what is the probability of concreate cardinality of intersection of k subsets of set?

Question

Given set $S$, where $|S| = N$, and drawn randomly $k$ subsets of $S$: $a_1, a_2, a_3, ..., a_k$ with cardinalities: $n_1, n_2, n_3, ..., n_k$. What is the probability that the intersection of $a_1, a_2, a_3, ..., a_k$ will have the cardinality equal (or equal and greater) to p (which is parameter).

The problem I encountered is that when I try to count all possible subsets of given cardinality, for the denominator of probability I get:

$$\prod_{i=1}^k\binom{N}{n_i}$$

which I believe is correct. But when i try to count such examples that has cardinality of intersection at least $p$ (to put in in numerator) then I can come up only with this: $\binom{N}{p}$ which is number of possibilities to draw the intersection, and then $\prod_{i=1}^k\binom{N-p}{n_i-p}$ representing number of possibilities of drawing rest of the subsets. Unfortunately this is not correct approach because resultig formula:

$$\frac{\binom{N}{p}\prod_{i=1}^k\binom{N-p}{n_i-p}}{\prod_{i=1}^k\binom{N}{n_i}}$$

is sometimes bigger than 1.0 for some $p=0, 1, 2, 3, ..., N$. I think the reason is that when i first draw intersection (eg. {1,2,3}), and then draw rest of the set (eg. {4,5}) I get a result ({1,2,3,4,5}) which I count multiple times (eg. when I draw intersection {3,4,5} and rest {1,2}). How to correct the numerator of formula to count every case only once?

Answer 1

What you've done thus far is fine, but you need to use inclusion/exclusion in your numerator. So instead of:

$$\frac{\binom{N}{p}\prod_{i=1}^k\binom{N-p}{n_i-p}}{\prod_{i=1}^k\binom{N}{n_i}}$$ you instead need:

$$\frac{\displaystyle\sum_{j=p}^N (-1)^{j-p} \binom{j}{p} \binom{N}{j} \prod_{i=1}^k\binom{N-j}{n_i-j}}{\displaystyle\prod_{i=1}^k\binom{N}{n_i}}$$

with the convention that ${a \choose b}=0$ for all $b < 0$.

Edit:

DANGIT! Realize I made a mistake, but cannot fix it right now. Inclusion/exclusion is the right way to go, but there is a missing factor in the numerator. I will fix later this evening, if another less confused combinatorialist does not fix it first.

Edit:

Shoot, that's what I get for trying to do this too quickly. Apologies. For each element $i$ of the index set $S$, define the set $A_i$ to be the set of all selections such that $i$ is contained in $a_1 \cap a_2 \cdots \cap a_k$. We are seeking the number of selections that are contained in exactly $p$ of these sets $A_i$.

We use a standard generalization of inclusion/exclusion (see for example joriki's excellent description) for this count. Using his notation, fulfilling a condition $A_i$ means having a selection of $a_1, a_2, \ldots a_k$ in the set $A_i$. The number of selections fulfilling $j$ conditions is uniform regardless of which conditions are chosen, namely $$\prod_{i=1}^k\binom{N-j}{n_i-j}$$

Therefore, we can use the last of joriki's formulae to give the number of ways to have an intersection size of exactly $p$ (note: The $j$ in joriki's formula is our $p$, and his $k$ is our $j$): $$\displaystyle \sum_{j=p}^N (-1)^{j-p} \binom{j}{p} \binom{N}{j} \prod_{i=1}^k\binom{N-j}{n_i-j}$$

This is the numerator of our probability.