A man is known to speak truth 4 out of 5 times. A coin is tossed. He reports that a head appears. The probability that actually there was a head, is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 4/5
Let \( E_1 \): head appears, \( E_2 \): tail appears, and A: the man reports a head.
\( \displaystyle P(E_1)=P(E_2)=\frac{1}{2} \)
\( \displaystyle P(A|E_1)=\frac{4}{5}\ \text{(he tells the truth)} \)
\( \displaystyle P(A|E_2)=\frac{1}{5}\ \text{(he lies)} \)
By Bayes' theorem:
\( \displaystyle P(E_1|A)=\frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1)+P(E_2)P(A|E_2)} \)
\( \displaystyle =\frac{\frac{1}{2}\cdot\frac{4}{5}}{\frac{1}{2}\cdot\frac{4}{5}+\frac{1}{2}\cdot\frac{1}{5}} \)
\( \displaystyle =\frac{\frac{4}{10}}{\frac{5}{10}} \)
\( \displaystyle =\frac{4}{5} \)
Shortcut: Use the ratio method: head and report-head has weight \( \frac{1}{2}\cdot\frac{4}{5} \), tail and report-head has weight \( \frac{1}{2}\cdot\frac{1}{5} \). The ratio is 4 : 1, so the probability is \( \frac{4}{4+1}=\frac{4}{5} \).
NCERT (2026-27 edition):
Bayes' Theorem
