Past exam of the mathematics course of the University of Cambridge / 2002 / ia / Paper 2 / 9F / b / Solution

Every urn contains $n-1$ balls. Let $k=n-r$ be its number of blue balls; choosing an urn uniformly makes $k$ uniform on $0,\dots ,n-1$. Conditional on that urn, the probability of a blue first draw is $k/(n-1)$, so the law of total probability givesDrawing twice without replacement requires $n\ge 3$. Given $k$, the probability that both draws are blue is $k(k-1)/((n-1)(n-2))$, and henceTaking the ratio defining conditional probability gives the second requested probabilityAlthough a blue first draw removes a blue ball from its urn, it also makes an urn rich in blue balls more likely. Averaging with this changed conditional distribution is essential.