A man is known to speak truth 3 out of 4 times. He throws a die and reports that it is four. The probability that it is actually four is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{3}{8}\)
Let \(E_1\) be the event that four occurs, \(E_2\) that four does not occur, and R that the man reports a four.
\( \displaystyle P(E_1) = \frac{1}{6},\quad P(E_2) = \frac{5}{6} \)
He reports four when four occurs and he speaks the truth, or when four does not occur and he lies:
\( \displaystyle P(R|E_1) = \frac{3}{4},\quad P(R|E_2) = \frac{1}{4} \)
By Bayes’ theorem:
\( \displaystyle P(E_1|R) = \frac{P(E_1)P(R|E_1)}{P(E_1)P(R|E_1) + P(E_2)P(R|E_2)} \)
\( \displaystyle = \frac{\frac{1}{6} \cdot \frac{3}{4}}{\frac{1}{6} \cdot \frac{3}{4} + \frac{5}{6} \cdot \frac{1}{4}} \)
\( \displaystyle = \frac{\frac{3}{24}}{\frac{3}{24} + \frac{5}{24}} = \frac{3}{8} \)
Shortcut: By the ratio method, the two ways of reporting four have weights \(1 \times 3 : 5 \times 1 = 3 : 5\), so the probability of the first is \(\frac{3}{3 + 5} = \frac{3}{8}\).
NCERT (2026-27 edition):
Bayes' Theorem
