Solution
= Solution
With equally many balls of each colour, uniform <sampling without replacement> gives
$$
p_w=\frac{N/2}{N}\frac{N/2-1}{N-1}\frac{N/2-2}{N-2}=\frac{N-4}{8(N-1)}.
$$
Here an admissible population producing positive <probability> for three white balls has even $N\geq6$, so the cancellation is legitimate. Equating this to $2/19$ gives $19(N-4)=16(N-1)$, hence
$$
\boxed{N=20.}
$$