3.1.12 problem 14

(a)

$$1-P(X=0)=1-e^{-\lambda }$$

$$P(X\ge 2)=1-P(X=0)-P(X=1)=(1-e^{-\lambda })-e^{-\lambda }\lambda$$

(b)

$$P(X=k|X>0)=\frac{P(X=k)}{P(X>0)}=\frac{{\lambda }^k}{(e^{\lambda }-1)k!}$$