A man is known to speak truth 3 out of 4 times. He throws a coin three times and reports that a single head appears. The probability that actually a single head appears is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{9}{14}\)
Let \(E_1\): exactly one head appears, \(E_2\): it does not, and A: the man reports a single head.
Three tosses have 8 equally likely outcomes, and exactly one head occurs in HTT, THT, TTH:
\( \displaystyle P(E_1)=\frac{3}{8},\quad P(E_2)=\frac{5}{8} \)
He reports a single head if it happened and he speaks the truth, or if it did not happen and he lies:
\( \displaystyle P(A|E_1)=\frac{3}{4},\quad P(A|E_2)=\frac{1}{4} \)
By Bayes’ theorem:
\( \displaystyle P(E_1|A)=\frac{\frac{3}{8}\times\frac{3}{4}}{\frac{3}{8}\times\frac{3}{4}+\frac{5}{8}\times\frac{1}{4}} \)
\( \displaystyle =\frac{\frac{9}{32}}{\frac{9}{32}+\frac{5}{32}} \)
\( \displaystyle =\frac{9}{14} \)