You have \(1000\) coins, one of which is faulty: it has a head on both sides. You randomly draw a coin, and, without examining it, toss it \(10\) times. As it happens, you get \(10\) heads in a row. What’s the probability that it’s the faulty one?
Bayes theorem is useful for solving this problem. Let \(P(F)\) represent the prior probability that the drawn coin is faulty. Let \(P(H)\) represent the prior probability of the coin landing on heads \(10\) times in a row.
We can calculate \(P(H)\) by considering the two possible cases for the drawn coin—it is either faulty or not:
\[P(H) = \frac{1}{1000}(1)+\frac{999}{1000} \left( \frac{1}{2^{10}} \right) = \frac{2023}{1024000}\]
We trivially know that \(P(F) = \frac{1}{1000}\), since only one coin is faulty. Additionally, we know \(P(H|F) = 1\), since the faulty coin always lands on heads.
Inserting these values into Bayes’ theorem gives the following solution:
\[\begin{split} P(F|H) &= \frac{P(F)P(H|F)}{P(H)} \\ &= \left( \frac{1}{1000}\right) \left(\frac{1024000}{2023} \right) \\ &\approx 0.506 \end{split}\]