TITLE: What is the use of the poisson distribution in this exercise when the question seems to have no relation
QUESTION [0 upvotes]: In one exercise they give me
1) The probability function of having k failures in a machine: $$P_k=\dfrac{1}{ek!}\\$$ where $$k=1,2,3,... (infinity)\\$$
2) The probability that the machine stop working having k failures: $$1-\left(\dfrac{1}2\right)^k\\$$ (here they dont give me a $$P_k=\\$$)
And then, they ask the question: What is the probability that the machine wont stop working?
For me solution is a simple SumP(X=x)=1, then $$1-\left(1-\left(\dfrac{1}2\right)^k\right)\\$$ and the final solution is $$\left(\dfrac{1}2\right)^k\\$$
I dont understand why I need the first poisson probability distribution. 
Thanks for your time and help.

REPLY [2 votes]: What you need to sum is the probability of $k$ failures times the probability that the machine stops given $k$ failures.  The final solution cannot depend on $k$.  The chance the machine stops is then
$$\sum_{k=0}^\infty \frac 1{ek!}\left(1-\frac 1{2^k}\right)$$
and you need to evaluate the sum.
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