# Supposedly easy hypergeometric probability question? Are my answers right to parts i and ii and how to do iii?

Fix positive integers $r_{1} \leq K_1$ and $r_{2} \leq K2$ and let $N=K_1+K_2$ and $n=r_1+r_2$. A subset $S \subseteq \{1,2, \ldots, N\}$ having $n$ elements is chosen.

i) How many possibilities are there for $S$?

ii) How many possibilities are there for $S$ that satisfy $|S \cap \{1,2,3,\ldots,K_1\}|=r_1$ and $|S \cap \{K_1+1,K_1+2,K_1+3, \ldots,N\}|=r_2$? Explain. ($\cap$ here means intersection)

iii) If $S$ is chosen uniformly at random from amongst all subsets of $\{1,2,...,N\}$ having $n$ elements what is the probability that it satisfies $|S \cap \{1,2,3,\ldots,K_1\}|=r_1$?

For i) I got $2^n$ possibilities because each element is either in or out so two choices. For ii) I got that $S$ can only equal the full set $\{1,2,3,\ldots,N\}$ in order to satisfy both the given equations.

I'm unsure about part (iii)$\ldots$

Your analysis is not quite right. Essentially the integers $1$ through $N = K_1+K_2$ have been divided into two buckets: $1$ through $K_1$, and $K_1+1$ through $N$.
Part (i) essentially asks how many different ways are there to draw $n$ numbers out of the whole collection of $N$ numbers. The answer to that is the simple binomial $\binom{N}{n}$.
Part (ii) essentially asks how many different ways are there to draw $r_1$ numbers out of the $K_1$ numbers in the first bucket, and also $r_2$ numbers out of the $K_2$ numbers in the second bucket. The answer to that is the product of two binomials, $\binom{K_1}{r_1}\binom{K_2}{r_2}$.
Part (iii) essentially asks for the conditional probability that, given that one picks $n$ numbers out of the whole collection of $N$ numbers (the answer to part (i)), what is the probability that the selection has $r_1$ out of the first bucket (the answer to part (ii), since there must be $n-r_1 = r_2$ numbers in the second bucket). The answer here is therefore the ratio of your two answers:
$$\frac{\binom{K_1}{r_1}\binom{K_2}{r_2}}{\binom{N}{n}}$$