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) = \[\frac{1}{6}\] \[P\left( \overline{ E } \right) = 1 - \frac{1}{6} = \frac{5}{6}\]

Also, 

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

\[P\left( \frac{A}{\overline{ {(E)}} \right)\]  = 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 =  \[P\left( \frac{E}{A} \right) = \frac{P\left( E \right)P\left( \frac{A}{E} \right)}{P\left( E \right)P\left( \frac{A}{E} \right) + P\left(\overline{ (E) } \right)P\left( \frac{A}{\overline{(E)}} \right)} = \frac{\frac{1}{6} \times \frac{4}{5}}{\frac{1}{6} \times \frac{4}{5} + \frac{5}{6} \times \frac{1}{5}} = \frac{4}{9}\]