I'm struggling with figuring out the formula for the following:
There are 8 red cards in a deck that contains 40 cards total. I want to know how to calculate the odds of drawing a minimum of 2 red cards if I get 5 draws without replacing the cards in the deck after drawing them.
Also, I am curious to know how that formula differs from the one that describes drawing exactly two red cards.
First, let's denote by $p_k$ the probability that you draw $k$ red cards out of your hand of $5$ cards. Obviously, $p_k$ is only non-zero for $0 \leq k \leq 5$.
The usual approach with these problems is to find the number of ways to draw $k$ red cards out of the $8$ available, and the number of ways to draw $5-k$ non-red cards out of the $32$ available. The product of these is the number of $5$-card hands that have exactly $k$ red cards:
$$ N_k = \binom{8}{k} \binom{32}{5-k} $$
If you aren't familiar with the binomial coefficient,
$$ \binom{n}{k} = \frac{n!}{k!(n-k)!} $$
is read "$n$ choose $k$" and denotes the number of ways to select $k$ objects out of a set of $n$ (with order immaterial). Then the total number of $5$-card hands is just
$$ N = \binom{40}{5} $$
Now, your desired ratio is just
$$ p_k = \frac{N_k}{N} $$
One could obtain the first answer by summing $p_2+p_3+p_4+p_5$, but it is slightly less work to compute $1-p_0-p_1$. And of course the second answer is just $p_2$ itself.