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Thread: Proof of geometric probability distribution

  1. Feb 14th 2010, 05:40 PM #1
    Matharch
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    Proof of geometric probability distribution

    I would need help with this proof. It seems any way I try to prove I get stuck or mess up somewhere, I know it's easy but basically I suck at proving stuff. So a bit help would be appreciated.

    \sum_{ y=1}^{ \infty } {q}^{ y-1 }p=1
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  2. Feb 14th 2010, 06:51 PM #2
    Focus
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    Well I assume that q=1-p, then you have that
    \sum_{n\geq 0} q^n p=p\sum_{n\geq 0} q^n=\frac{p}{1-q}=\frac{p}{p}.

    If you want to prove what the sum of a geometric series is then you can look it up on wikipedia.
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