Consider a geometric distribution X∼Geom(p)X \sim \text{Geom}(p)X∼Geom(p) where P(X=k)=(1−p)k−1pP(X=k) = (1-p)^{k-1}pP(X=k)=(1−p)k−1p. Find P(X>4∣X>2)P(X > 4 | X > 2)P(X>4∣X>2).
(1−p)2(1-p)^2(1−p)2
(1−p)4(1-p)^4(1−p)4
1−p1-p1−p
p2p^2p2