CHAPTER Accidents and incidence – discrete probability distributions and simulation 12 Chapter objectives This chapter will help you to: ■ work out the probabilities for a basic discrete probability distribution ■ calculate the mean and standard deviation of a discrete prob- ability distribution ■ model business processes with the binomial distribution ■ model business processes with the Poisson distribution ■ simulate simple random business processes with random numbers ■ use the technology: discrete probability distributions in EXCEL, MINITAB and SPSS, random number generation in EXCEL ■ become acquainted with business uses of discrete probability distributions and simulation Chapter 12 Accidents and incidence – discrete probability distributions and simulation 375 In this chapter we will bring together two key concepts from earlier chapters. The first of these is the idea of a frequency distribution, which shows the frequency or regularity with which the values of a variable occur, in other words how they are distributed across their range. The second key concept is that of probability, which we considered in Chapter 9. Here we will be looking at probability distributions which por- tray not the frequency with which values of a distribution actually occur but the probability with which we predict they will occur. Probability distributions are very important tools for modelling or representing processes that occur at random, such as customers visit- ing a website or accidents on a building site. These are examples of dis- crete random variables as they vary in a random fashion and can have only certain values, whole numbers in both cases; we cannot conceive of half a customer visiting a website or 0.3 of an accident happening. We use discrete probability distributions to model these sorts of variables. In studying probability distributions we will look at how they can be derived and how we can model or represent the chances of different combinations of outcomes using the same sort of approach as we use to arrange data into frequency distributions. Following that we will exam- ine two standard discrete probability distributions; the binomial and the Poisson. Lastly we will look at how random numbers and discrete probability distributions can be used to simulate the operation of ran- dom business processes. 12.1 Simple probability distributions In section 4.4.2 of Chapter 4 we looked at how we could present data in the form of a frequency distribution. This involved defining cat- egories of values that occurred in the set of data and finding out how many observed values fell into each category, in other words, the fre- quency of each category of values in the set of data. The results of this process enabled us to see how the observations were distributed over the range of the data, hence the term frequency distribution. A probability distribution is very similar to a frequency distribution. Like a frequency distribution, a probability distribution has a series of categories, but instead of categories of values it has categories of types of outcomes. The other difference is that each category has a prob- ability instead of a frequency. In the same way as a frequency distribution tells us how frequently each type of value occurs, a probability distribution tells us how probable each type of outcome is. 376 Quantitative methods for business Chapter 12 In section 5.2.1 of Chapter 5 we saw how a histogram could be used to show a frequency distribution. We can use a similar type of diagram to portray a probability distribution. In Chapter 6 we used summary measures, including the mean and standard deviation, to summarize distributions of data. We can use the mean and standard deviation to summarize distributions of probabilities. Just as we needed a set of data to construct a frequency distribution, we need to identify a set of compound outcomes in order to create a probability distribution. We also need the probabilities of the simple outcomes that make up the combinations of outcomes. Example 12.1 Imported Loobov condoms are sold in packets of three. Following customer complaints the importer commissioned product testing which showed that due to a randomly occur- ring manufacturing fault 10% of the condoms tear in use. What are the chances that a packet of three includes zero, one, two and three defective condoms? The probability that a condom is defective (D) is 0.1 and the probability it is good (G) is 0.9. The probability that a packet of three contains no defectives is the probability that a sequence of three good condoms were put in the packet. P(GGG) ϭ 0.9 * 0.9 * 0.9 ϭ 0.729 The probability that a packet contains one defective is a little more complicated because we have to take into account the fact that the defective one could be the first or the second or the third condom to be put in the packet. so P(1 Defective) ϭ P(DGG or GDG or GGD) Because these three sequences are mutually