= Solution
There are ten balls of each colour. Under uniform <sampling without replacement>, all $\binom{20}{12}$ 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 $\binom{10}{4}\binom{10}{8}$ choices. The <hypergeometric distribution> therefore gives
$$
\boxed{P(\text{four white})=\frac{\binom{10}{4}\binom{10}{8}}{\binom{20}{12}}.}
$$
Thus the requested integers are $\boxed{N_1=N_2=10,\ m_1=4,\ m_2=8}$, with $N=20$ and $k=12$.
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