For a infinite coin flip consider the probability of success $ p \in (0,1) $.
For $n \in \mathbb{N}_0 $ and $ r \in \mathbb{N} $ define the probability $f(k,r,p)$ ( $k$ failures then the $r$th sucess ).
Show:
$ \sum_{k=0}^{\infty} f(k,r,p) = 1 $.
So I've already have a possible solution but I want to know if my solution is ok. So can you please check my attempt?
I want to ask a tiny question before I try to prove the equation: $f$ is a generalization of a geometric distribution, right? Anyways:
So $ \sum_{k=0}^{\infty} f(k,r,p) = \sum_{k=0}^{\infty} {k+r-1 \choose k}p^r(1-p)^k = p^r \sum_{k=0}^{\infty} {k+r-1 \choose k}(-1)^k(-(1-p))^k =p^r \sum_{k=0}^{\infty}{-r \choose k}(-(1-p))^k = p^r(1-(1-p))^{-r} = p^rp^{-r} = 1.$
What did I do in the third equation? : ${k+r-1 \choose k}(-1)^k = {-r \choose k} $ .