Probability & Statistics Ch.2 Section 3 Question 18

Solutions for “Probability and Statistics: The Science of Uncertainty” (Second Edition). These are solutions I have come up with; I offer no guarantee of accuracy.

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 $\text{Poisson}(\lambda t)$ distribution and is such that the number of arrivals in nonoverlapping periods are independent.

Suppose telephone calls arrive at a help line at the rate of two per minute. A Poisson process provides a good model.

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

Solution

From Example 2.3.6 in the textbook:

p_Y(y) = P(Y=y) = \frac{\lambda^y}{y!}e^{-\lambda}

(a) What is the probability that five calls arrive in the next 2 minutes?

λ = (calls per minute) × (number of minutes) = 2 × 2 = 4

y = 5 (desired number of calls)

P(Y=5) = \frac{4^5}{5!}e^{-4} = \frac{1024}{120}e^{-4} \approx 0.1563

(b) What is the probability that five calls arrive in the next 2 minutes and then five more calls arrive in the following 2 minutes?

Since non-overlapping periods are independent, this is part (a) occurring twice in succession:

P(\text{both}) = P(Y=5)^2 \approx (0.1563)^2 \approx 0.0244

(c) What is the probability that no calls will arrive during a 10-minute period?

λ = 2 × 10 = 20

y = 0 (desired number of calls)

P(Y=0) = \frac{20^0}{0!}e^{-20} = e^{-20} \approx 2.06 \times 10^{-9}