Question

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

Answer 1

Probability of that man speaking truth
$p(T) = \frac{3}{4}$
Probability of that man speaking lies
$p(F) = \frac{1}{4}$

Probability of occurring 4 on dice p(4)= 1/4,

Probability of not occurring 4 on dice p(N4) = 5/4

He throw a die,and reports that it is a four;

1) 4 comes on dice,man speak truth

$p(1) =p(4) p(T)=\frac{1}{6} \times \frac{3}{4} = \frac{3}{24}$

2) 4 comes on dice,man speak lie

3) 4 does not come,man lies that 4 comes

$p(3)= p(N4) p(F)=\frac{5}{6} \times \frac{1}{4} = \frac{5}{24}$

4) 4 does not come,man speak truth

So,there are cases 1 and 3 in favour

So,Probability that four actually comes, apply Bay's theorem
$= \frac{p(1)}{p(1) + p(3)} \\ \\ = \frac{ \frac{3}{24} }{ \frac{3}{24} + \frac{5}{24}} \\ \\ = \frac{3}{8} \\ \\ = \frac{3}{8}$

Thus,Probability that four actually comes is 3/8.

Hope it helps you