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.
Best Answer
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 $\approx0.98981$.