exclusive, according to the addition rule of probability (you may like to refer back to section 10.3.1 of Chapter 10): P(1 Defective) ϭ P(DGG) ϩ P(GDG) ϩ P(GGD) The probability that the first of these sequences occurs is: P(DGG) ϭ 0.1 * 0.9 * 0.9 ϭ 0.081 The probability of the second: P(GDG) ϭ 0.9 * 0.1 * 0.9 ϭ 0.081 It is no accident that the probabilities of these sequences are the same. Although the exact sequence is different the elements that make up both are the same. To work out the compound probabilities that they occur we use the same simple probabilities but in Chapter 12 Accidents and incidence – discrete probability distributions and simulation 377 In Example 12.1 the probability distribution presents the number of defectives as a variable, X, whose values are represented as x. The vari- able X is a discrete random variable. It is discrete because it can only take a limited number of values – zero, one, two or three. It is random because the values occur as the result of a random process. The symbol ‘P(x)’ represents the probability that the variable X takes a particular value, x. For instance, we can represent the probability that the number of females is one, as P(X ϭ 1) ϭ 0.243 a different order, and the order does not affect the result when we multiply them together. If you work out P(GGD) you should find that it also is 0.081. The probability of getting one defective is therefore: P(1 Defective) ϭ 0.081 ϩ 0.081 ϩ 0.081 ϭ 3 * 0.081 ϭ 0.243 We find the same sort of thing when we work out the probability that there are two defectives in a packet. P(2 Defectives) ϭ P(DDG or DGD or GDD) P(DDG) ϭ 0.1 * 0.1 * 0.9 ϭ 0.009 P(DGD) ϭ 0.1 * 0.9 * 0.1 ϭ 0.009 P(GDD) ϭ 0.9 * 0.1 * 0.1 ϭ 0.009 So P(2 Defectives) ϭ 3 * 0.009 ϭ 0.027 Finally P(3 Defectives) ϭ 0.1* 0.1* 0.1 ϭ 0.001 We can bring these results together and present them in the form of a probability distribution. Note that the sum of these probabilities is 1 because they are mutually exclusive (we cannot have both one defective and two defectives in a single packet of three) and collec- tively exhaustive (there can be only none, one, two or three defectives in a packet of three). Number of defectives (x) P(x) 0 0.729 1 0.243 2 0.027 3 0.001 1.000 378 Quantitative methods for business Chapter 12 Figure 12.1 shows the probability distribution we compiled in Example 12.1 in graphical form. We can find summary measures to represent this distribution, in the same way as we could use summary measures to represent distributions of data. However, as we don’t have a set of data to use to get our sum- mary measures we use the probabilities to ‘weight’ the values of X, just as we would use frequencies to obtain the mean from a frequency dis- tribution. You work out the mean of a probability distribution by multi- plying each value of x by its probability and then adding up the products: ␮ ϭ ∑xP(x) Notice that we use the Greek symbol ␮ to represent the mean of the distribution. The mean of a probability distribution is a population mean because we are dealing with a distribution that represents the prob- abilities of all possible values of the variable. Once we have found the mean we can proceed to find the variance and standard deviation. We can obtain the variance, ␴ 2 , by squaring each x value, multiplying the square of it by its probability and adding the products. From this sum we subtract the square of the mean. ␴ 2 ϭ ∑x 2 P(x) Ϫ ␮ 2 The standard deviation, ␴, is simply the square root of the variance. Again you can see that we use a Greek letter in representing the vari- ance and the standard deviation because they are population measures. 0 0.0 0.1 0.2 0.3 0.5 0.6 0.7 0.8 0.4 x P(x) 123 Figure 12.1 The probability distribution of X, the number of defectives Chapter 12 Accidents and incidence – discrete probability distributions and simulation 379 The mean of a probability distribution is sometimes referred to as the expected value of the distribution. Unlike the mean of a set of data, which is based on what the observed values of a variable actually were, the mean of a probability distribution tells us what the values of the variable are likely, or expected to be. We may need to know the probability that a discrete random variable takes a particular value or a lower value. This is known as a cumulative probability because in order to get it we have to add up or accumulate other probabilities. You can calculate cumulative probabilities directly from a probability distribution. Example 12.2 Calculate the mean and the standard deviation for the probability distribution in Example 12.1. The mean, ␮, is 0.300, the total of the xP(x) column. The variance, ␴ 2 , is 0.360, the total of the x 2 P(x) column minus the square of the mean: ␴ 2 ϭ 0.360 Ϫ 0.300 2 ϭ 0.360 Ϫ 0.090 ϭ 0.270 The standard deviation: ␴ ϭ √␴ 2 ϭ √0.270 ϭ 0.520 xP(x) xP(x) x 2 x 2 P(x) 0 0.729 0.000 0 0.000 1 0.243 0.243 1 0.243 2 0.027 0.054 4 0.108 3 0.001 0.003 9 0.009 0.300 0.360 Example 12.3 Calculate a set of cumulative probabilities from the probability distribution in Example 12.1. Suppose we want the probability that X, the number of defectives, is either two or less than two. Another way of saying this is the probability that X is less than or equal to two. We can use the symbol ‘р’ to represent ‘less than or equal to’, so we are looking for P(X р 2). (It may help you to recognize this symbol if you remember that the small end of the ‘Ͻ’ part is pointing at the X and the large end at the 2, implying that X is smaller than 2.) 