Probability that a man speaks truth is \(\displaystyle\frac{3}{4}\). He throws a die and reports that it is a six. The probability that it is actually a six is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{3}{8}\)
Let \(E_1\) be the event that a six occurs, \(E_2\) the event that a six does not occur, and A the event that the man reports a six.
\(\displaystyle P(E_1) = \frac{1}{6},\quad P(E_2) = \frac{5}{6}\)
When a six occurs, he reports a six if he speaks the truth. When a six does not occur, he reports a six if he lies (as usual in this problem, a lie about a non-six is taken to be the report “six”):
\(\displaystyle P(A|E_1) = \frac{3}{4},\quad P(A|E_2) = \frac{1}{4}\)
By Bayes’ theorem,
\(\displaystyle P(E_1|A) = \frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)}\)
\(\displaystyle = \frac{\frac{1}{6}\cdot\frac{3}{4}}{\frac{1}{6}\cdot\frac{3}{4} + \frac{5}{6}\cdot\frac{1}{4}}\)
\(\displaystyle = \frac{\frac{3}{24}}{\frac{3}{24} + \frac{5}{24}} = \frac{3}{8}\)
NCERT (2026-27 edition):
Bayes' Theorem
