Assuming $n$ independent Bernoulli trials with different probabilities, the Poisson binomial distribution is the discrete probability distribution that describes the number of $X$ successes.
A Hypergeometric distribution is the discrete probability distribution that describes the probability of $k$ successes in $n$ draws, in trials without replacement.
What is the distribution that describes the number of $X$ successes of a Bernoulli trial without replacement and with different probabilities?
Edit: More specifically, I am trying to reason directly from probabilities.
Hypergeometric distribution and the Multivariate hypergeometric distribution does not explicitly deals with probabilities (i.e deals with number of success in a number of draws). In the urn with balls of different colors example, is there a way for me to access the number of success knowing the probability of drawing each color, but in a scenario where I do not know how many balls of each color the urn contains?
In other words: assuming an urn containing balls with n colors; and assuming I do not know the number of balls with each color, but I know that the probabilities of sampling each color is p=(p1,p2,…pn). I want to know, for example, the probability of sampling, without replacement, one X1, one X2 and one X5.