Is anybody able to provide a high level explanation of the solution to April 23 Qn 1ii / iii please? Even with the examiners report I just can't follow the logic / get my head around it. Thanks!
Hi Michael I'd approach this by trying to think about the different possible transitions. Say we are currently in state i. This means there are i white balls in box A. As there are n balls in each box and n total white and black balls, this means there are n-i black balls in box A, i black balls in box B and n-i white balls in box B. If we select a black ball from box A and a while ball from box B, then this means the process goes from state i to state i + 1. The probability of selecting a black ball from box A is (n-i) / n. The probability of selecting a white ball from box B is (n-i) / n. So, the probability of this happening is (n-i)^2 / n^2. Note that the above holds as long as i = 0, 1, ... n-1. The probability formula still makes sense for i = n (in that it gives 0) but I'd be careful here. There is no state n+1 and so the probability of going from state n to state n + 1 is really undefined rather than being 0. See if this helps get you started on the other possible transitions also. Andy
