Question

A man is known to speak the 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. $\frac{3}{8}$
• B. $\frac{1}{8}$
• C. $\frac{1}{4}$
• D. $\frac{1}{2}$

Answer 1

The correct option is A $\frac{3}{8}$

Probability that man speaks the truth is $\frac{3}{4}$.
The probability that man lies is $\frac{1}{4}$.
Probability of getting a 6 $=\frac{1}{6}$
Probability of not getting a six $=\frac{5}{6}$
Applying Baye's theorem, we get the required probability as
$\frac{\frac{1}{6}.\frac{3}{4}}{\frac{1}{6}.\frac{3}{4}+\frac{5}{6}.\frac{1}{4}}$
$=\frac{3}{8}$.