac{A}{\overline{\mathit{E}}}}\right)}=\frac{{\displaystyle \frac{1}{6}}\times {\displaystyle \frac{4}{5}}}{{\displaystyle \frac{1}{6}}\times {\displaystyle \frac{4}{5}}+{\displaystyle \frac{5}{6}}\times {\displaystyle \frac{1}{5}}}=\frac{4}{9}$