Probability of an Intersecton being non-empty Let U be a set s.t. $|U|=n$.
Let $X_1, X_2, ..., X_{10}\subseteq U$ and $|X_i|=m_i$.
Given the cardinality of the $|X_i|$ and the cardinality of the pairwise intersections
$$|X_i \cap X_j| = l_{ij}, \text{ for } i\not=j,\forall i,j\in \{1,2,...,10\}
$$
What is the probability that $|X_1\cap X_2 \cap ... \cap X_{10}|>0$ ?
And how can I calculate it? 

It's quite some time ago that I learned about probability theory, and it wasn't my strongest subject. So I would be really grateful for any help.

All the best, Luca

Update:
I found this question that gave me some new ideas.
Expected value of probability of intersection. 
I want to use the following formula

In my case, I change the equation to get
$$\big|\bigcap\limits_{i=1}^{10} X_i\big| = ...$$
The first two sums are given. Then I want to calculcate the expected value for each of the missing sums.
For the intersection of three sets I want to use:
$$E[|X_i\cap X_j|\mid |X_i|, |X_j|]=\frac{|X_i||X_j|}{|U|}=\frac{m_i m_j}{n}$$
but applied for three sets.
I thought about doing this:
Let's say I want to calculate the expected value for $|X_i\cap X_j \cap X_k|$. Therefore I calculate:
$$E[|X_i\cap X_j| \cap |X_i\cap X_k| | |X_i \cap X_j|, |X_i \cap X_k] = \frac{l_{ij}\cdot l_{ik}}{(l_{ij}+l_{ik})}$$
$$E[|X_i\cap X_j| \cap |X_j\cap X_k| | |X_i \cap X_j|, |X_j \cap X_k] = \frac{l_{ij}\cdot l_{jk}}{(l_{ij}+l_{jk})}$$
$$E[|X_i\cap X_k| \cap |X_j\cap X_k| | |X_i \cap X_k|, |X_j \cap X_k] = \frac{l_{ik}\cdot l_{jk}}{(l_{ik}+l_{jk})}$$
I divide by the $l_{ij}+l_{jk}$ since I know that
$$(X_i\cap X_j) \cap (X_j\cap X_k) \subseteq (X_i\cap X_j) \cup (X_j\cap X_k) \text{ and } |(X_i\cap X_j) \cup (X_j\cap X_k)|\leq l_{ij}+l_{jk}$$
And then I want to choose the middle of the three values as my expected value for the intersection of three sets.

Does that make any sense? I could then go on for 4 sets, then 5, ... until calculating my expected value for 10 sets.
 A: I see.  Yes, if one were to select a point $x\in U$ at random, the probability of selecting a point in $X_3$ for example is, as you say, $\frac{m_3}{n}$.  Similiarly, the probability of selecting a point in both $X_3$ and $X_7$ would be $\frac{l_{37}}{n}$.
It is certainly possible that the intersection of all $10$ sets is empty, in which case the probability of selecting a point lying in all $10$ of the sets would be $0$.  However, assuming the intersection of all 10 sets is nonempty, the probability of selecting a point in the intersection of all $10$ sets is $\frac{|X_1 \cap X_2 \cap ... \cap X_{10}|}{n}$.  (However, it is not clear to me we have enough information to determine the size of this intersection.)
A: If we do not know the distribution of subsets $X_i$ chosen from $U$, then we cannot determine this probability.  For example, suppose we know that $U = \{a, b, c, d\}$ and sets $X_1, X_2, X_3$ are chosen such that $|X_i| = 2$ for all $i$, and $|X_i \cap X_j| = 1$ for all $i \not= j$.
Then we could have $X_1 = \{a, b\}, X_2 = \{a, c\}, X_3 = \{a, d\}$, and then
$$
X_1 \cap X_2 \cap X_3 = \{a\} \not= \emptyset
$$
Or, alternatively, we could have $X_1 = \{a, b\}, X_2 = \{b, c\}, X_3 = \{a, c\}$, and then
$$
X_1 \cap X_2 \cap X_3 = \emptyset
$$
To evaluate the probability that we have one of the collections equivalent to the former, as opposed to one that is equivalent to the latter, we need to know some kind of distribution for the $X_i$.  Without it, the probability cannot be determined.

If we know that all subsets of magnitude $k$ are a priori equally likely, then we can answer this question.  For instance, in the scenario above, the probability that the mutual intersection is non-empty is $1/2$: Wlog, the first subset is $\{a, b\}$ and the second subset is $\{a, c\}$, and then for the third subset, the only choices are $\{a, d\}$ and $\{b, c\}$, which are equally likely, but only one of which gives a non-empty mutual intersection.
But I think coming up with a closed-form answer for general $m_i$ and $I_{i, j}$ is likely to be challenging.  Let me think about that.
