A company buys a policy to insure its revenue in the event of major snowstorms that shut down business. The policy pays nothing for the first such snowstorm of the year and $\mathrm{\$}9300$ for each one thereafter, until the end of the year. The number of major snowstorms per year that shut down business has a Poisson distribution with mean 1.9.
Find the expected amount paid to the company under this policy during a one-year period.

Step 1 ：Let X be the number of major snowstorms per year. X follows a Poisson distribution with $\lambda =1.9$. Then the expected number of snowstorms is equal to $E[X]=1.9$.

Step 2 ：Define Y as the number of major snowstorms with payments, so $Y=max(0,X-1)$. Calculate the expected value of Y: $E[Y]=\sum _{y=0}^{\mathrm{\infty }}y\cdot P(Y=y)=\sum _{y=1}^{\mathrm{\infty }}(y-1)\cdot P(X=y)$

Step 3 ：Calculate E[Y]: $E[Y]=\sum _{y=2}^{\mathrm{\infty }}(y-1)\cdot \frac{e^{-1.9}\cdot {1.9}^y}{y!}=e^{-1.9}\cdot \sum _{y=2}^{\mathrm{\infty }}\frac{{1.9}^y}{(y-1)!}$

Step 4 ：Recognize that the remaining sum is the sum of a Poisson distribution with mean ${\lambda }^{\prime }=1.9$ shifted by one unit: $E[Y]=e^{-1.9}\cdot (\sum _{y=1}^{\mathrm{\infty }}\frac{{1.9}^y}{y!}-\frac{1.9}{1!}-\frac{1}{0!})=e^{-1.9}\cdot (e^{1.9}-1.9-1)$

Step 5 ：Calculate the expected amount paid under the policy during a one-year period: $E[Paid]=\mathrm{\$}9300\cdot E[Y]=\mathrm{\$}9300\cdot (e^{-1.9}\cdot (e^{1.9}-1.9-1))=\mathrm{\$}12127.5764$