380 Quantitative methods for business Chapter 12 The cumulative probabilities like those we worked out in Example 12.3 are perfectly adequate if we want the probability that a variable takes a particular value or a lower one, but what if we need to know the probability that a variable is higher than a particular value? We can use the same cumulative probabilities if we manipulate them using our knowledge of the addition rule of probability. If, for instance, we want to know the probability that a variable is more than two, we can find it by taking the probability that it is two or less away from one. P(X Ͼ 2) ϭ 1 Ϫ P(X р 2) We can do this because the two outcomes (X being greater than two and X being less than or equal to two) are mutually exclusive and collec- tively exhaustive. One and only one of them must occur. There are no other possibilities so it is certain that one of them happens. We can find the cumulative probabilities for each value of X by taking the probability that X takes that value and adding the probability that X takes a lesser value. You can see these cumulative probabilities in the right-hand column of the following table: The cumulative probability that X is zero or less, P(X р 0), is the probability that X is zero, 0.729, plus the probability that X is less than zero, but since it is impossible for X to be less than zero we do not have to add anything to 0.729. The second cumulative probability, the probability that X is one or less, P(X р 1), is the probability that X is one, 0.243, plus the probability that X is less than one, in other words, that it is zero, 0.729. Adding these two probabilities together gives us 0.972. The third cumulative probability is the probability that X is two or less, P(X р 2). We obtain this by adding the probability that X is 2, 0.027, to the probability that X is less than 2, in other words that it is one or less. This is the previous cumulative probability, 0.972. If we add this to the 0.027 we get 0.999. The fourth and final cumulative probability is the probability that X is three or less. Since we know that X cannot be more than three (there are only three condoms in a packet), it is certain to be three or less, so the cumulative probability is one. We would get the same result arithmetically if we add the probability that X is three, 0.001, to the cumu- lative probability that X is less than three, in other words that it is two or less, 0.999. Number of defectives (x) P(x) P(X ഛ x) 0 0.729 0.729 1 0.243 0.972 2 0.027 0.999 3 0.001 1.000 Chapter 12 Accidents and incidence – discrete probability distributions and simulation 381 In the expression P(X Ͼ 2), which represents the probability that X is greater than two, we use the symbol ‘Ͼ’ to represent ‘greater than’. (It may help you to recognize this symbol if you remember that the larger end of it is pointing to the X and the smaller end is pointing to the 2, implying than X is bigger than 2.) Although the situation described in Example 12.1, considering packets of just three condoms, was quite simple, the approach we used to obtain the probability distribution was rather laborious. Imagine that you had to use the same approach to produce a probability distribution if there were five or six condoms in a packet instead of just three. We had to be careful enough in identifying the three different ways in which there could be two defectives in a packet of three. If the packets contained five condoms, identifying the different ways that there could be say two defectives in a packet would be far more tedious. Fortunately there are methods of analysing such situations that do not involve strenuous mental gymnastics. These involve using a type of probability distribution known as the binomial distribution. At this point you may find it useful to try Review Questions 12.1 and 12.2 at the end of the chapter. 