Chapter 6 The Poisson Distribution Counting Rules When the outcomes of a chance experiment are equally likely, one way to determine the probability of some event E is to calculate number of outcomes for which E occurs P(E ) total number of possible outcomes For example, if a chance experiment consists of rolling two dice, there are 36 possible outcomes, shown in the accompanying table. Possible Outcomes for Rolling Two Dice 1,1 1,2 1,3 1,4 1, 1,6 2,1 2,2 2,3 2,4 2, 2,6 3,1 3,2 3,3 3,4 3, 3,6 4,1 4,2 4,3 4,4 4, 4,6,1,2,3,4,,6 6,1 6,2 6,3 6,4 6, 6,6 If the dice are fair, each of these 36 outcomes is equally likely. Because 6 of these outcomes result in a total of 7 (the highlighted outcomes in the table), P (total on two dice is 7 ) 6 36 Listing the possible outcomes and then counting the total number of outcomes and the number of outcomes favorable to the event of interest is one way to determine the two counts needed to calculate the desired probability. However, if the number of possible outcomes is large, this can be tedious. In many situations, a quicker way to count is to use one of the following approaches: 1. Fundamental Counting Rule 2. Permutations 3. Combinations Fundamental Counting Rule When an outcome can be thought of as being generated by following a sequence of steps and the number of different ways each step can be completed is known, the fundamental counting rule specifies the total number of possible outcomes. Fundamental Counting Rule Suppose an outcome can be generated by following a sequence of k steps. If Step 1 can be completed in n 1 ways, Step 2 can be completed in n 2 ways, and so on, then total number of outcomes (n 1 )(n 2 ) (n k )
CHAPTER 6 3 For example, suppose that you have purchased a new combination lock. The lock has three wheels, each of which can be set to one of ten positions labeled, 1,,. How many different ways are there to set the combination for the lock? In this case, the answer should be obvious because each combination is a three-digit number, and there are 1, different three-digit numbers (, 1,, ). But the fundamental counting rule could have also been used. Setting a combination can be viewed as consisting of three steps: Step 1: Selecting a digit for the first wheel Step 2: Selecting a digit for the second wheel Step 3: Selecting a digit for the third wheel Step 1 can be completed in ten different ways, Step 2 can be completed in ten different ways, and Step 3 can be completed in ten different ways. Using the fundamental counting rule, total number of possible combinations (1 )(1 )(1 ) (1,) Example 6.34 Selecting a Password Suppose that the password for a particular debit card must consist of four characters. It must begin and end with a letter, and the two middle characters must be digits. How many different passwords are possible? You can view selecting a password as consisting of four steps, with the first step being selection of the first password character, and so on. Because there are 26 letters and ten digits, using the fundamental counting rule results in Permutations and Combinations In some situations, outcomes can be thought of as making selections from a fixed set of items. For example, an instructor who assigned ten homework problems may want to select three of these problems to put on a quiz. There are a large number of different possible quizzes that might result. If the homework problems are labeled H 1, H 2,, and H 1, one possible quiz might consist of H 1, H, and H 6 and another quiz might consist of H 2, H 7, and H 1. Each subset of three problems selected from the ten homework problems represents a different possible quiz. Before looking at a way to determine how many different quizzes are possible, you need to think about one more thing does order make a difference? That is, would you consider a quiz consisting of problems H 1, H, and H 6 in that order a different quiz than a quiz that had the same three problems, but in a different order (such as H 1, H 6, and H )? If different orders are considered as different outcomes, you want to count the number of different ordered subsets, which are called permutations. If different orders of the same items are considered as the same outcome, you want to count the number of unordered subsets, which are called combinations. Definition total number of passwords (26 )(1 )(1 )(26 ) 67,6 A permutation is an ordered subset of k distinct items selected from a set of n distinct items. A combination is an unordered subset of k distinct items selected from a set of n distinct items. Before considering formulas that will allow you to calculate the number of different permutations or the number of different combinations, let s look at a simple example.
