Often it is taken that a truthful person commands, more respect in the society. A man is known to speak the truth 4 out of 5 times. He throws a die and reports that it is a six. Find the probability that it is actually a six.

$\frac{4}{9}$

The correct answer is Option (1) → $\frac{4}{9}$ ##

Let $H_1$ be the event that 6 appears on throwing a die, $H_2$ be the event that 6 does not appear on throwing a die. Let $E$ be the event that he reports it is six.

$P(H_1) = \frac{1}{6}, P(H_2) = 1 - \frac{1}{6} = \frac{5}{6}$

$P(E/H_1) = \frac{4}{5}, P(E/H_2) = \frac{1}{5}$

$P(H_1/E) = \frac{P(H_1) \cdot P(E/H_1)}{P(H_1) \cdot P(E/H_1) + P(H_2) \cdot P(E/H_2)}$

$= \frac{\frac{1}{6} \cdot \frac{4}{5}}{\frac{1}{6} \cdot \frac{4}{5} + \frac{5}{6} \cdot \frac{1}{5}} = \frac{4}{9}$