A man is known to speak the truth \(3\) out of \(4\) times. We throws a die and reports that it is a six. The probability that it is actually a six is

\(\frac{3}{8}\)

Given:

Speaks the truth 3 out of 4 times

Throw a die and reports that it is six.

The probability that the man speaks the truth is \(P(A) = \frac{3}{4}\)

The probability that the man lies is \(P(B) = 1 − P(A) = 1 − \frac{3}{4} = \frac{1}{4}\)

Probability getting a six \(= \frac{1}{6}\)

Probability of not getting a six \(= 1 − \frac{1}{6} = \frac{5}{6}\)

\(P(A/B) = \frac{P(B/A)P(A)}{P(B)}\)

\(P(A/B) = \)Probability of A given B is true

\(P(B/A) = \)Probability of B given A is true

Applying Baye's theorem, the required probability will be

\(= \frac{\frac{1}{6}\frac{3}{4}}{\frac{1}{6}\frac{3}{4} + \frac{5}{6}\frac{1}{4}} = \frac{3}{8}\)