A newspaper dealer gets $n$ newspapers every day for sale. The number $X$ of papers sold is a random variable following the Poisson distribution with parameter $\lambda.$
I am interested in computing $P(X=k)$ which I think should be: $$P(X=k) = \frac{e^{-\lambda} (\lambda)^k}{k!}$$ for $k=0,1,2,...,n.$ But one can immediately notice that the probabilities will not sum up to $1.$ They could if one did the following: $$P(X=k) = \frac{e^{-\lambda} (\lambda)^k}{k!}/\sum_{i=0}^{n}\frac{e^{-\lambda} (\lambda)^i}{i!}.$$ But if one were to come up with this expression from first principles then I am not sure how one would do so. Perhaps someone can explain how derive $P(X=k)?$
Aren't you over thinking it?
I would model this as $X \sim Pois(\lambda)$ is demand for papers and $Y$, the number of papers sold, as being equal to $Y = \min (X,n)$. Since if demand exceeds $n$, then you sell all $n$ papers, but no more. I'm not sure it makes sense to model the papers sold as being Poisson if there is a cap on the number sold, but it certainly makes sense to model the demand for papers as Poisson, since it is essentially a queuing problem.
Under this you would have
$P(Y = k) = \frac{\lambda^k}{k!}e^{-\lambda}$ for $k = 0,1,2,\dots, n-1$
But
$P(Y = n) = \sum_{k=n}^\infty \frac{\lambda^k}{k!}e^{-\lambda}$