4 CHAPTER 6 Example 6.3 Selecting Club Officers Suppose that five members of a service club (Allan, Betty, Cindy, David, and Eric) have volunteered to serve as officers for the upcoming year. There are three offices: president, vice-president, and treasurer. How many different ways are there to fill the three offices? Here, order is important, because the outcome A, B, C (where Allan is president, Betty is vice-president, and Cindy is treasurer) is a different outcome than C, B, A. To evaluate the number of different outcomes, you want to know how many different permutations of three members selected from the five volunteers are possible. Think of generating an outcome by first selecting a president ( choices), then selecting a vice-president (4 choices remaining), and then selecting a treasurer (3 choices). The fundamental counting rule can be used to determine that There are formulas that will allow you to calculate the number of permutations or the number of combinations directly, without having to go through the work illustrated in Example 6.3. These formulas use factorial notation. Recall that for any positive integer k, k factorial is denoted by k! and is defined as and! is defined to be 1. Permutations number of permutation ( )(4 )(3 ) 6 What if the club was managed by a three-person executive board and that these individuals did not hold specific offices, such as president? In this situation, order doesn t matter you would consider the executive board consisting of A, B, and C to be the same as the board consisting of B, A, and C. In this situation, you want to calculate the number of different combinations of three members selected from the five volunteers. Consider the executive board consisting of A, B, and C. These three members can be listed in six different orders: ABC, ACB, BAC, BCA, CAB, CBA. Each of these six ordered arrangements was counted as different when you determined that there were 6 permutations. The count of permutations over-counted combinations by a factor of six. This means that number of combinations 6 6 1 k! (k)(k 2 1)(k 2 2) (1) The number of ordered subsets of k items selected from a set of n items, denoted by n P k, is calculated using the following formula: P n! n k (n 2 k )! Example 6.36 ipod Shuffle Suppose that Chris has ten songs in a playlist on his ipod. A shuffle is a random ordering of songs from a playlist. How many different shuffles of four songs can be created from songs in this playlist? Each possible shuffle is a permutation of four songs chosen from the ten in the playlist. The number of different shuffles is P 1! 1 4 (1 2 4)! 1! 6! 3,628,8,4 72 Most graphing calculators and statistics software packages can calculate the number of permutations (or the number of combinations) given n and k.
CHAPTER 6 Combinations The number of unordered subsets of k items selected from a set of n items, denoted by n C k, is calculated using the following formula: C n! n k k!(n 2 k )! Example 6.37 ipod Revisited Suppose you want to create a playlist of 1 songs on an ipod. A playlist is a collection of songs chosen from all the songs on an ipod. How many different playlists are possible if you have 3 songs on your ipod? Because a playlist is an unordered collection of songs, each playlist is a combination of 1 songs chosen from the 3 songs on the ipod. The number of different playlists is C 3! 3 1 1!(3 2 1)! 3! 1! 2! This can be evaluated using a graphing calculator or software to obtain C 3,4,1 3 1 That is over 3 million different playlists! And with more songs on the ipod, this number would be even larger. Example 6.38 Figure Skating Competitions In figure skating competitions, skaters perform both a short program and a long program. For the short program, skaters are divided at random into groups of 6 skaters. The skating order within each group is assigned at random. Suppose that 24 skaters have entered a competition. How many different possibilities are possible for the program listing for the first group of skaters to perform the short program? It is often the case with counting problems that there are multiple ways to solve them. Let s consider two different ways to answer this question. Solution 1 You can think of determining the program listing for the first group as consisting of two steps: Step 1: Select the six skaters who will be in the first group. Step 2: Determine the skating order for these six skaters. Step 1 can be completed in 24 C 6 ways, and then for each of these possible groups of skaters, there are 6 P 6 ways to order the skaters. The total number of different possible program listings is (using the fundamental counting rule) ( 24 C 6 ) ( 6 P 6 ) ( 24! 6!18! )( 6!! ) (134,6)(72) 6,,12 Solution 2 Another way to think about this problem is to recognize that even though the description of how the first group and skating order are determined was described as a two-step process (selecting the group, then determining the order), this is equivalent to just asking how many different ordered arrangements of six can be formed from the set of 24 skaters a permutation. So, total number of program listings 24 P 6 6,,12
