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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.

gbi1977
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    That last step looks wrong. Do you think $\sum a_ib_i=(\sum a_i)(\sum b_i).?$ – Thomas Andrews Apr 23 '18 at 00:05
  • What does $X$ represent? Is it a random variable or just the number of balls drawn? – Henry Apr 23 '18 at 00:13
  • @Henry $X$ represents the number of balls drawn. It's not actually random, because for every $m$ it has probability of 1... this is where I'm wrong maybe? – gbi1977 Apr 23 '18 at 00:21
  • @ThomasAndrewsT you're absolutely right. Edited and changed the question. – gbi1977 Apr 23 '18 at 00:24
  • If you know there are $m$ balls, I would have thought you do not need to sum over $m$ from $1$ to $k$ – Henry Apr 23 '18 at 07:05
  • @Henry the formal explanation as to why $m$ shouldn't be defined as a random variable is because it determines different sample spaces? – gbi1977 Apr 23 '18 at 20:40

1 Answers1

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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

Henry
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    Incidentally, this tells you $\sum_{k=m}^n {k-1 \choose m-1} = {n \choose m}$ which can be rewritten as $\sum_{k=m}^{n} {k \choose m} = {n+1 \choose m+1}$ – Henry Apr 23 '18 at 07:11