12.2 The binomial distribution The binomial distribution is the first of a series of ‘model’ statistical distributions that you will meet in this chapter and the two that follow it. The distribution was first derived theoretically but is widely used in dealing with practical situations. It is particularly useful because it enables you not only to answer a specific question but also to explore the consequences of altering the situation without actually doing it. You can use the binomial distribution to solve problems that have what is called a binomial structure. These types of problems arise in situ- ations where a series of finite, or limited number of ‘experiments’, or ‘trials’ take place repeatedly. Each trial has the same two mutually exclu- sive and collectively exhaustive outcomes, as the bi in the word binomial might suggest. By convention one of these outcomes is referred to as ‘success’; the other as ‘failure’. To analyse a problem using the binomial distribution you have to know the probability of each outcome and it must be the same for every trial. In other words, the results of the trials must be independent of each other. Words like ‘experiment’ and ‘trial’ are used to describe binomial situations because of the origins and widespread use of the binomial 382 Quantitative methods for business Chapter 12 distribution in science. Although the distribution has become widely used in many other fields, these scientific terms have stuck. The process in Example 12.1 has a binomial structure. Putting three condoms in a packet is in effect conducting a series of three trials. In each trial, that is, each time a condom is put in a packet, there can be only one of two outcomes: either it is defective or it is good. In practice, we would use tables such as Table 2 in Appendix 1 on page 618 to apply the binomial distribution. These are produced using an equation, called the binomial equation, which you will see below. You won’t need to remember it, and you shouldn’t need to use it. We will look at it here to illustrate how it works. We will use the symbol X to represent the number of ‘successes’ in a certain number of trials, n. X is what is called a binomial random variable. The probability of success in any one trial is represented by the letter p. The probability that there are x successes in n trials is: You will see that an exclamation mark is used several times in the equa- tion. It represents a factorial, which is a number multiplied by one less than itself then multiplied by two less itself and so on until we get to one. For instance four factorial, 4!, is four times three times two times one, 4 * 3 * 2 * 1, which comes to 24. PX x n xn x pp xnx () () ! !( )! *ϭϭ Ϫ Ϫ Ϫ 1 Example 12.4 Use the binomial equation to find the first two probabilities in the probability distribution for Example 12.1. We will begin by identifying the number of trials to insert in the binomial equation. Putting three condoms in a packet involves conducting three ‘trials’, so n ϭ 3. The variable X is the number of defectives in a packet of three. We need to find the probabilities that X is 0, 1, 2 and 3, so these will be the x values. Suppose we define ‘success’ as a defective, then p, the probability of success in any one trial, is 0.1. We can now put these numbers into the equation. We will start by working out the probability that there are no defectives in a packet of three, that is X ϭ 0. This expression can be simplified considerably. Any number raised to the power zero is one, so 0.1 0 ϭ 1. Conveniently zero factorial, 0!, is one as well. We can also carry out the subtractions. PX() ( ) 0 3! 0!(3 0)! * 0.1 0.1 00 ϭϭ Ϫ Ϫ Ϫ 1 3 Chapter 12 Accidents and incidence – discrete probability distributions and simulation 383 Finding binomial probabilities using printed tables means we don’t have to undertake laborious calculations to obtain the figures we are looking for. We can use them to help us analyse far more complex problems than Example 12.1, such as the problem in Example 12.5. If you look back at Example 12.1, you will find that this is the same as the first figure in the probability distribution. The figure below it, 0.243, is the probability that there is one defective in a packet of three, that is X ϭ 1. Using the binomial equation: Look carefully at this expression. You can see that the first part of it, which involves the factorials, is there to reflect the number of ways there are of getting a packet with one defective, 3(DGG, GDG and GGD). In the earlier expression, for P(X ϭ 0), the first part of the expression came to one, since there is only one way of getting a packet with no defectives (GGG). You may like to try using this method to work out P(X ϭ 2) and P(X ϭ 3). PX() ( ) (!) 1 3! 1!(3 1)! * 0.1 0.1 *2*1 * 0.1(0.9) 6 1(2* 1) * 0.1(0.81) 3* 0.081 0.243 1 2 ϭϭ Ϫ Ϫ ϭ ϭϭϭ Ϫ 1 3 12 31 PX() 0 3! 1(3)! * 1(0.9) 3*2*1 *2*1 * (0.9* 0.9 * 0.9) 1* 0.729 0.729 3 ϭϭ ϭ ϭϭ 3 Example 12.5 Melloch Aviation operates commuter flights out of Chicago using aircraft that can take ten passengers. During each flight passengers are given a hot drink and a ‘Snack Pack’ that contains a ham sandwich and a cake. The company is aware that some of their pas- sengers may be vegetarians and therefore every flight is stocked with one vegetarian Snack Pack that contains a cheese sandwich in addition to ten that contain ham. If 10% of the population are vegetarians, what is the probability that on a fully booked flight there will be at least one vegetarian passenger for whom a meat-free Snack Pack will not be available? This problem has a binomial structure. We will define the variable X as the number of vegetarians on a fully booked flight. Each passenger is a ‘trial’ that can be a ‘success’, [...]