6 CHAPTER 6 Exercises 6.11 How many different passwords are possible if a password must begin with a letter and must consist of eight characters that must be letters or numbers? 6.116 California automobile license plate numbers begin with a non-zero digit, followed by three letters, followed by three digits (which can be zero). How many different license plate numbers are possible using this format? 6.117 Jeanie has a list of seven errands she needs to complete, but she only has time to do three of them today. How many ways are possible for Jeanie to select the three errands she will complete today? 6.118 A professor plans to ask 3 different students in her math class to participate in a class demonstration. How many different ways to select these three students are possible if there are students in the class? 6.11 A professor assigned ten homework problems. He plans to select four of these problems to be graded. a. Suppose you only have time to complete six of these problems. How many different ways are there to choose which six problems you will do? b. How many different ways are there for the professor to select four problems to be graded? c. How many different ways are there to select four problems from the six that you completed? d. If the professor selects the four problems to be graded at random, what is the probability that all four problems selected for grading are from the six that you completed? 6.12 How many different sequences of three different characters can be formed using the letters NEWYORK? 6.121 A baseball team has 1 players. The coach must submit a starting batting order, which consists of players and the order in which they will bat. How many different batting orders are possible? The Poisson Distribution In Section 6.7 you learned about two discrete probability distributions the binomial distribution and the geometric distribution. In this section, you will see another widely used discrete probability distribution called the Poisson distribution. A Poisson distribution is used to describe the behavior of a random variable that counts the number of occurrences of some random event during a fixed time interval or over a fixed space. For example, you might be interested in the number of calls to a customer service center during a three-hour period or the number of accidents occurring along a particular stretch of highway. If the event you are interested in occurs at random, but with a known overall rate, and the properties in the following box are satisfied, then the random variable x number of occurrences in a fixed interval has a Poisson distribution. If x = number of occurrences in a fixed interval and the following conditions are met, x has a Poisson distribution. Conditions for a Poisson Distribution 1. The probability of more than one occurrence in a very small interval is. 2. The probabilities of an occurrence in each of two intervals of the same length are equal. 3. The number of occurrences in any particular interval is independent of the number of occurrences prior to that interval. These conditions are satisfied when occurrences are random but with a fixed overall rate. For example, consider calls coming in to a customer service center, and define x to be x number of calls received during a 1-hour period From past experience, you might know that calls come in at random times but with an overall rate of about two per minute. It is then reasonable to think that
CHAPTER 6 7 1. The chance of more than one call in any very small time interval (for example,.1 seconds) is equal to. 2. The chances of a call coming in during each of two different intervals of the same length are equal. 