Let XXX follow a geometric distribution P(X=k)=(1−p)k−1pP(X=k) = (1-p)^{k-1}pP(X=k)=(1−p)k−1p. What is P(X is even)P(X \text{ is even})P(X is even)?
1−p2−p\frac{1-p}{2-p}2−p1−p
p2−p\frac{p}{2-p}2−pp
1−p2+p\frac{1-p}{2+p}2+p1−p
pp+1\frac{p}{p+1}p+1p