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I am returning to learning some probability after a 5-6 year hiatus. I am looking at a problem from my old college book, which I think I have solved, but I am surprised doesn't have a simpler answer. I want to see if anyone here can give a better answer. I see this form of problem come up a bunch, since its just a standard discrete PMF, but it is capped at some arbitrary spot.

Here is the problem.

An internet service provider uses 50 modems to serve the needs of 1000 customers. It is estimated that at a given time each customer will need a connection with probability $0.01$ independent of the other customers.

(a) What is the PMF of the number of modems in use at the given time?

(b) Repeat part (a) by approximating the PMF of the number of customers that need a connection with a Poisson PMF.

(c) What is the probability that there are more customers needing a connection than there are modems? Provide an exact as well as an approximate formula based on the Poisson approximation of part (b).

My answer is as follows.

(a) Using $p$ for the PMF. $p(k) = \binom{1000}{k}(0.01)^k(0.99)^{1000-k}$ for $k < 50$ and $p(50) = 1-\sum_{i=0}^{49}\binom{1000}{i}(0.01)^i(0.99)^{1000-i}$.

(b) $p(k) = \frac{10^ke^{-10}}{k!}$ for $k < 50$ and $p(50) = 1 - \sum_{i = 0}^{49} \frac{10^ie^{-10}}{i!}$

(c) Just have to remove the case where there are exactly 50 customers, so we get using the binomial PMF $$1-\sum_{i=0}^{50}\binom{1000}{i}(0.01)^i(0.99)^{1000-i}$$ and using the Poisson PMF I get this $$1 - \sum_{i = 0}^{50} \frac{10^ie^{-10}}{i!}$$ To compute those last two I needed to use wolfram alpha. Is there another form that makes these easier to compute by hand?

Thank you.

EDIT: I noticed I forgot the case where $i = 0$ in all of my summations. Ian's solution in the comments key'd me into that since my solutions were not smaller than his upper bound given by the Lagrange error. WolframAlpha had a hard time giving me approximations when I typed in my formulas as I did before. But I could establish that the values were much smaller as expected by the Lagrange error.

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    I'm not sure I see the point of doing a Poisson approximation with an actual cutoff; the math is easier if you just replace Bin(n,p) with Poisson(np) without introducing any cutoff. For large enough n (depending on how close p is to 1), the probability that a Poisson(np) is bigger than n will be negligibly small anyway. – 2017-01-12
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    So here the number of customers that you see is Bin(1000,0.01) which is approximately Poisson(10). So the Poisson estimate for your desired quantity would really be $e^{-10} \sum_{k=51}^\infty 10^k/k!$. The Lagrange error estimate shows this is less than $10^{51}/51!$. – 2017-01-12

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I do not see anything wrong with what you wrote but for part A, but you could approximate the distribution using the normal distribution. It would make the math a whole lot easier. The same applies to part C.

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    I did Poisson distribution since that is what the questions asked for. I am not sure how to approximate a binomial distribution with a normal distribution at this point unfortunately. – 2017-01-12