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Question
A is known to speak truth 3 times out of 5 times. He throws a die and reports that it is one. Find the probability that it is actually one.
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Solution
Let A, E1 and E2 denote the events that the man reports the appearance of 1 on throwing a die, 1 occurs and 1 does not occur, respectively.
\[\therefore P\left( E_1 \right) = \frac{1}{6} \]
\[ P\left( E_2 \right) = \frac{5}{6}\]
\[\text{ Now } , \]
\[P\left( A/ E_1 \right) = \frac{3}{5}\]
\[P\left( A/ E_2 \right) = \frac{2}{5}\]
\[\text{ Using Bayes' theorem, we get }\]
\[\text{ Required probability } = P\left( E_1 /A \right) = \frac{P\left( E_1 \right)P\left( A/ E_1 \right)}{P\left( E_1 \right)P\left( A/ E_1 \right) + P\left( E_2 \right)P\left( A/ E_2 \right)}\]
\[ = \frac{\frac{1}{6} \times \frac{3}{5}}{\frac{1}{6} \times \frac{3}{5} + \frac{5}{6} \times \frac{2}{5}}\]
\[ = \frac{3}{3 + 10} = \frac{3}{13}\]
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