The problem is as follows:
Let S be a set of n elements. At the first stage each element in S is in- dependently removed with probability p. Those elements not removed constitute the set S1. If S1 is not the empty set then each of its ele- ments is independently removed with probability p, with the remaining elements constituting the set S2, and so on. Let N be the least k such that Sk is empty. Give a formula for the probability that N = r. Hint: N can be expressed as the maximum of a set of independent geometric random variables.
Here's my thought so far: let 1st element attain the max, then $P(N = r) = P(X_1 = r) \prod_{i = 2}^n P(X_i <= r)$, where the $X_i \sim Geom(p)$. However, I'm unsure of the above as there isn't a rigorous chain of steps that I'm following.
EDIT: The following is my latest thought which seems more correct than the above. Let $k$ denote the number of elements that survive until the last round. Then, $P(N = r) = \sum_{k = 1}^n {n \choose k} [P(X = r)]^k [P(X < r)]^{n - k}$, where $X \sim Geom(p)$.
Let the elements of $S$ be $s_1$ to $s_n$. For $i=1$ to $n$, define random variable $X_i$ by $X_i$ is the number of steps until $s_i$ is removed. Then $X_i$ has geometric distribution with parameter $p$. The number $N$ of steps until all the $s_i$ are gone is then $\max(X_1,\dots,X_n)$.
The probability that $X_i$ is $\gt r$ is $(1-p)^r$, so the probability that $N\gt r$ is $(1-p)^{rn}$, which is $((1-p)^n)^r$. It follows that $N$ has geometric distribution with parameter $1-(1-p)^n$.