Let $X$ be distributed as $Bin(n,p)$ , a binomial random variable with parameters $n$ and $p$ . Suppose $$\lim_{n\rightarrow\infty}np=\lambda,$$ where $\lambda$ is a positive real constant, then $X$ is asymptotically distributed as $Poisson(\lambda)$ , a Poisson distribution with parameter $\lambda$ .

Basically, when the size of the population $n$ is very large and the occurrence of certain event $A$ is rare, where $p$ , the probability of $A$ is very small, the binomial random variable $X$ can be approximated by a Poisson random variable.

Sketch of Proof. Let $X\sim Bin(n,p)$ . So \begin{eqnarray*} P(X=m) &=& \frac{n!}{m!(n-m)!}p^m(1-p)^{n-m} \\ &=& \frac{n!}{n^m(n-m)!}\frac{(np)^m}{m!}(1-\frac{np}{n})^{n-m} \\ &=& \frac{n!}{n^m(n-m)!}\frac{(np)^m}{m!}(1-\frac{np}{n})^n(1-\frac{np}{n})^{-m}. \end{eqnarray*}As $n\rightarrow\infty$ , $$\frac{n!}{n^m(n-m)!}=\frac{n}{n}\frac{n-1}{n}\cdots\frac{n-m+1}{n}\approx 1,$$ $$(1-\frac{np}{n})^{-m}\approx (1-\frac{\lambda}{n})^{-m}\approx 1,$$ $$(1-\frac{np}{n})^n\approx (1-\frac{\lambda}{n})^n\approx e^{-\lambda},$$ and $$\frac{(np)^m}{m!}\approx \frac{\lambda^m}{m!}.$$ Therefore, $$P(X=m)\approx \frac{\lambda^m}{m!}e^{-\lambda} = Poisson(\lambda).$$

Example. Suppose in a given year, the number of fatal automobile accidents has a binomial distribution for a particular insuarance company with five hundred automobile insurance policies. On average, there is one policy out of the five hundred that will be involved in a fatal crash. What is the probability that there will be no fatal accidents (out of five hundred policies) in any particular year?

Solution. If $X$ be the number of fatal accidents in a year from a population of 500 auto policies, then $X\sim Bin(n,p)$ with $n=500$ and $p=1/500$ . $\lambda=500\times 1/500=1$ and so $$P(X=0)\approx e^{-1}\approx 0.368.$$ Using the binomial distribution, we have $$P(X=0)=(1-\frac{1}{500})^{500}\approx 0.367.$$