A man speaks the truth 3 out of 5 times. He draws a card at random from a well-shuffled pack of 52 playing cards and reports that it is an ace. The probability that it is actually an ace is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{1}{9}\)
Let \(E_1\) be the event that the card is an ace and \(E_2\) the event that it is not, and let R be the event that the man reports an ace.
\( \displaystyle P(E_1) = \frac{4}{52} = \frac{1}{13},\quad P(E_2) = \frac{12}{13} \)
He reports an ace when the card is an ace and he speaks the truth, or when it is not an ace and he does not speak the truth:
\( \displaystyle P(R|E_1) = \frac{3}{5},\quad P(R|E_2) = \frac{2}{5} \)
By Bayes’ theorem:
\( \displaystyle P(E_1|R) = \frac{\frac{1}{13}\times\frac{3}{5}}{\frac{1}{13}\times\frac{3}{5}+\frac{12}{13}\times\frac{2}{5}} \)
\( \displaystyle = \frac{3}{3+24} = \frac{3}{27} = \frac{1}{9} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{1}{13}\) — Incorrect. This is the chance of an ace before his report is taken into account.
Option (2) \(\displaystyle\frac{3}{65}\) — Incorrect. This is \(\frac{1}{13}\times\frac{3}{5}\), only the numerator of Bayes’ formula.
Option (3) \(\displaystyle\frac{8}{9}\) — Incorrect. This is \(P(E_2|R)\), the probability that the card is not an ace, given his report.