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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.

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    $\begingroup$ It's easier than that; you have drawn $k$ balls; the probability that the largest of these was drawn on any particular draw is the same... $\endgroup$ Commented Feb 4, 2018 at 7:03

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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 $${\sum ^n _{i=k-1} {i!\over (i-k+1)!}}\over {n!\over (n-k)!}$$

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If you sort all your random balls $b_1,...b_k$ in increasing order the probability that the last one is $b_k$ is $1/k$.

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