Question

A speaks the truth 4 out of 5 times. He throws a die and reports that there was a 6, the
probability that actually there was a 6 is
(A) 2/9
(B) 4/9
(C) 5/9
(D) none of these​

Answer 1

Answer:

yar waise to option number (D ) HOGA

Answer 2

Answer:

I know I am late, but this might help u

The problem arises in the case where the guy didn't get a six, but decides to lie.

So the probability of not getting a six is 5/6

and the probability that he lies is 1/4

Fair enough but we still don't know what is the faulty number that he replied with, it might be any number but the actual one.

So the probability that he replies with 6 is going to be 1/5.

And on solving using Bayes' theorem we get 3/4.

This should be the correct answer for the exact wording of the question as you have provided.