What is the expected waiting time to have at least k members, knowing that each member is drawn from a population size N with the probability of p? Consider a population of size N.
Each day, each member of the population is drawn without replacement with the probability of p and put to the pool. The pool is initially empty. 
So the first day, maybe no members are selected. In the 2nd day, there are maybe 2 are selected...
What is the expected time when the size of the pool is k?
 A: Two ways to approach the problem:
Geometric order statistics  For each member of the population, define (independent) random variables $T_i$, which is the day that member exits to the pool. $T_i$ has geometric distributions,
$$ \DeclareMathOperator{\P}{\mathbb{P}}
   \P(T_i=k)=(1-p)^k \cdot p, \quad k=1,2,3,\dotsc
$$
Then define the order statistics
$$
  T_{(1)} \le T_{(2)} \le \dotsm \le T_{(N)}
$$
and the interest is in the order statistic $T_{(k)}$, which is the day the pool reaches size $k$.  So you need to find the expected value of geometric order statistics ...
Markov chains Let the chain be $X_t$, the number of members in the pool on day $t$, starting with $X_0=0$. The transition probabilities wil be binomial,
$$
   \P(X_{t+1}=u+k \mid X_t=u)= \binom{N-u}{k} p^k (1-p)^{N-u-k} 
$$
But my Markov chains are too rusty to try to develop this any more now. Maybe the first approach is more hopeful ... but this points to an interesting relation between (discret) order statistics and Markov chains!
