How to compute hypergeometric distribution probabilities for complex events? How would I calculate the sums of two (or more) hypergeometric distributions. If, using a standard deck of cards, I want to determine the probability of draw 2 red cards and one Black Queen. I cannot just change my "good card" size and use one formula cause that wouldn't tell me what I need.
So, given 52 cards, 26 of which are red and given a draw of 4 cards, the hypergeometric probability of drawing 2 red cards is 0.32. With Black Queens, there are 2 "successes" in the population and I want to draw 1, the probability is 0.17. 
In Python using Scipy I can use the hypergeom function
[M,n,N] = [52,26,5] # M=Population,n=Successes,N=drawn
rv = hypergeom(M,n,N)
pR = rv.pmf(2) # probability of two reds in a hand of 5

n1 = 2
rv = hypergeom(M,n1,N)
pBQ = rv.pmf(1) # probability of 1 Black Queen in a hand a 5

How do I then calculate the probability that out of a 5 card draw I get 2 reds and 1 black Queen?
 A: Since drawing more red cards implies you have drawn fewer black cards, the red card count and black queen count are not independent.  That makes it difficult to combine the probabilities of each event in any simple way to obtain the answer.
Instead, do it the old-fashioned way: count every hand having two red cards, one black queen, and (presumably) two other black cards.  Divide that by the count of all possible five-card hands, because each such hand has the same probability (under a fair draw, anyway).
The number of such hands is counted by taking the number of two-card subsets of all $26$ cards, written $\binom{26}{2}$, multiplying that by the number of one-card subsets of the two black queens, written $\binom{2}{1},$ and multiplying the result by the number of two-card subsets of the remaining $24$ black cards, written $\binom{24}{2}.$  This will be divided by $\binom{52}{5}$ to obtain the probability.
Applying the formula
$$\binom{n}{k} = \frac{n!}{(n-k)!k!} = \frac{n(n-1)\cdots(n-k+1)}{k(k-1)\cdots(1)}$$
yields the answer
$$\frac{\binom{26}{2}\binom{2}{1}\binom{24}{2}}{\binom{52}{5}} = \frac{5\times 23}{2\times 7^2 \times 17}=\frac{115}{1666}\approx 6.9\%.$$

This reasoning applies, mutatis mutandis, to any event that specifies how many elements of each non-overlapping subset of a population must appear in a sample (without replacement) of a given size.
