Probability of having $k$ similar elements in two subsets. Given a distinct set of elements X and two randomly selected subsets of it $X_1,X_2$ (selected with equal distribution), I would like to find the probability that $|X_1 \cap X_2|\ge m$ where $$0 \le m \le \min{\left(|X_1|,|X_2|\right)}$$
It has been a long time since I have tackled any probability related math problem and I am unsure of how to approach this one. As far as I understand the number of combinations where $|X_1 \cap X_2|=0$ is
$$\dbinom{|X|}{|X_1|}\cdot \dbinom{|X|-|X_1|}{|X_2|}$$
and the number of combinations where $|X_1 \cap X_2|=1$  is
$$\binom{|X|}{|X_1|}\cdot\binom{|X_1|}{1}\cdot \binom{|X|-|X_1|}{|X_2|-1}$$
So I can try to sum up all the combinations and divide them by $$\binom{|X|}{|X_1|}\cdot\binom{|X|}{|X_2|}$$
Is this approach correct? Is there a better one?
I would use this in order to find relations between words in a large text corpus, so I would prefer to avoid unnecessary calculations.
Thanks
 A: The approach is correct, but you don't need to sum up all the combinations, just $|X_1 \cap X_2| = m$ and those after, all the way till $|X_1 \cap X_2| = min\{|X_1|, |X_2|\}$.
For example the number of combinations where $|X_1 \cap X_2| = m$ is 
$$ \binom{|X|}{|X_1|} \binom{|X_1|}{m} \binom{|X|-|X_1|}{|X_2|-m} $$
Hence the resulting probability should be 
$$ \frac{\sum_{i = m}^{min\{|X_1|, |X_2|\}}\binom{|X|}{|X_1|} \binom{|X_1|}{i} \binom{|X|-|X_1|}{|X_2|-i}}{\binom{|X|}{|X_1|} \binom{|X|}{|X_2|}} $$
A: If you have a set with $n$ elements and intersect two randomly 
selected subsets, the probability that the intersection will have 
exactly $k$ elements is: $4^{-n} 3^{n-k} \binom{n}{k}$ (the $4^{-n}$ 
is because there are $4^n$ ways to choose a pair of subsets from $n$ 
elements). 
I couldn't actually prove this, but I'm pretty sure it's true and that 
I'm just being lazy. Perhaps someone else could provide a proof? 
If you accept the formula above, the mean for a given value of $n$ is 
$\frac{n}{4}$, the variance is $\frac{3 n}{16}$, and the standard 
deviation is thus $\frac{\sqrt{3 n}}{4}$ 
For large values of $n$, the distribution is essentially normal with 
the parameters above. 
