. A man is known to speak the truth 3 out of 4 times

Best Answer

Explanation:

Let R,S be the event that the man reports that is a six and the throw is a six respectively.  

P(R|S)= the probability that the man reports it is a six when it is a six = probability of speak truth =3/4

P(R|S')= the probability that the man reports it is a six when it is not a six = probability of not speak truth =1/4

P(S)= the probability that throw is six =1/6

P(S')= the probability that throw is not a six =1-1/6=6-1/6=5/6

Now we have to find the probability that the throw is actually six when the man reports it is a six i.e. P(S|R)

By the Bayes theorem,

Final Answer:

Hence the correct option is (A).i.e 3/8