Conditional probability of intersection of multiple hypergeometric distributions What's given:


*

*I have an urn with with a set $S$ of balls where $|S| = N$. Each ball $b_i$ has a unique id and can either be red or blue. There are $m$ red balls in the urn.

*$d$ times I randomly draw a subset $D_j \subseteq S$ of the balls from the urn. $|D_j|$ is predetermined for each trial. After each trial, I put all the balls I drew back, so the population is always the same for each trial.

*I know that $D_j$ has $r_j$ red balls and I know the id of the balls that I drew, but I don't know which of the balls I drew were red.

*Let $E_j$ be the event that $D_j$ contains $r_j$ red balls.


Find an expression for: $$P(E_n|\bigcap_{j = 1}^{n - 1}E_j)$$
Here's a more concrete example:
$N = 5, m = 3, d = 2$
$D_1 = \{b_1, b_2\}, r_1 = 1$
$D_2 = \{b_1, b_2, b_3\}, r_2 = 2$  
I think the answer is supposed to be 2/3 because I wrote out all the combinations, but I don't know how to generalize it. Because I'm drawing from a finite population without replacement, I think I'm supposed to use a hypergeometric distribution. Something like $X_j \sim HypG(|D_j|, N, m)$. So I'm looking for $P(X_n = r_n | \bigcap_{j = 1}^{n - 1} X_j = r_j)$.
Since we gain information about how many red balls a particular subset of the population has, the $X_j$s are not independent. Once trial 1 happens, we know that exactly 1 of ball 1 and ball 2 are red, so for trial 2 we can eliminate any possibilities that both ball 1 and 2 are red. If they were independent, I think we would have a 3/5 chance of getting trial 2. Knowing which balls were in which trial affects knowledge about the outcome.
I've tried solving using
$$P(X_n = r_n | \bigcap_{j = 1}^{n - 1} X_j = r_j) = \frac{P(\bigcap_{j = 1}^{n} X_j = r_j)}{P(\bigcap_{j = 1}^{n - 1} X_j = r_j)}$$
But I run into trouble trying to interpret the outcome of the previous trials, especially when I'm getting the intersection of more than two trials. How do I account for the information that exactly $r_j$ balls from $D_j$ are red?
 A: Your presentation here is very complicated, but the problem appears to be quite trivial.  Unless I am missing something here, you are effectively just drawing a sequence of independent values from the hypergeometric distribution.  (You have not specified any source of statistical dependence between the trials.)  Letting $d_i \equiv |D_i|$ be the number of balls drawn on trial $i$, you have:
$$\mathbb{P}(E_i) = \mathbb{P}(D_i = r_i) = \frac{{m \choose r_i} {N-m \choose d_i-r_i}}{{N \choose d_i}}.$$
Since you replace the balls each time, assuming random sampling of the balls in each trial, there should be no dependence between the trials, so you would have:
$$\mathbb{P}(E_n | E_1 \ \cap \ ... \ \cap \ E_{n-1} ) = \mathbb{P}(E_n) = \frac{{m \choose r_n} {N-m \choose d_n-r_n}}{{N \choose d_n}}.$$
Now, unless I have misunderstood something here, most of what you have added to this question is an unnecessary complication.  Since all balls are replaced after each trial, and the numbers of balls, etc., is taken as known, there is nothing in your question that would induce any statistical dependence between the trials.
