I have an odd problem which can be phrased in a general way, and a more specific way. I'm curious about the answers to both. Although, really, it's the k=0 case that I'm really interested in - deriving the answer with respect to some properties of the distribution of m.

These may be totally basic, so, apologies if they are.

1) I have several different hypergeometric distributions, H(k; N, m, n) where k is the number of 'success' draws, N is the population size, m is the number of possible success draws, and n is the total number of draws. Each distribution has a different value for m, but all else is the same. Is there an easy way to either sum them up or provide a more compact notation for them?

Or, to phrase it as a ball and urn question, I have many boxes, each with N balls. In each urn i, m(i) balls are white and the rest are black. If I take n draws from each urn in turn, what is the average probability of k white balls drawn in each urn? (indeed, is there a distribution for this - there must be)

2) More specifically, I'm interested in the case where k=0. This actually reduced quite nicely to


But...again, I want to sum over a lot of different values of m from different members of a population. Is there a way to get at this with, say, an average value of m or otherwise?

Again, in box urn terms, this would be the average probability of drawing NO white balls from any of the urns.

p.s. I have actually always wondered a similar thing for the binomial distribution and suspect the answers may be related. Sure, B(a,p)+B(b,p) = B(a+b,p) where a and b are the number of trials and p is the probability of success. But what about B(a,p1)+B(a,p2)?

  • $\begingroup$ Your binomial problem is essentially the same one as the hypergeometric one with $m_i=Np_i$, as $Hyp(k|N,Np_i,n)\to Bin(k|n,p_i)$ as $N\to\infty$ $\endgroup$ Commented Sep 25, 2011 at 10:20

1 Answer 1


I don't think there will be a simple or general form for the distribution of the sum of independent hypergeometric distributions. You'd need to work with convolutions. It would be relatively easy to do the calculations numerically if there aren't too many urns.

However, the specific case in (2), of the probability that they are all 0, is rather simple. You just take $\prod_i \Pr(k_i = 0).$

Regarding your postscript, on the sum of binomials with different proportions: this is not binomial. Consider the extreme case with $p_1 = 0$ and $p_2 = 1$.

  • $\begingroup$ Yup, I had fallen on something similar myself. My question is more how to get the average probability across all urns when k=0. I'm trying to see if I can come up with an expression that uses the average value (or any other distributional properties) of m, really. It's really m I'm interested in here. See my edit above. $\endgroup$
    – jebyrnes
    Commented Sep 25, 2011 at 9:42
  • $\begingroup$ I don't think you'll be able to simplify the average of $\Pr(k=0)$ to something that involves the average of $m$. You're likely stuck with numerical results. Consider the binomial case. The average of $p_i^n$ for fixed $n$, varying $p_i$, won't reduce to something that involves just the average of the $p_i$. $\endgroup$
    – Karl
    Commented Sep 25, 2011 at 12:24
  • $\begingroup$ That's my fear - although, even if N and n are constant? There's no way to at least derive a distribution of Pr(k=0) if you know the distribution of m - or all values of m? $\endgroup$
    – jebyrnes
    Commented Sep 25, 2011 at 13:38
  • $\begingroup$ @jebyrnes - You can write it down or calculate it numerically, but it won't simplify much. $\endgroup$
    – Karl
    Commented Sep 25, 2011 at 14:07
  • $\begingroup$ On the other hand, what if you had a probability distribution for m? Say it was poisson distributed? Or, better, geometrically distributed? Would one be able to work from there? $\endgroup$
    – jebyrnes
    Commented Oct 3, 2011 at 16:58

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