Question

A man is known to speak truth 3 out of 4 times. He throws a die and reports that it is a six. Find the probability that it is actually a six.

Answer 1

Let the events E be the event that the man speaks truth and event \(\bar E\) be the event that the man speaks lie. 

And the event A be the event that die is shown a six. 

Therefore, the probability that man speaks truth is P(E) = \(\frac{3}{4}\) (Given) 

And the probability that man speak lie is P(\(\bar E\)) = 1 - P(E) = 1 - \(\frac{3}{4} = \frac{1}{4}\).

Now, the probability that die is shown a six when given that man is speak truth about appearing 6 is same as the probability of getting 6 is P(A|E) = \(\frac{1}{6}\)

The probability that die is shown a six given that the man is speak lie about appearing 6 is same as the probability of not getting a six is P(A|\(\bar E\)) = \(\frac{5}{6}\).

The probability that die is actually showing a six is same as the probability that the man is speak truth given that the die is showing a six is P(E|A).

Now, by Bayes’ theorem, we know that P(E|A) = \(\frac{p(A|E)p(E)}{p(A|E)p(E) + p(A|\bar E)p(\bar E)}\)

Hence, the probability that die is actually showing a six is \(\frac{3}{8}\).