Infinite coin flipping Show $ sum_{k=0}^{infty} f(k,r,p) = 1 $
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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} $ . ...