Question

A man is known to speak truth 2 out of 3 times. He throws a die and reports that number obtained is a four. Find the probability that the number obtained is actually a four.

• A. $\frac{2}{7}$
• B. $\frac{3}{7}$
• C. $\frac{4}{7}$
• D. $\frac{3}{4}$

Answer 1

The correct option is A $\frac{2}{7}$

Let A be the event that the man reports that number four is obtained.
Let $E_1$ be the event that four is obtained and $E_2$ be its complementary event.
Then, $P\left(E_1\right)$ = Probability that four occurs = $\frac{1}{6}$
$P\left(E_2\right)$ = Probability that four does not occur = $1-P\left(E_1\right)=1-\frac{1}{6}=\frac{5}{6}$
Also, $P\left(\frac{A}{E_1}\right)$ = Probability that man reports four and it is actually a four = $\frac{2}{3}$
$P\left(\frac{A}{E_2}\right)$ = Probability that man reports four and it is not a four = $\frac{1}{3}$
By using Bayes’ theorem, probability that number obtained is actually a four,
$P\left(\frac{E_1}{A}\right)=\frac{P\left(E_1\right)\left(\frac{A}{E_1}\right)}{P\left(E_1\right)\left(\frac{A}{E_1}\right)+P\left(E_2\right)\left(\frac{A}{E_2}\right)}=\frac{\frac{1}{6}\times \frac{2}{3}}{\frac{1}{6}\times \frac{2}{3}+\frac{5}{6}\times \frac{1}{3}}=\frac{2}{7}$