The number of daily flaws of the machines of a factory is given by a Poisson variable with mean 3. Each flaw is fixed by one maintenance employee in a whole day. How many maintenance employees are needed if they want to fix every flaw with probability 0.9?
So far I've thought that if $X$ is the number of daily flaws then $X \sim Poisson(3)$
And so $X$ is also the number of employees needed.
So we would be looking for an n (number of employees) such that $\mathbb{P}(\mbox{ fixing every flaw }) =0.9$
Then, fixing every flaw only happens if X=n.
So we would be looking for a n such that $\mathbb{P}(X=n) =0.9$ i.e. $\frac{e^{-3}3^n}{n!}=0.9$
But that does not make sense, Or does it?