Question

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

(a) 1/2 

(b) 1/3

(c) 3/8

(d) 3/7

Answer 1

Correct option (c) 3/8

Explanation:

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

 E₁ = six occurs 

 E₂ = six does not occurs 

 A = the man reports that it is a six. 

We have,

Now, P (A ⁄ E₁) = Probability that the man reports that there is a six on the die given then 6 has occured on die = Probability the man speaks truth =3/4 

P(A/E₂) = Probability that the man reports that there is six is on die given that six has not occured on 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. By Baye’s rule, we have