Probability distribution of size of union of two sets I have a set $S$ with $M$ elements in it. Given $N_A$ and $N_B$, I randomly pick two sets, $A \subseteq S$, $B \subseteq S$, so that $|A| = N_A$ and $|B| = N_B$. What is the probability $P(L)$ that $|A \cup B| = L$?
Example
$S = \{1, 2, 3, 4, 5, 6\}$, $N_A = 3$, $N_B = 3$
I pick at random $A = \{1, 4, 5\}$ and $B = \{2, 4, 6\}$. Then $L = |\{1, 2, 4, 5, 6\}| = 5$
 A: W.L.O.G. suppose $N_A \ge N_B$ 
If $L<N_A$ the answer is zero!
So suppose that we pick $N_A$ elements randomly. You must pick $L-N_A$ elements in $S-A$ and $N_B-(L-N_A)$ elements in A.
Setting $|S|=N$ So the answer is $$\frac{{N-N_A \choose L-N_A} \times {N_A \choose N_B+N_A-L}}{{N \choose N_A}\times{N \choose N_B}}$$
A: *

*See, that $\min{(N_A, N_B)}\leq L \leq \min{(|S|,N_A+ N_B)}$

*The set $X$ with $L$ elements (there are ${|S|\choose L}$ sets like that) can be represented as the sum of two sets with $N_A$ and $N_B$ elements in $K(L)$ ways, where
$$K(L)={L\choose N_A}{N_A \choose N_A+N_B-L}$$
First we choose $N_A$ elements that belongs to set $A$, and then among these elements we choose elements that belongs also to $B$.
We have then the number 

*Total number of sets $A$ and $B$ is
$$|\Omega|={|S|\choose N_A}{|S|\choose N_B}$$

*We have then:
$$P(L)=\begin{cases}\frac{{|S|\choose L}{L\choose N_A}{N_A \choose N_A+N_B-L}}{{|S|\choose N_A}{|S|\choose N_B}} &, \min{(N_A, N_B)}\leq L \leq \min{(|S|,N_A+ N_B)}\\
0&, \text{in other cases}\end{cases}$$

A: The second answer is incorrect. We can see this in the special case $N_B=0$, $N_A=L$, for which the resulting probability must be 1, since in this case $A\cup B=A$. However, the formula from the second answer would give us instead $\frac{1}{\binom{N}{N_A}}$ which is only correct for $A=\emptyset$.
I did some testing with small numbers, i.e. generating the powerset of $\{1,\ldots,n\}$, take all subsets with $N_A$ and $N_B$ elements, form their unions and then count; this suggests that the first answer is probably correct. For example, for $N_A=5, N_B=3, N=10, L=6$ we get the answer $\frac{5}{12}$ from both the formula and pure counting.
