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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.

zhuli
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Limanido
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    I think you may have misunderstood the problem. For every $k$, the probability that the machine has $k$ failures is given. For every $k$ the probability that the machine having $k$ failures will stop working is given. Now they are asking - how likely is it that the machine will stop working altogether. In other words, the final result is just a number between $0$ and $1$, inclusive - it does not depend on any parameter $k$. –  Dec 27 '19 at 17:52
  • Can I look at it as a conditional probability?: P(the machine stop working / the failure)=P(the machine has failures and stop working)/P(the machine has failures). So if I look it as two independent events, P(the machine has failures and stop working) = P(the machine stop working)*P(the machine has failures) ? – Limanido Dec 27 '19 at 18:32
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    Sure, and then you sum over all $k$'s to get the total probability - and you end up with the sum as in the answer below. –  Dec 27 '19 at 18:33

1 Answers1

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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.

Ross Millikan
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