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.

Explanation

Solution:
probability of speaking truth 3 out of 4 times `P(T)=frac{3}{4}`;
probability of not speaking truth `P(F)=1-P(T)=1-frac{3}{4}=frac{1}{4}`;
probability of getting `=frac{1}{6}`;
probability pf not getting `6=1-frac{1}{6}=frac{5}{6}`
So,`P(T|F)=frac{P(F|T)*P(T)}{P(F)}`

where,P(T|F)=probabillity of T when F is true
P(F|T)=probability of F when T is true;
apply Baye's theorem `=frac{frac{1}{6}*frac{3}{4}}{frac{1}{6}*frac{3}{4}+frac{5}{6}*frac{1}{4}}= frac{3}{8}`