With/out replacement ball drawing distribution So the question asks if an urn contains n balls labeled 1,2,..., n. We draw m balls from the urn, one at a time. Let Y be the largest label of a ball in the draw. Find the distribution of Y (including the range) in case:
(a) The balls are drawn without replacement (and m ≤ n).
(b) The balls are drawn with replacement.
So the first thing I am confused about is, what does it mean by "find the distribution of Y"? Am I supposed to do like, when Y <1, p=0, when 1≤Y≤n, p is XXX and when Y=n, p=1 such things? 
So what I have is, 
Given that urn contains n balls labeled 1,2,3,.....n
I have to select m balls out of n balls and Y be the largest label of a ball in the draw
The distribution of Y in case the balls are drawn without replacement:
P(X=Y) = $\frac{1}{n} + \left ( (\frac{m}{n})(\frac{1}{n-1}) \right ) + \left ( (\frac{m}{n}) (\frac{m-1}{n-1})(\frac{1}{n-2})\right ) + ........+\left ( (\frac{m}{n})(\frac{m-1}{n-1})(\frac{m-2}{n-2}) \right )+ ..........=1$
The distribution of Y in case the balls are drawn with replacement:
P(X=Y) = $\frac{1}{n} + \left ( (\frac{m}{n}) (\frac{1}{n})\right ) + \left ( (\frac{m}{n})(\frac{m}{n}) (\frac{1}{n})\right ) + ....+ \left ( \frac{m}{n} . \frac{m}{n}.\frac{m}{n}....\frac{m}{n}\right ).\frac{1}{n}$
       = $\frac{1}{n}\left [ 1 + \frac{m}{n} +(\frac{m}{n}) ^{2} +(\frac{m}{n}) ^{3} +.....+ (\frac{m}{n}) ^{m} \right ]$
Do these look right? But I still do not have a clue about how to write the distrution. Any clue will be appreacited!
 A: Finding the distribution of Y would be finding the values of $Pr(Y\le y)$ for any value $y$. You know from the problem these probabilities are determined by the values at $y=1,\ldots,n$. For the case of drawing with replacement, you can think of $Pr(Y\le y)$ as the probability of selecting $m$ balls out of $n$ all of which are labeled less than $y$. That is ${y\choose m}/{n\choose m}$, all the choices of $m$ balls among those labeled $1,\ldots,y$ divided by all possible m draws with replacement. For the case of drawing with replacemnt, you can think of it as the probability that the maximum value of $m$ independent samples is less than $y$. The probability that a single draw is less than $y$ is $y/n$, so by independence the probability that $m$ are less than $y$ is $(y/n)^m$.
A: When drawing $m$ balls with replacement, the probability that the maximum is equal to a value ($k$) is: the probability that all balls are at most that value minus the probability that all balls are less than that value.
$$\mathsf P(Y=k) = \mathsf P(\max\{X_i\}_m\leq k)-\mathsf P(\max\{X_i\}_m\leq k-1)$$
When drawing with replacement, you need to find the probability of selecting a ball of that value and $m-1$ balls of the $k-1$ less than that value, out of all the ways to select any $m$ of $n$ balls.
