91Ó°ÊÓ

Suppose that ${10}^6$people arrive at a service at times that are independent random variables, each of which is uniformly distributed over $(0,{10}^6)$.

Let N denote the number that arrives in the first hour.

the number of people, 10⁶ that come to the station within the first hour has an exact binomial distribution with parameters n=${10}^6$and the probability of success,localid="1647158070597" $p=\frac{1}{{10}^6}$.

that is,localid="1647428159709" $N~Binom\left({10}^6,\frac{1}{{10}^6}\right)$

The size of the population is very large, $n={10}^6$.

The constant probability of success is very small, $p=\frac{1}{{10}^6}$.

Since n is large and p is small, use the Poisson approximation.

thus,

$$\begin{gathered} np={10}^6×\frac{1}{{10}^6} \\ =1 \end{gathered}$$

Which is nothing but, $\mathrm{λ}=1$(parameter of Poisson distribution).

hence, $N~Pois(\mathrm{λ}=1)$

therefore,

$$\begin{gathered} P\left(N=i\right)=\frac{e^{-\mathrm{λ}}{\mathrm{λ}}^i}{i!} \\ P(N=1)=\frac{e^{-\left(1\right)}{\left(1\right)}^i}{i!} \\ =\frac{e^{-1}{\left(1\right)}^i}{i!} \\ =\frac{e^{-1}}{i!} \end{gathered}$$