I was discussing the following question with a friend but we are stuck, this is probably obvious for someone with a mathematical background.
Say we have an urn with $N$ balls of which a $M$ are blue. We now take $n$ balls out of the urn without putting them back. What is the probability of having at least $k$ blue balls.
Wikipedia tells me the probability of getting exaktly k blue balls is
$ P(X=k)=\frac{{n\choose k}{N-n\choose M-k}}{{N\choose M}}$
Is there a closed form solution for the probability for drawing i or more blue balls?
$\sum_i^N P(X=i)$
Yes and no. You're asking for $$\sum\limits_{i=k}^n P(X=i)=1-\sum\limits_{i=0}^{k-1}P(X=i)=1-P(X\leq k-1),$$which is the cif of the hypergeometric distribution (see. wikipedia). If k=0, so you're asking the probability of having zero or more blue balls, the answer is obviously 1.