Solution
= Solution
Let $R_1,R_2$ be the two red-card <events>. Sampling without replacement gives
$$
\mathbb P(R_1\cap R_2)=\frac mN\frac{m-1}{N-1}.
$$
This differs from the product $m^2/N^2$, since
$$
\mathbb P(R_1\cap R_2)-\mathbb P(R_1)\mathbb P(R_2)
=-\frac{mn}{N^2(N-1)}<0.
$$
Thus \b[the events are not independent], for every positive $m,n$. A first red card reduces the remaining red fraction, a basic instance of how <sampling without replacement need not be independent>.