Chapter 4.6, Problem 99E

If λ > 0 and α is a positive integer, the relationship between incomplete gamma integrals and sums of Poisson probabilities is given by

$\frac{1}{\text{Г(}\alpha \text{)}}{\displaystyle {\int }_{\lambda }^{\infty }{y}^{\alpha -1}{e}^{-y}dy={\displaystyle \underset{x=0}{\overset{\alpha -1}{\sum }}\frac{{\lambda }^x{e}^{-\lambda }}{x!}}}$

a If Y has a gamma distribution with α = 2 and β = 1, find P(Y > 1) by using the preceding equality and Table 3 of Appendix 3.

b    Applet Exercise If Y has a gamma distribution with α = 2 and β = 1, find P(Y > 1) by using the applet Gamma Probabilities.