# Proof of geometric probability distribution

• February 14th 2010, 06:40 PM
Matharch
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.

$\sum_{ y=1}^{ \infty } {q}^{ y-1 }p=1$
• February 14th 2010, 07:51 PM
Focus
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.