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
Answer & explanation
Correct answer: option 3
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
\(= \frac{\frac{1}{6}\frac{3}{4}}{\frac{1}{6}\frac{3}{4} + \frac{5}{6}\frac{1}{4}} = \frac{3}{8}\)