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.

Let E₁, E₂ and A be the events defined as follows:

E₁ = six occurs, E₂ = six does not occur, and A = the man reports that it is a six.

Let E₁, E₂ and A be the events defined as follows:

E₁ = six occurs, E₂ = six does not occur, and A = the man reports that it is a six.

We have, P(E₁) = 1/6, P (E₂) = 5/6 

Now, P(A/E₁) = Probability that the man reports that there is a six on the die given that six has occurred on the die.

And, P(A/E₂) = Probability that the man reports that there is a six on the die given that six has not occurred on the die.

= Probability that the man does not speak truth = 1 – 3/4 = 1/4

We have to find P(E1/A) i.e., the probability that there is six on the die given that the man has reported that there is six.