Q:

Probabilities of same number occurrence in choosing two set of integers

$A$ has a set of integers, $\{1, \ldots , 1000\}$.
$B$ has another set of integers (identical to $A$'s), $\{1, \ldots , 1000\}$.
A draws randomly $50$ integers from $A$'s set without replacement.
B draws randomly $100$ integers from $B$'s set without replacement.
What is the probabilities that A and B have $1,2,\ldots,49,50$ same number(s) respectively?

I feel this problem is somewhat related to hypergeometric distribution, but have no idea how to fit it into this case.

A:

Let random variable $X$ be the number of numbers that A and B have in common. The distribution of $X$ is indeed hypergeometric.
Imagine (it makes no difference) that A draws her numbers first.
Call the $50$ numbers drawn by A good. We want the probability that $X=k$. So we want the probability that among B's numbers, there are exactly $k$ good, and therefore $100-k$ bad.
There are $\binom{1000}{100}$ equally likely choices for B. There are $\binom{50}{k}\binom{950}{100-k}$ choices in which $k$ are good and $100-k$ are bad. Divide.

