Chapter 2 Section 3 Question 18

Question

(Poisson processes and queues) Consider a situation involving a server, e.g., a cashier at a fast-food restaurant, an automatic bank teller machine, a telephone exchange, etc. Units typically arrive for service in a random fashion and form a queue when the server is busy. It is often the case that the number of arrivals at the server, for some specific unit of time t, can be modeled by a $Poisson(\lambda t)$ distribution and is such that the number of arrivals in nonoverlapping periods are independent. In Chapter 3, we will show that $\lambda t$ is the average number of arrivals during a time period of length $t$, and so $\lambda$ is the rate of arrivals per unit of time. Suppose telephone calls arrive at a help line at the rate of two per minute. A Poisson process provides a good model.

- What is the probability that five calls arrive in the next 2 minutes?
- What is the probability that five calls arrive in the next 2 minutes and then five more calls arrive in the following 2 minutes?
- What is the probability that no calls will arrive during a 10-minute period?

Answer

Author: Mohammad-Ali Bandzar| Date:Oct 19 2020

- (a) What is the probability that five calls arrive in the next 2 minutes?
- From Example 2.3.6 in the book we know: $$p\scriptsize{Y}\normalsize{(y)=P(Y=y)=\frac{\lambda^y}{y!}e^{-\lambda}}$$ $\lambda$ represents the mean number of occurrences in the given interval in this case it will be calls per minutes multiplied by number of minutes $\lambda=2*2=4$

$y$ in this case will be the desired number of calls in this case $y=5$ $$P(Y=y)=\frac{\lambda^y}{y!}e^{-\lambda}$$ $$P(Y=5)=\frac{4^5}{5!}e^{-4}$$ $$P(Y=5)=\frac{1024}{120}e^{-4}$$ $$P(Y=5)\approx 0.15629$$ - (b) What is the probability distribution of the number of white balls observed?
- This is the probability that part a occurs twice in succession. $$P(Y=5)^2$$ $$\approx (\frac{4^5}{5!}e^{-4})^2$$ $$\approx 0.02442$$
- (c) What is the probability that no calls will arrive during a 10-minute period?
- From Example 2.3.6 in the book we know: $$p\scriptsize{Y}\normalsize{(y)=P(Y=y)=\frac{\lambda^y}{y!}e^{-\lambda}}$$ $\lambda$ represents the mean number of occurrences in the given interval in this case it will be calls per minutes multiplied by number of minutes $\lambda=2*10=20$

$y$ in this case will be the desired number of calls in this case $y=0$ $$P(Y=y)=\frac{\lambda^y}{y!}e^{-\lambda}$$ $$P(Y=0)=\frac{20^0}{0!}e^{-20}$$ $$P(Y=0)=e^{-20}$$ $$P(Y=5)\approx 0.000000002061$$