A man is known to speak truth 4 out of 5 times. He throws a die and reports that five appears. Then the probability that actual five appears on the dice is

\(\frac{4}{9}\)

Given, $P(T) = \frac{4}{5}$

required probability = $\frac{P(T) = \frac{1}{6}}{P(T) = \frac{1}{6}+P(F) = \frac{5}{6}}$

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