Question

Costumeers arrive at a checkout counter in a department store according to a Poisson distribution

at an average of seven per hour. During a given hour, what are the probabilities that

(a) no more than three customers arrive?

(b) at least two customers arrive?

(c) exactly four customers arrive?

Answer 1

Let $X=$ the number of costumers arrive at a checkout counter in a department store: $X\sim Po(\lambda).$

$$P(X=x)=\dfrac{e^{-\lambda}\cdot\lambda^x}{x!}$$

Given $\lambda=7.$

(a)

$$P(X\leq3)=P(X=0)+P(X=1)+P(X=2)$$

$$+P(X=3)=\dfrac{e^{-7}\cdot7^0}{0!}+\dfrac{e^{-7}\cdot7^1}{1!}+\dfrac{e^{-7}\cdot7^2}{2!}$$

$$+\dfrac{e^{-7}\cdot7^3}{3!}=e^{-7}(1+7+\dfrac{49}{2}+\dfrac{343}{6})$$

$$\approx0.081765$$

(b)

$$P(X\geq2)=1-(P(X=0)+P(X=1))$$

$$=1-\dfrac{e^{-7}\cdot7^0}{0!}-\dfrac{e^{-7}\cdot7^1}{1!}=1-8\cdot e^{-7}$$

$$\approx0.992705$$

(c)

$$P(X=4)=\dfrac{e^{-7}\cdot7^4}{4!}\approx0.091226$$