Do set intersection cardinalities depend on one another? Let sets $A,B,C,D$ be sets of numbers without repetitions in the range $[1,…,n]$. I'm assuming that the sizes of sets were pre-determined ($|A|=a$, $|B|=b$, $|C|=c$, $|D|=d$, and $a,b,c,d$ can't be $0$), and the sets were chosen uniformly on $1,...,n$. So, given the sizes of the sets, are the sizes of the pair-wise intersections dependent on one another? And what about the sizes of the triple-wise intersections? In other words, knowing that $|A|=a,|B|=b,|C|=c,|D|=d$, are the following equations correct?
$P(|A\cap B|=k \land |A\cap C|=w) = P(|A\cap B|=k)\cdot P(|A\cap C|=w)$
$P(|A\cap B\cap C|=k \land |A\cap B \cap D|=w) = P(|A\cap B\cap C|=k)\cdot P(|A\cap B\cap D|=w)$
I tried to think about this in terms of d-separation and it seems that maybe they are independent of one another (so the equations are correct), but I'm not sure.
I would appreciate any input, thanks in advance.
 A: 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
$$A=A'\cap B'\\
B=C'\\
C=D'
$$
