combinatorics - Probability to choose two disjoint subsets, of fixed length, from a set with specific properties

I came through the following combinatoric problem.
Say we have the set $[n]=\{0,1,...,n\}$ and we choose two subsets $I_1,I_2$ randomly from $[n]$ such that $I_1\cap I_2=\emptyset.$ Also $|I_1|=|I_2|=b<(n+1)/2.$ Further, we fix a subset $P$ of $[n]$ with $|P|=h_1+h_2< b.$
I want to compute
$$Pr=Pr(I_1,I_2\xleftarrow{R} [n]: |I_1\cap P|=h_1, |I_2\cap P|=h_2, I_1\cap I_2=\emptyset)$$
By inspection this is equal to
$$Pr=\frac{\binom{b}{h_1}\binom{b}{h_2}}{\binom{n+1}{h_1+h_2}}.$$
Also we can see that $$|\{I_i\subset [n]:|I_i\cap P|=h_i\}|=\binom{n+1-h_i}{b-h_i}\ (i=1,2).$$

Any hint how to prove the formula for $Pr$?

Answer

Assuming you meant unbiasedly selecting $b$ and $b$ elements from $[n]$ into $I_1, I_2$ without replacement (so $I_1\cap I_2=\emptyset$ is given).

Let $Q=[n]\smallsetminus P$

You seek the probability that the $b$ elements of $I_1$ are a selection of $h_1$ from the elements in $P$ and $b-h_1$ from the elements in $Q$, and that the $b$ elements in $I_2$ are a selection of the remaining $h_2$ elements in $P$ and $b-h_2$ from the remaining elements in $Q$.

That is, the probability of partitioning a subset of size $h_1+h_2$ (ie $P$) into parts of size $h_1$ and $h_2$, and a subset of size $n+1-h_1-h_2$ (ie $Q$) into parts of size $b-h_1,b-h_2,$ and $n+1-2b$, when partitioning the whole set of size $n+1$ into parts of size $b, b,$ and $n+1-2b$.

$\require{cancel}\begin{align}\mathbb P\big(\lvert I_1\cap P\rvert = h_1, \lvert I_2\cap P\rvert=h_2\big) ~&=~ \dfrac{\dbinom{h_1+h_2}{h_1, h_2} \dbinom{n+1-h_1-h_2}{b-h_1,b-h_2,n+1-2b}}{\dbinom{n+1}{b,b,n+1-2b}} \\[2ex] &=~\dfrac{\dfrac{(h_1+h_2)!~(n+1-h_1-h_2)!}{h_1!~h_2!~(b-h_1)!~(b-h_2)!~\cancel{(n+1-2b)!}}}{\dfrac{(n+1)!}{b!~b!~\cancel{(n+1-2b)!}}}\\[2ex] &=~ \dfrac{\dfrac{b!~b!}{h_1!~(b-h_1)!~h_2!~(b-h_2)! }}{\dfrac{(n+1)!}{(h_1+h_2)!~(n+1-h_1-h_2)!}} \\[2ex] &=~ \dfrac{\dbinom{b}{h_1, b-h_1 }\dbinom{b}{h_2, b-h_2 }}{\dbinom{n+1}{h_1+h_2,n+1-h_1-h_2}} \\[2ex] &=~\dfrac{\dbinom{b}{h_1}\dbinom{b}{h_2}}{\dbinom{n+1}{h_1+h_2}}\end{align}$

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Remark: This is also the probability for a selection of $h_1$ from the $b$ elements in a fixed $I_1$ and $h_2$ from the $b$ elements in $I_2$, when selecting, without bias or replacement, $h_1+h_2$ elements for $P$ from the $n+1$ elements in $[n]$.   An equivalent task.