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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?

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Solution

## 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 = $\mathrm{P}\left(\frac{E}{A}\right)=\frac{\mathrm{P}\left(E\right)\mathrm{P}\left(\frac{\mathit{A}}{\mathit{E}}\right)}{\mathrm{P}\left(E\right)\mathrm{P}\left(\frac{A}{E}\right)+\mathrm{P}\left(\overline{)\mathit{E}}\right)\mathrm{P}\left(\frac{A}{\overline{)\mathit{E}}}\right)}=\frac{\frac{1}{6}×\frac{4}{5}}{\frac{1}{6}×\frac{4}{5}+\frac{5}{6}×\frac{1}{5}}=\frac{4}{9}$

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