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.
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.