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. $\frac{2}{5}$
• C. $\frac{4}{7}$
• D. $\frac{3}{24}$

Answer 1

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

Let $E_1$ be the event it is actually $6$ and $E_2$ be the event of not actually $6$ on the die.
Let $A$ be the event that the man speaks truth
$P\left(E_1\right)=\frac{1}{6},P\left(E_2\right)=\frac{5}{6}$
$P\left(\frac{A}{E_1}\right)=\frac{3}{4},P\left(\frac{A}{E_2}\right)=\frac{1}{4}$
Using Bayes' Theorem,
$P\left(E_1/A\right)=\left(\frac{P\left(E_1\right).P\left(A/E_1\right)}{P\left(E_1\right).P\left(A/E_1\right)+P\left(E_2\right).P\left(A/E_2\right)}\right)=\left[\frac{(\frac{1}{6}\times \frac{3}{4})}{(\frac{1}{6}\times \frac{3}{4})+(\frac{5}{6}\times \frac{1}{4})}\right]=\frac{3}{8}$