Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2024 / Paper 1 / 16 / b

The probability that the first ball is white is $x$. Partitioning that event according to the second colour shows that $P(WB)=x-p_w$. Uniform sampling without replacement makes the two draw positions exchangeable, so $P(BW)=P(WB)$. ThereforeFor a direct count, the same value is $2(xN)((1-x)N)/[N(N-1)]$.

The probability of two black balls is the complement of the other two possibilities:The probability of equal colours is $P(WW)+P(BB)=1-2x+2p_w$. Thus the required conditional probability isThis is defined when the conditioning event has positive probability. If $N=2$ with one ball of each colour, that event is impossible and the conditional probability is undefined.