A urn containing $n$ balls, numbered $1,2,...,n$, and $k$ balls are chosen at random without replacement. I have a homework question with specific $n$ and $k$ given for the below question, but I would rather understand how this works for any given $n$ or $k$ to build my intuition for these questions.
A urn containing $n$ balls, numbered $1,2,...,n$, and $k$ balls are chosen at random without replacement. Let $X$ be the largest-numbered ball removed. Determine the probability function of $X$.
$X$ takes values from $S$
My thoughts: In the number set from $1,2,...,k-1$, all of these numbers have zero probability of being the greatest number. We then have probability of not drawing a specific ball(say ball $x\in S$) $\frac{n-1}{n}*\frac{n-2}{n-1},...,\frac{n-k}{n-k+1}$ = $Pr(Q)$. Then the probability thus of drawing that specific ball is merely $1 - Pr(Q)$
The probability of each number being the greatest goes up towards number $n$. So $1 - Pr(Q)$ should actually be the probability $n$ is the greatest number drawn. But that doesn't workout mentally, because it would seem the probability that ball numbered $n-1$ is the largest number drawn, would be $1 - pr(Q)$ take away the probability that ball numbered $n$ is drawn, which has the same probability, and therefore the probability of getting ball numbered $n-1$ would be zero.
Any tips would be greatly appreciated!

The largest, as you pointed out, is one of $k,k+1,\dots,n$. 
There are $\binom{n}{k}$ equally likely ways to choose $k$ balls.
Let us count the number of "favourables," that is, the number of ways to choose $k$ balls with $m$ the largest. 
So we must choose $k-1$ balls from the balls $1,2,\dots,m-1$ to keep $m$ company. Or perhaps to be bullied by $m$.  There are $\binom{m-1}{k-1}$ ways to do this.
Now you should be able to write down an expression for $\Pr(X=m)$.
