6.25 - A First Course in Probability

Question

Suppose that${10}^6$people arrive at a service station at times that are independent random variables, each of which is uniformly distributed over$\left(0,{{10}^6}_{}\right)$. Let N denote the number that arrive in the first hour. Find an approximation for$P\left\{N=i\right\}$.

Step-by-Step Solution

Answer

Approximation for P[N=i] $=\frac{e^{-1}}{i!}$

Step 1: Given information

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,$p=\frac{1}{{10}^6}$.

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

Step 2: Explanation

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,

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

Which is nothing but, $\lambda =1$(parameter of Poisson distribution).

hence, $N~Pois (\lambda =1)$

therefore,

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