= Solution
When only a tiny fraction of either species is removed, the trout fraction among the remaining fish stays nearly constant at $p=K/N$. Successive catches are then approximately independent trials with that fixed success <probability>. The count therefore approaches the <binomial distribution>
$$
\boxed{\mathbb P(X=j)\approx\binom Mj p^j(1-p)^{M-j},\qquad p=K/N.}
$$
The approximation comes from negligible depletion, not from an exact <independence> property of <sampling without replacement>.
Back to article page
