Do set intersection cardinalities depend on one another?

Solution 1:

We have a a set of $n$ elements and a subset $A$ of $a$ elements, so $|A|=a$. Now we choose randomly a subset $B$ with $b$ elements, 'Randomly' means that all subsets have the same probability to be chosen. There are $$\binom n b$$ ways to choose $B$. If we want that the set $B$ has exactly $k$ elements in common with $A$, there are $\binom a k$ ways to select the common elements from $A$, and $$\binom {n-a} {b-k}$$ the remaining elements such that they are not from $A$- So it is necessary that $$a\ge k \\ n+k\ge a+b$$ otherwise we cannot find such sets. These remaining element are independently chosen from the elements that we have chosen from $A$. So the number of sets $B$ is $$\binom a k \binom {n-a} {b-k}$$ The number of ways we can chose $B$ without tke into account $A$ is $$\binom n b$$ From this follows $$P(|A\cap B|=k \;\Big |\;|B|=b)=\frac {\binom a k \binom {n-a} {b-k}} {\binom n b}\tag 1$$

If we select $B$ and $C$ such that $|B|=b$ and $|C|=c$ and $A$ and $B$ have $k$ elements in common and $A$ and $C$ have $w$ elements i common we have $$\binom a k \binom {n-a} {b-k}$$ ways to choose $B$ $$\binom a w \binom {n-a} {c-w}$$ ways to choose $C$ and so $$\binom a k \binom {n-a} {b-k} \binom a w \binom {n-a} {c-w}$$ to select a pair $B,C$ that satisfy the condition, because choosing the elements of $B$ does not influence the selection of the elements of $C$ So all in all we have

$$P(|A\cap B|=k \land |A\cap B|=w \;\Big |\;|B|=b \land |C|=c )=\frac {\binom a k \binom {n-a} {b-k} \binom a w \binom {n-a} {c-w}} {\binom n b \binom n c} \tag 2$$

From $(1)$ follows $$P(|A\cap B|=k \;\Big |\;|B|=b) \cdot P(|A\cap C|=w \;\Big |\;|C|=c) =\frac {\binom a k \binom {n-a} {b-k}} {\binom n b} \cdot \frac {\binom a w \binom {n-a} {c-w}} {\binom n c}$$ which is equal to $(2)$.

The case of for sets $A',B',C',D'$ an be reduced to this case of three sets if we set $$A=A'\cap B'\\ B=C'\\ C=D' $$