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, E₁ and E₂ denote the events that the man reports the appearance of 5 on throwing a die, 5 appears and 5 does not appear, respectively.

\[\therefore P\left( E_1 \right) = \frac{1}{6}\]

\[ P\left( E_2 \right) = \frac{5}{6}\]

\[\text{ Now } , \]

\[P\left( A/ E_1 \right) = \frac{8}{10}\]

\[P\left( A/ E_2 \right) = \frac{2}{10}\]

\[\text{ Using Bayes' theorem, we get } \]

\[\text{ Required probability }  = P\left( E_1 /A \right) = \frac{P\left( E_1 \right)P\left( A/ E_1 \right)}{P\left( E_1 \right)P\left( A/ E_1 \right) + P\left( E_2 \right)P\left( A/ E_2 \right)}\]

\[ = \frac{\frac{1}{6} \times \frac{8}{10}}{\frac{1}{6} \times \frac{8}{10} + \frac{5}{6} \times \frac{2}{10}}\]

\[ = \frac{8}{8 + 10} = \frac{8}{18} = \frac{4}{9}\]