# Poisson vs Binomial distribution.

by Jcampuzano2
Tags: binomial, distribution, poisson
 Mentor P: 18,346 The problem with the binomial distribution is that it is very hard to calculate. So the second question would be $$\sum_{k=4}^6 \binom{1000}{k} (0.005)^k0.995^{1000-k}$$ This is the correct answer. But computing those binomial coefficients is not very fun. However, we can show that if we are working with binomial(n,pn) distributions and if $np_n\rightarrow \lambda$ for some $\lambda$, then $$\binom{n}{k} p^k (1-p)^{n-k} \rightarrow e^{-\lambda} \frac{\lambda^k}{k!}$$ So, if n is very large and p is very small, then the Binomial(n,p) distribution is very close to the Poisson(np) distribution. So, in our case, p=0.005 is small and n=1000 is large. The product is medium: 5. So we can approximate the answer by $$\sum_{k=1}^6 e^{-5} \frac{5^k}{k!}$$ And we are also rid of that pesky binomial coefficient. This approximation is also theoretically interesting. The sum of two (independent) Poisson distributions is always a Poisson distribution, for example. But the sum of two (independent) binomial distributions is not binomial.