For a Geometric distribution P(X=k)=(1−p)k−1pP(X=k) = (1-p)^{k-1}pP(X=k)=(1−p)k−1p for k=1,2,…k=1, 2, \dotsk=1,2,…, what is the probability P(X>2)P(X > 2)P(X>2)?
(1−p)2(1-p)^2(1−p)2
1−p21 - p^21−p2
(1−p)p(1-p)p(1−p)p
p2p^2p2