POISSON RANDOM VARIABLES Suppose a random phenomenon occurs with a mean rate of occurrences or happenings per unit of time or length or area or volume, etc. Note: >. Eamples: 1. Cars passing through an intersection per minute. 2. Customers arriving per 5 minutes at a bank to visit a teller. 3. Telephone calls arriving at company s service department per hour. 4. Flaws per 1 metres in electrical wire. 5. Weeds per square metre of crop land. 6. Stars per segment of the night sky. 7. Chocolate chips per cookie. Let X be a random variable that counts the number of occurrences of this random phenomenon over one unit of given size (e.g. time, length, area, volume). Then X ~ P () with Value Space V X = {, 1, 2, 3, } and probability function p( ) e = for =,1,2,3, Note: > is the mean rate of occurrence per specified unit of time, length, space, etc. Properties: (1) < ( ) = < 1 p e e (2). = 1 = Property (2) implies that = = e e since e = = 1. = = Page 1 of 5
Eample Telephone calls arrive at a company s service department at an average rate of 6 per hour. 1. What is the probability that eactly 8 calls arrive during the net hour? Let X be a random variable counting the number of calls arriving in the net hour. Then X ~ Po (6). P[8 calls during net hour] = P[X = 8] = 6 8 e 6 8! = (.2478752)( 1,679,616) 4,32 =.133 2. What is the probability of 4 calls arriving during the net half hour? Let Y be a random variable counting the number of arrivals in 3 minutes. Then Y ~ Po(3). e 3 P[4 arrivals during net 3 minutes] = P[Y = 4] = 4! 3 4 = (.4978768)( 81) 24 =.168 Comment: Because occurrences are assumed to happen uniformly over intervals, changing the size of an interval can be handled by changing the mean rate. 3. What is the probability of at least 2 calls arriving in the net hour and a half? Let Z be a random variable counting the number of calls arriving in the net 9-minutes. Then Z ~ Po(9). P[at least 2 arrivals during net 15 minutes] = P[Z 2] = 1 P[Z = or 1] = 1 - e 9! - e 9 1! 1 = 1-1 e = 1 -.123498 =.9988 Page 2 of 5
TWO SUGGESTIONS FOR CONDITIONS FOR A POISSON PROCESS POISSON PROCESS: 1 1. Occurrences happen AT RANDOM; 2. Occurrences are INDEPENDENT of one another; 3. Occurrences are UNIFORMLY or evenly distributed over the interval being used. POISSON PROCESS: 2 The process is such that occurrences happen at a constant average rate over each unit or interval of time, length, area, volume, etc. such that 1. the probability of a single occurrence in a short interval is proportional to the length of the interval; 2. the probability of more than one occurrence during such a short interval is negligible (i.e. effectively ); and 3. occurrences in disjoint intervals happen independently of each other. Page 3 of 5
Poisson Approimation to the Binomial If the number n of trials in a Binomial eperiment is large, calculating probabilities can be difficult. One approimation available is provided by the Poisson distribution. Suppose that, in a Binomial eperiment, the number n of trials is large, that the probability p of Success at each trial is small, and that the product np is "moderate" in size. Recall that the mean of a Binomial random variable is = np. If X ~ B(n, p) and Y ~ Po (np), then P[X = k] P[Y = k] especially for smaller values of k. Eample: A certain disease occurs once in every 5 people living in a certain country. If a random sample of 1 residents of this country is obtained and tested for this disease, what is the probability that at most one has the disease? Solution: Let X be a random variable counting the number of those sampled who have the disease. Then X ~ B(1,.2) with = np = 1(.2) = 2. Let Y ~ Po (2). P[X 1] = P[X = or 1] P[Y = or 1] 2 2 1 e 2 e 2 = + =.1353 +.277 =.46! 1! Note: The eact answer using the Binomial distribution is 1 1 1 1 999.2.998 +.2.998 1 ( ) ( ) ( ) ( ) =.1356 +.2767 =. 4573 To four places of decimal, the approimate answer is.3 higher than the correct value, off by.3 only 1 =.74%..457 Page 4 of 5
Some Problems 1. During the past summer, a Saskatchewan farmer found that weeds in his barley field occurred at a rate of five per square metre. What is the probability of 3 weeds being present in a randomly chosen square metre of this field? What is the probability of 25 weeds being present in a 2 metre by 5 metre piece of this field? 2. Flaws occur in electrical wire at a rate of one per 1 metres. What is the probability of more than one flaw in a 1 metre roll of this wire? What is the probability of eactly one flaw in a 75 metre roll of this wire? 3. A batch of cookie dough contains 8 chocolate chips and is used to produce 1 cookies. What is the probability that one of these cookies contains eactly 8 chocolate chips? What is the probability that one of these cookies contains fewer than 4 chocolate chips? 4. A certain golfer has a 1% chance of getting a hole-in-one on a par 3 hole. If this golfer plays 25 par 3 holes in his lifetime, what is the probability that he gets at least one hole-in-one? Page 5 of 5