The number of times X that machines of a certain kind need rebooting in a month has Poisson distribution with parameter z. This parameter depends on the actual machine and therefore it is itself a random variable Z, which we are told by the manufacturer is exponentially distributed with parameter 1.
a) If your company has bought one of these machines, what is the probability that it will require to be rebooted exactly n times next month? You should simplify your answer until the only letter that appears in it is n.
b) What is the conditional probability that your company’s machine will need to be rebooted 3 times next month, given it didn’t need to be rebooted during the current month?
For a), I found $$P(X=n) = \int_{0}^\infty P(X=n|Z=z)f_z(z)dz = \frac{1}{n!}\int_{0}^\infty e^{-2z}z^ndz = \frac{\Gamma(n+1)}{n!2^{n+1}} = \frac{1}{2^{n+1}} $$
However, I can't quite conceptualize whether I'm done. Did I just calculate the marginal distribution of X, and now I need to sum this function over the possible values of n, or would that be the CDF?
Also, for b) , are draws from a Poisson distribution independent? In that case it would just be $P(X=3)$, but my intuition tells me that this is not true, since if the machine did not need to be rebooted this month, that would make it more likely the company has a machine that requires fewer reboots. Not sure how to conceptualize part b.
Best Answer
Your treatment of a) is correct; note in particular your probabilities sum to $1$.
For b) let $X_1,\,X_2$ denote the numbers of reboots required in consecutive months, so$$P(X_2=3|X_1=0)=\frac{P(X_2=3\land X_1=0)}{P(X_1=0)}=\frac{\int_0^\infty\frac16z^3e^{-3z}dz}{\int_0^\infty e^{-2z}dz}=\frac{2}{81}.$$You may wish to consider the similar problem of verifying $P(X_2\ge3|X_1=0)=\frac{1}{27}$; it comes down to summing a geometric series.