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I've this issue with this Poisson distribution. I need to calculate the probability that there will not be more than one failure during a particular week.

Given electricity power failures occur according to a Poisson distribution with an average of 3 failures every twenty weeks.

I've found this solution

The average number of failures per week is μ = 3/20 = 0.15 "Not more than one failure" means we need to include the probabilities for 0 failures plus 1 failure

P(X0) + P(X1) = (e-0.15 * 0.150) / 0! + (e-0.15 * 0.151) / 1!

= 0.98981

Nevertheless, I don't get where 0,150 and 0.151 come from ? Besides, I always get 1 for a probability of P(xO)!

Thank you if you can help.

Carlos Carvalho
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1 Answers1

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What you have written as answer does not make much sense and is hard to read.

Have a look at MathJax basic tutorial.

If $X$ is a random variable having Poisson-distribution with parameter $\mu$ then:$$P(X=k)=e^{-\mu}\frac{\mu^k}{k!}\text{ for }k=0,1,2,\dots$$

So the answer is: $$P(X=0)+P(X=1)=e^{-\mu}\frac{\mu^0}{0!}+e^{-\mu}\frac{\mu^1}{1!}=e^{-\mu}(1+\mu)$$

Just substitute $\mu=\frac3{20}=0.15$ and you will find $\approx0.98981$.

drhab
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  • I agree, the answer was not very clear, but I still don't get how that value comes from. When I plug the values into excel, I get a completely different results, (i.e. P(X=0) is 1 ? and P(X=1) = 0,16. SO my question still, how find the P(X=0) ? – Carlos Carvalho Jun 10 '19 at 11:19
  • "So my question still is how to find $P(X=0)$". In my answer you are told how to get that answer. I have no sight on your excel sheet, so of course I cannot tell you what is wrong with that. – drhab Jun 10 '19 at 12:48