Question

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.

• A. $\frac{5}{8}$
• B. $\frac{3}{8}$
• C. $\frac{7}{8}$
• D. $\frac{1}{8}$

Answer 1

Answer: (B)

Let $E_1$, $E_2$ and $A$ be the events defined as follows :
$E_1 =$ die shows six i.e. six has occured,
$E_2 =$ die does not show six i.e. six has not occurred and
$A =$ the man reports that six has occurred.
We wish to calculate the probability that six has actually occurred given that the man reports that six occurs i.e. $P(E_1|A)$.
Now, $P(E_1) =\frac{1}{6}$, $P(E_2) = \frac{5}{6}$
$P(A|E_1) =$ probability that the man reports that six occurs given that six has occurred
i.e.y probability that the man is telling the truth $=\frac{3}{4}$
$P(A |E_2) =$ probability that the man reports that six occurs given that six has not occurred
i.e., probability that the man does not speak truth $= \frac{1}{4}$
By Bayes' theorem,

$P\left(E_{1}|A\right) = \frac{P\left(E_{1}\right)P\left(A |E_{1}\right)}{P \left(E_{1}\right)P\left(A|E_{1}\right) + P\left(E_{2}\right)P\left(A|E_{2}\right)}$
$= \frac{\frac{1}{6}\cdot\frac{3}{4}}{\frac{1}{6}\cdot\frac{3}{4}+\frac{5}{6}\cdot\frac{1}{4}} = \frac{3}{8}$