Joint Distr.—3 independent R.V with Poisson distr. X,Y,Z...E(X)=3, E(Y)=1, E(Z)=4, P(X+Y+Z<=1)?

Question

Let X,Y,Z be independent Poisson random Variables with E(X)=3, E(Y)=1, and E(Z)=4. What is P(X+Y+Z<=1)?

What I figured:

(e^(-3)e^(-1)e^(-4)*3^x*1^y*4^z)/(x!y!z!)

Given the discrete PMFS we could just do the sums of (1,0,0)+(0,1,0)+(0,0,1)

e^(-8)*3^x*1^y*4^z/(x!y!z!) e^(-8)*(3+1+4).....which is 8e^-8, but my answer is slightly off.

The Correct Answer is 9e^(-8).

Can someone explain how to get that answer?

Answer 1

You can use Probability Generating Function of Poisson Distribution $$\left\{ \begin{align} & {{M}_{X}}(t)={{e}^{{{-\lambda }_{1}}(1-t)}} \\ & {{M}_{Y}}(t)={{e}^{{{-\lambda }_{2}}(1-t)}} \\ & {{M}_{Z}}(t)={{e}^{{{-\lambda }_{3}}(1-t)}} \\ \end{align} \right.\Rightarrow {{M}_{X+Y+Z}}(t)={{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})(1-t)}}$$ Indeed $X+Y+Z$ has a Poisson($\lambda_1+\lambda_2+\lambda_3$) distribution.We have $$P(X+Y+Z\le 1)=P(X+Y+Z=0)+P(X+Y+Z=1)=\frac{{{8}^{0}}{{e}^{-8}}}{0!}+\frac{{{8}^{1}}{{e}^{-8}}}{1!}=9{{e}^{-8}}$$