1. The problem statement, all variables and given/known data I have a bag of n coins, and 1 is fake - it has 2 heads. a) Determine the probability that if I flip a coin and it comes up heads, the coin is fake. b) If a coin is flipped k times and comes up heads k times, what is the probability that the coin is fake? 2. Relevant equations Bayes Theorem: P(A|B) = [P(B|A) P(A)] / P(B) 3. The attempt at a solution For part a) I think I have this correctly. P(fake) = 1/n P(!fake) = (n-1) / n P(heads) = P(heads|fake) P(fake) + P(heads|!fake) P(!fake) = 1 * 1/n + 1/2 * (n-1) / n = 1/n + (n-1) / 2n = 2/2n + (n-1) / 2n = (n+1) / 2n P(fake|heads) = [P(heads|fake) P(fake)] / P(heads) = 1 * 1/n / (n+1) / 2n = 1/n / (n+1) / 2n = 1/n * 2n / (n+1) = 2n / n(n+1) = 2 / (n+1) For part b), I'm a little stuck. I think the probability of the coin coming up heads k times = P(heads)k or [(n+1) / 2n]k. My reasoning here is that ignoring the fake coin the probability of getting 5 heads in a row would be 1/2 * 1/2 * 1/2 * 1/2 * 1/2 = (1/2)k. So P(fake|(k heads)) = [P((k heads) | fake) * P(fake)] / P(k heads) and P((k heads) | fake) = 1, so P(fake|(k heads)) = [1 * 1/n] / [(n+1) / 2n]k = [1/n] / [(n+1) / 2n]k = 1/n * [2n / (n+1)]k = 1/n * [2nk/(n+1)k] = 2nk / n(n+1)k But I'm unsure of my reasoning.