hypergeometric distribution with a priori probability Is there a way to calculate drawing without replacement m from N elements s.t. r out of m are red balls and g out of m are green balls, following hypergeometric distribution, but when there's a given probability for drawing an element (as oppose to uniform in the case of hypergeometric) ?
for example, red balls are uniformly distributed and green balls are uniformly distributed.
There're in total G green balls.
The way I've been thinking to accomplish this, is following: 
$$
\prod_{i=0}^{m} \cdot \binom{m}{i}\cdot(\frac{G-i}{N})^i\cdot(1-\frac{G-i}{N})^{(m-i)}
$$
the binomial coefficient is for number of combinations choosing i green balls from m. The second term is the probability of choosing a green ball, and the third term is the probability for not choosing a green ball (choosing a red ball).
All three terms are multiplied m times, since the choice of green ball is dependent on previous choice (without replacement).
If green balls are distributed in some other distributions, than the second and third term would be sampled according to this distribution.
Thank you
 A: As far as I can understand, you are asking two different problems:
1) when the number of the balls for each colours is known. Then it is a combinatorial problem, and not the easiest one.
You may find the formula and one of the possible proof:
http://arxiv.org/pdf/1511.06142.pdf
2) If you don't know the exact number of the balls of each colour in the box, but only their distribution, then the problem seems harder. And I can't help that much... I think the answer is not combinatorial, but is a probability density function, described a weighted sum of the distribution of each colour in the box.
Edit:
After talking to a much wiser colleague, the formula that provides the answer should be:
$\mathbb{P}(C_1 = n_, C_2 = n_2, ... , C_c = n_c) = \binom{n}{n_1, n_2, ... n_c} \binom{N-n}{N_1 - n_1, N_2-n_2,... , N_c-n_c}/\binom{N}{N_1, N_2, ..., N_c}$
Where $\binom{a}{b, d}$ is the multinomial coefficient, $n$ is the total number of balls extracted, $N$ the total number of balls in the box, $C_j$ is the random variable that count the number of balls of color $j$, $n_j$ is the number of balls extracted of colour $j$, and $N_j$ is the total number of balls of color $j$ initially in the box. 
