I have the following problem
A large airline company has observed that on average $3$ of its airplanes present some kind of damage and put out of service over a week. In order to preserve its fame, the company has aways airplanes available which use to replace the one that presents some kind of damage. What is the minimum number of airplanes the airline must have, so with probability $0.95$, there is an available airplane to replace some airplane that presents some kind of damage.
If I let $\mathbf X$ to be the number of airplanes that present some kind of damage then I know that $\mathbf X$ follows a Poisson distribution with parameter $3$. But I do not know how to find the minimum number of airplanes.
$X$ has a distribution that follows $e^{-3}(3^k)/k!$ where $k$ is the number of planes that get damaged.
Needing a replacement plane is the same as the plane breaking down. So, we need to find the probability that we need 0 planes, 1 plane, 2 planes, 3 planes, etc.. So, we are looking for $P(X < x) = .95$. So, we sum up the probabilities, until we reach .95 as follows (I disregarded the first few sums, as they were too small) \begin{array}{ccc} k =\_ & Probability & Sum \\ k=0 & 0.04978 &\\ k =1 & 0.14936& \\ k =2 & 0.22404& \\ k =3 & 0.22404& 0.64722 \\ k =4 & 0.16803& 0.81525\\ k=5 & 0.10081& 0.91606\\ k = 6 & 0.05041& 0.96647\\ \end{array}
If we keep 6 replacement planes around, the probability that we will need all of them is .95. This ensure that there is a replacement plane with probability .95.