Probability that one of the cells of a random Venn Diagram exceeds a given threshold. I apologize if this has been answered before; I found a couple similar questions on Math.SE, but none were exactly the same.
Suppose we are given a universe $\Omega$ and integers $b_1, b_2, S$ satisfying $0 < b_1, b_2 < S < |\Omega|$, and we choose two subsets $A,B \subset \Omega$ satisfying $|A|,|B| < S$ uniformly at random (with replacement).
What is the probability that at least one of the following conditions will be satisfied?


*

*$|A \cap B| \geq b_1$

*$|A^c \cap B| \geq b_2$

*$|A \cap B^c| \geq b_2$

*$|A^c \cap B^c| \geq b_1$
Edit: I would be interested in lower bounds if there's no exact solution!
 A: Now I don't have time to investigate your problem and I don't know how far I'll be able to advance in it, so I provide the calculations below in a hope that it can be helpful to somebody to proceed further and finally to win the bounty. 
Put $T=|\Omega|$. A total number of possible choices of a pair $(A,B)$ is $$N=\left(\sum_{i=0}^{S-1} {T\choose i}\right)^2.$$ Given a set $a=|A|$, a number of possibilities to choose a set $B$ not satisfying any of the conditions is
$$M(a)=\sum\left\{ {a\choose i} {{T-a}\choose j}:i,j\ge 0, a-b_2<i<b_1, T-b_1<j<b_2, i+j<S\right\}.$$
So the total number of possibilities to choose a set $B$ not satisfying any of conditions is 
$$M=\sum_{a=0}^{S-1} {T\choose a} M(a).$$
With these formulae we can estimate the required probability $1-\tfrac MN$ for values of parameters in which we are interested by a computer. Since with fixed $n$ binomial coefficients ${n\choose k}$ grows very fast, I expect that $M$ is almost equal to its largest summand $${T\choose a}{a\choose i} {{T-a}\choose j}=\frac{T!}{i! j!(a-i)!(T-a-j)!},$$
that is
$$M\approx \max\left\{\tfrac{T!}{i!j!(a-i)!(T-a-j)!}: 0\le a<S, i, j\ge 0, a-b_2<i<b_1, T-b_1<j<b_2, i+j<S\right\}.$$