3. The number of calls during any particular interval is independent of the number of calls that were received prior to this interval. These observations are consistent with calls coming in at random times, but with an overall rate of two per minute. Because the conditions are met, the behavior of the random variable x can be described by a Poisson distribution. The Poisson Distribution Let the overall rate of occurrence of the random event in 1 unit of time or space (such as 1 minute or 1 mile) t a fixed interval of time or space (such as 2 minutes or miles) Then if the conditions for a Poisson distribution are met x number of occurrences in an interval of length t has a Poisson distribution. The Poisson distribution is defined by the following formula: P(x) (x occurences in an interval of length t) ( t ) x e 2 t x, 1, 2, x! (e in this formula is a mathematical constant that is approximately equal to 2.718.) Example 6.3 Paint Bubbles Sometimes bubbles form in paint as it dries, creating flaws in the paint surface. Suppose that for a particular type of paint, bubbles form at random, with an overall rate of.3 per square foot. What is the probability of observing five bubbles in a 1-square-foot wall that has been painted with this paint? In this example, the occurrence you are interested in is a paint bubble. Bubbles occur at random with a known rate of.3 per square foot. This mean that.3, the unit of space for this rate is 1 square foot, and the interval of interest is t 1 square feet. The desired probability is then p() ( (.3)(1) ) e 2(.3)(1)! 3 (2.718) 23 (234)(.48) 11.64! 12 12.7 Graphing calculators and statistical software can also be used to compute Poisson probabilities. For example, Minitab was used to compute p (x # ) p() 1 p(1) 1 p(2) 1 p(3) 1 p(4) 1 p().4787 1.14361 1.22442 1.22442 1.16831 1.181.1682 The difference in the value of p() from the value in the hand calculations (.181 versus.7) is a result of Minitab using more decimal accuracy in the computations and in the value of e. Example 6.4 Calls to -1-1 Suppose that in a particular county, calls come in to the -1-1 emergency number at random with an overall rate of three per hour. What is the probability of observing eight calls in a two-hour period? To answer this question, you would need to evaluate
8 CHAPTER 6 ( t) x e 2 t p(x) x! where x 8, 3, and t 2. This results in p (8) (6) 8 (2.718) 26 (1,67,616)(.248) 4,16.448 8! 4,32 4.32.13 Exercises 6.122 An automobile club reports that requests for roadside assistance come in at random times with an overall rate of four per hour. a. What is the probability that exactly ten requests are received in a particular two-hour time period? b. What is the probability that there are no calls in a particular one-hour time period? 6.123 A cookie manufacturer makes chocolate chip cookies. When the cookie dough is prepared, chocolate chips are added to the dough in a way that distributes them at random throughout the dough. Because about 12, chips are added to a batch of dough that makes 1, cookies, the company knows that, on average, there are 12 chocolate chips per cookie. a. What is the probability that a randomly selected cookie will have four or fewer chocolate chips? b. What is the probability that the total number of chocolate chips in three randomly selected cookies will be exactly 36? 6.124 Suppose that the random variable x has a Poisson distribution with rate 6 per hour. For a particular one-hour time interval, is it more likely that x or that x 7? Support your answer with the appropriate probability calculations. 6.12 Suppose that the random variable x has a Poisson distribution with rate. per hour and the random variable y has a Poisson distribution with rate 2 per hour. a. Calculate the probabilities needed to complete the following tables. Round your answers to four decimal places. x 1 2 3 4 6 7 8 p(x) b. Possible values for a random variable that has a Poison distribution are, 1, 2, 3, Even though possible values are all nonnegative integers, what do you notice about p(x) for t 1 as the value of x gets large relative to? 6.126 Draw a probability histogram for each of the Poisson distributions for which you calculated probabilities in the previous exercise (., t 1 and 2, t 1 ) using possible values to on the horizontal axis. How do the two probability distributions compare with respect to shape, center, and variability? y 1 2 3 4 6 7 8 p(y)
CHAPTER 6 Counting Rules Answers for Selected Exercises Counting Rules 6.126 6.11 (26) (36) 6 6,6,34,736 6.116 () (26) 3 (1) 3 18,184, 6.117 7 C 3 3 6.118 C 3 1,6 6.11 (a) 1 C 6 21 (b) 1 C 4 21 (c) 6 C 4 1 (d) 1 21.714 6.12 7 P 3 21 6.121 1 P 1,816,214,4 The Poisson Distribution 6.122 (a) 4, t 2, p(1).3 (b) 4, t 1, p().183 6.123 (a) 12, t 1 p() 1 p(1) 1 p(2) 1 p(3) 1 p(4).61 1.737 1.4424 1.176 1.386.763 (b) 12, t 3, p(36).66336 6.124 6, t 1, p().16623, p(7).137677, p() is greater than p(7). 6.12 (a) x p(x).66 1.333 2.78 3.126 4. 16. 2 6. 7. 8.. y p(y).133 1.277 2.277 3.184 4.2.361 6.12 7.34 8.. 2 p(x) p(y).6..4.3.2.1.6..4.3.2.1 1 1 2 2 3 3 4 x 4 y 6 7 8 Both distributions are positively skewed. The distribution of x is centered at about., whereas the distribution of y is centered at about 2. There is more variability in the y distribution. 6 7 8 (b) As the value of x gets large compared to the value of, p(x) gets close to.