The probability that the intersection of two sets has a determinate size Suppose I have a finite set A and two subsets B, C $\subseteq$ A. I'd like to know what's the probability that the intersection of those two sets is greater than or equal to some number.
More specifically, I'd like to know the probability that the intersection of B and C equals some specific proportion of B (or C).
For instance, if I have the sets A = [0,100], B = [50,100] and C = [40,60], I know that the intersection of B and C takes up 20% of B and 50% of C. Now, supposing I have some definite set A, some definite subset B $\subseteq$ A and a random subset C $\subseteq$ A, what's the probability that C $\cap$ B takes up x% or more of B? And vice-versa?
I'm not even sure my question makes sense, but there it is.
 A: Suppose that B and C are subsets of A, |A| = a, |B| = b, |C| = c.  To answer the question, is the intersection of B and C exactly size d, for d $\leq$ b, c, fix B.  now, an arbitrary choice of C is a selection of c elements from A; then there are $_a C_c$ possible choices for C, and $_b C_d \cdot _{a-b}C_{c-d}$ of them give an intersection of size d.
A: Let $|A| = n$ and let $P(k)$ denote the probability that the intersection $D$ of arbitrary $B, C \subseteq A$ has cardinality exactly $k$ for $0 \le k \le n$. We get
$$P(k) = \frac{\text{#of ways of choosing $B$, $C$ with $|D| = k$}}{\text{# of ways of choosing $B$, $C$}}$$
Note first that there are $\binom nk$ choices for $D$. For $B$ we can choose arbitrarily from the $2^{n-k}$ sets containing $D$. For $C$, our choice is less free: we have to ensure the intersection with $B$ is exactly $D$, and not more. 
So we break it into cases: If $B$ has size $k$, then there is one choice for $B$, and we can choose $C$ arbitrarily from $2^{n-k}$ possibilities. If $B$ has size $k + 1$, then there are $\binom {n-k}1$ choices for $B$ and $2^{n-k-1}$ choices for $C$. Denoting $m = n-k$ and summing, there are
$$\sum_{j = 0}^m 2^{m-j} \binom mj = 3^m$$
possible choices for $B, C$ with intersection $D$. (Calculating the above sum is a fun exercise.) Thus there are 
$$3^{n-k} \binom nk$$
possible ways of choosing the subsets with the desired property. Since there are $(2^n)^2 = 4^n$ ways of choosing any two subsets of $A$, we get
$$P(k) = \frac{3^{n-k} \binom nk}{4^n}.$$
(If this solution is in error, please point it out! I wouldn't put my house on it just yet. A comforting thing is that $\sum_{k=0}^n P(k) = 1.$)
