I (think I) solved the problem of finding the distribution of random variable $Y=k$ denoting the largest out of $m\leq n$ balls, drwan out of an urn containing all balls labled $1,\dots,n$, without replacement.
Using combinatorics I'm quite sure that $\Pr\left(Y=k,X=m\right)=\begin{cases} 0 & m>k\\ \frac{{k-1 \choose m-1}}{{n \choose m}} & m\leq k \end{cases}$.
The thing is I got stuck when I tried to find the marginal distribution of $Y$: $ \Pr\left(Y=k\right)=\sum_{m=1}^{n}\Pr\left(Y=k,X=m\right)\\=\sum_{m=1}^{k}\Pr\left(Y=k,X=m\right)+0\\=\sum_{m=1}^{k}\frac{{k-1 \choose m-1}}{{n \choose m}}$
Assuming my method was correct (and please tell me if it isn't so), how can I simplify the last expression? Thanks in advance.
On the original question, my approach would be
$\Pr(Y=k)=\Pr(Y \le k)-\Pr(Y \le k-1) = \dfrac{k \choose m}{n \choose m }-\dfrac{k-1 \choose m}{n \choose m }= \dfrac{k-1 \choose m-1}{n \choose m }$
for $m \le k \le n$ (and zero otherwise).
This looks like your expression for $\Pr\left(Y=k,X=m\right)$, but I think it answers the question