Question

A man is known to speak truth is 75% cases. If he throws an unbiased die and tells his friend that it is a six, then the probability that it is actually a six, is

• A. $\frac{1}{6}$
• B. $\frac{1}{8}$
• C. $\frac{3}{4}$
• D. $\frac{3}{8}$

Answer 1

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

Let E denote the event that a six occurs and A be the event theat the man reports that it is a six. We have $P\left(E\right)=\frac{1}{6},P\left(E^{\prime }\right)=\frac{5}{6},P\left(\frac{A}{E}\right)=\frac{3}{4}$ and $P\left(\frac{A}{E^{\prime }}\right)=\frac{1}{4}$. By Baye’s theorem
$P\left(\frac{E}{A}\right)=\frac{P\left(E\right).P\left(\frac{A}{E}\right)}{P\left(E\right).P\left(\frac{A}{E}\right)+P\left(E^{\prime }\right).P\left(\frac{A}{E^{\prime }}\right)}=\frac{\left(\frac{1}{6}\right)\left(\frac{3}{4}\right)}{\left(\frac{1}{6}\right)\left(\frac{3}{4}\right)+\left(\frac{5}{6}\right)\left(\frac{1}{4}\right)}=\frac{3}{8}$