Question

. Consider a shop that sells mobile phones. The number of customers arriving per
hour is assumed to follow a Poisson distribution with mean 2. Each customer
visiting the shop will buy a phone with probability 0.25. Answer the following
questions:
(a) What is the probability that more than 3 customers will arrive during the next
hour?
(b) What is the probability that in a 2-hour period not more than 5 customers
will arrive?
2
(c) If 15 customers visit the shop during a day, what is the probability that they
will sell at least 5 phones?
(d) When the shop opens in the morning, what is the probability that they have
to wait more than half an hour for the first customer to arrive?

Answer 1

a) $\lambda=2.$

$P(X>3)=1-P(X=0)-P(X=1)-P(X=2)-P(X=3)=\\ =1-e^{-2}(\frac{2^0}{0!}+\frac{2^1}{1!}+\frac{2^2}{2!}+\frac{2^3}{3!})=0.1429.$

b) $\lambda=2*2=4.$

$P(X>5)=1−P(X=0)−P(X=1)−P(X=2)−P(X=3)+\\+P(X=4)+P(x=5)=\\ =1−e^{−4}(\frac{4^0}{0!}+\frac{4^1}{1!}+\frac{4^2}{2!}+\frac{4^3}{3!}+ \frac{4^4}{4!}+\frac{4^5}{5!})=0.2149.$

c) Binomial distribution with $n=15, p=0.25.$

$P(X\ge5)=1-P(X<5)=1-C_{15}^00.25^00.75^5-C_{15}^10.25^10.75^4-\\ -C_{15}^20.25^20.75^3-C_{15}^30.25^30.75^2-C_{15}^40.25^40.75^1=0.3135.$

d) There are no customers in the first half of hour $\lambda=\frac{2}{2}=1$ .

$P(X=0)=e^{-1}\frac{1^0}{0!}=0.3679.$