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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 $L = |\{1, 2, 4, 5, 6\}| = 5$

2 Answers 2

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  1. See, that $\min{(N_A, N_B)}\leq L \leq \min{(|S|,N_A+ N_B)}$
  2. 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
  3. Total number of sets $A$ and $B$ is $$|\Omega|={|S|\choose N_A}{|S|\choose N_B}$$
  4. We have then: $$P(L)=\begin{cases}\frac{{|S|\choose L}{L\choose N_A}{N_A \choose N_A+N_B-L}}{{|S|\choose N_A}{|S|\choose N_B}} &, \min{(N_A, N_B)}\leq L \leq \min{(|S|,N_A+ N_B)}\\ 0&, \text{in other cases}\end{cases}$$
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W.L.O.G. suppose $N_A \ge N_B$

If $L

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 $$\frac{{N-N_A \choose L-N_A} \times {N_A \choose N_B+N_A-L}}{{N \choose N_A}\times{N \choose N_B}}$$