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

With $N/2$ white balls, uniform sampling without replacement gives the probabilityRearranging, $4p_w(N-1)=N-2$, soA finite admissible population has even $N\ge 2$ and $0\le p_w<1/4$, so the denominator is nonzero.

Interchanging the two colours preserves the probability distribution when their population counts are equal. Thus the probability of two black balls is also $p_w$. The three mutually exclusive possibilities are two white, two black, or different colours, so

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.

With equally many balls of each colour, uniform sampling without replacement givesHere an admissible population producing positive probability for three white balls has even $N\ge 6$, so the cancellation is legitimate. Equating this to $2/19$ gives $19(N-4)=16(N-1)$, hence

There are ten balls of each colour. Under uniform sampling without replacement, all $\left(\genfrac{}{}{0ex}{}{20}{12}\right)$ subsets of twelve labelled balls are equally likely. A favourable subset chooses four of the ten white balls and eight of the ten black balls, giving $\left(\genfrac{}{}{0ex}{}{10}{4}\right)\left(\genfrac{}{}{0ex}{}{10}{8}\right)$ choices. The hypergeometric distribution therefore givesThus the requested integers are $\boxed{N_1=N_2=10,m_1=4,m_2=8}$, with $N=20$ and $k=12$.