There are n balls in an urn. They are labelled 1, 2, . . . , n. We randomly pick k balls (without replacement), one by one. Find the probability that the label on the kth ball is larger than on all previously picked balls.
I guess this is related to hypergeometric distribution but I do not know how to apply it here.
On
There are $n$ balls, out of which you pick $k$.
This means that you are asking for the probability that $(\forall x\in \Bbb N)[n_k\gt n_{k-x}]$.
If the highest ball is $k$, then there are $^{k-1}\text P _{k-1}=(k-1)!$ methods to arrange the balls before it. If the highest is $k+1$, then $^k\text P _{k-1}$, and so on till $n$ and $^{n-1}\text P _{k-1}$.
So the probability is $${\sum ^n _{i=k-1} {i!\over (i-k+1)!}}\over {n!\over (n-k)!}$$
If you sort all your random balls $b_1,...b_k$ in increasing order the probability that the last one is $b_k$ is $1/k$.