Question

Probability

Answer 1

@Ashleyisakitty

Answer 2

\[
E[h(X)] = \sum_{x}h(x)f(x)
\]In our case: \[
E[X^{-1}] = \sum_{x=1}^{\infty }\frac{(1-p)^{x-1}p}{x}
\]