... harmful bacteria? (b) more than two eggs contain harmful bacteria? (c) half the eggs contain harmful bacteria? (d) less than two eggs contain harmful bacteria? 400 Quantitative methods for business 12. 8 12. 9* 12. 10 12. 11 12. 12 12. 13 Chapter 12 Two in five people can associate a particular piece of music with the product it is used to promote in an advertising campaign Identify the probability that in... allocation for a certain outcome, for the purposes of the simulation that outcome is deemed to have occurred 390 Quantitative methods for business Chapter 12 Example 12. 8 The Munich company AT-Dalenni Travel specialize in organizing adventure holidays for serious travellers They run ‘Explorer’ trips to the Tien Shan mountain range in Central Asia Each trip has 10 places and they run 12 trips a year The business. .. trials 386 Quantitative methods for business Chapter 12 But what if we need to analyse how many things happen over a period of time? For this sort of situation we use another type of discrete probability distribution known as the Poisson distribution At this point you may find it useful to try Review Questions 12. 3 to 12. 8 at the end of the chapter 12. 3 The Poisson distribution Some types of business. .. the hotel is due to open for the summer season The builder has decided to operate the hotel himself for these 12 weeks To keep this simple he 404 Quantitative methods for business Chapter 12 has decided to accept only one-week bookings There are 8 double rooms and 3 single rooms in the hotel The cost of a double room for one week will be £300, and the cost of a single room for one week, £200 The numbers... 2 and 3 on pages 618–619 of Appendix 1 These tables 394 Quantitative methods for business Chapter 12 should be sufficient for your immediate requirements, but it is useful to know how to produce such information using software because space constraints mean that printed tables can only contain a limited number of versions of the distributions 12. 5.1 EXCEL You can obtain binomial probabilities one at... one probability you can click the button to the left of Input constant: and enter the x value in the space to the right of it 12. 5.3 SPSS For a single binomial probability ■ Choose the Compute option from the Transform pulldown menu 396 Quantitative methods for business Chapter 12 ■ In the Compute Variable window click on PDF.BINOM(q,n,p) in the long list under Functions: on the right-hand side of the... Example 12. 7 graphically See Figure 12. 3 At this point you may find it useful to try Review Questions 12. 9 to 12. 14 at the end of the chapter 12. 4 Simulating business processes Most businesses conduct operations that involve random variables; the numbers of customers booking a vehicle service at a garage, the number of products damaged in transit, the number of workers off sick etc The managers of these businesses... Example 12. 7 without the aid of tables we could calculate them using the formula for the distribution You won’t have to remember it, and probably won’t need to use it, but it may help your understanding if you know where the figures come from The probability that the number of incidents, X, takes a particular value, x, is: P (X ϭ x ) ϭ eϪ␮ * ␮x x! 388 Quantitative methods for business Chapter 12 The... fluctuations all have an effect The profit per customer 392 Quantitative methods for business Chapter 12 varies according to the following probability distribution: Profit per customer (€) Probability 400 500 600 700 0.25 0.35 0.30 0.10 Make random number allocations for this distribution and use the following random numbers to extend the simulation in Example 12. 9 and work out the simulated profit from the twelve... any use For more on simulation try Brooks and Robinson (2001) and Oakshott (1997) At this point you may find it useful to try Review Questions 12. 15 to 12. 19 at the end of the chapter 12. 5 Using the technology: discrete probability distributions and random number generation in EXCEL, discrete probability distributions in MINITAB and SPSS In analysing the examples in section 12. 2 and 12. 3 of this chapter . 0.243 2 0.027 3 0.001 1.000 378 Quantitative methods for business Chapter 12 Figure 12. 1 shows the probability distribution we compiled in Example 12. 1 in graphical form. We can find summary measures. ‘trial’ that can be a ‘success’, 384 Quantitative methods for business Chapter 12 We can show the binomial distribution we used in Example 12. 5 graphically. In Figure 12. 2 the block above 0 represents. Example 12. 7 graphically. See Figure 12. 3. At this point you may find it useful to try Review Questions 12. 9 to 12. 14 at the end of the chapter. 12. 4 Simulating business processes Most businesses