A man is known to speak the truth 3 out of 4 times. He throws a die and reports that it is five then the probability that it is actually five is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{3}{8} \)
Let \( E_{1} \): the die shows five, \( E_{2} \): it does not show five, and A: he reports a five.
\( \displaystyle P(E_{1})=\frac{1}{6} \)
\( \displaystyle P(E_{2})=\frac{5}{6} \)
\( \displaystyle P(A|E_{1})=\frac{3}{4} \)
If it is not five, he reports five only when he lies, which is taken to mean reporting five:
\( \displaystyle P(A|E_{2})=\frac{1}{4} \)
By Bayes’ theorem,
\( \displaystyle P(E_{1}|A)=\frac{\frac{1}{6}\cdot\frac{3}{4}}{\frac{1}{6}\cdot\frac{3}{4}+\frac{5}{6}\cdot\frac{1}{4}} \)
\( \displaystyle =\frac{\frac{3}{24}}{\frac{3}{24}+\frac{5}{24}} \)
\( \displaystyle =\frac{3}{8} \)
NCERT (2026-27 edition):
Bayes' Theorem
