{"id":863,"date":"2017-06-13T23:14:29","date_gmt":"2017-06-13T23:14:29","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-hccc-ma-124-1\/?post_type=chapter&#038;p=863"},"modified":"2018-08-14T20:30:17","modified_gmt":"2018-08-14T20:30:17","slug":"applications-with-probability","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-hccc-ma-124-1\/chapter\/applications-with-probability\/","title":{"raw":"Applications With Probability","rendered":"Applications With Probability"},"content":{"raw":"In the next section, we will explore more complex conditional probabilities and ways to compute them. Conditional probabilities can give us information such as the likelihood of getting a positive test result for a disease without actually having the disease. If a doctor thinks the chances that a positive test result nearly guarantees that a patient has a disease,they might begin an unnecessary and possibly harmful treatment regimen on a healthy patient. If you were to get a positive test result, knowing the likelihood of getting a false positive can guide you to get a second opinion.\r\n<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\nApplications With Probability\r\n<ul>\r\n \t<li>Compute a conditional probability for an event<\/li>\r\n \t<li>Use Baye\u2019s theorem to compute a conditional probability<\/li>\r\n \t<li>Calculate the expected value of an event<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Bayes' Theorem<\/h2>\r\nIn this section we concentrate on the more complex conditional probability problems we began looking at in the last section.\r\n\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/28211759\/800px-Bayes_Theorem_MMB_01.jpg\"><img class=\"aligncenter size-full wp-image-996\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/28211759\/800px-Bayes_Theorem_MMB_01.jpg\" alt=\"Blue neon sign of Bayes' Theorem equation\" width=\"800\" height=\"513\" \/><\/a>\r\n\r\nFor example, suppose a certain disease has an incidence rate of 0.1% (that is, it afflicts 0.1% of the population). A test has been devised to detect this disease. The test does not produce false negatives (that is, anyone who has the disease will test positive for it), but the false positive rate is 5% (that is, about 5% of people who take the test will test positive, even though they do not have the disease). Suppose a randomly selected person takes the test and tests positive.\u00a0 What is the probability that this person actually has the disease?\r\n\r\nThere are two ways to approach the solution to this problem. One involves an important result in probability theory called Bayes' theorem. We will discuss this theorem a bit later, but for now we will use an alternative and, we hope, much more intuitive approach.\r\n\r\nLet's break down the information in the problem piece by piece as an example.\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\n<strong>Suppose a certain disease has an incidence rate of 0.1% (that is, it afflicts 0.1% of the population)<\/strong><em>.<\/em>\u00a0The percentage 0.1% can be converted to a decimal number by moving the decimal place two places to the left, to get 0.001. In turn, 0.001 can be rewritten as a fraction: 1\/1000. This tells us that about 1 in every 1000 people has the disease. (If we wanted we could write <em>P<\/em>(disease)=0.001.)\r\n\r\n<strong>A test has been devised to detect this disease.\u00a0 The test does not produce false negatives (that is, anyone who has the disease will test positive for it)<\/strong><em>.<\/em>\u00a0This part is fairly straightforward: everyone who has the disease will test positive, or alternatively everyone who tests negative does not have the disease. (We could also say <em>P<\/em>(positive | disease)=1.)\r\n\r\n<strong>The false positive rate is 5% (that is, about 5% of people who take the test will test positive, even though they do not have the disease)<\/strong><em>.<\/em>\u00a0This is even more straightforward. Another way of looking at it is that of every 100 people who are tested and do not have the disease, 5 will test positive even though they do not have the disease. (We could also say that <em>P<\/em>(positive | no\u00a0disease)=0.05.)\r\n\r\n<strong>Suppose a randomly selected person takes the test and tests positive.\u00a0 What is the probability that this person actually has the disease?<\/strong>\u00a0Here we want to compute <em>P<\/em>(disease|positive). We already know that <em>P<\/em>(positive|disease)=1, but remember that conditional probabilities are not equal if the conditions are switched.\r\n\r\nRather than thinking in terms of all these probabilities we have developed, let's create a hypothetical situation and apply the facts as set out above. First, suppose we randomly select 1000 people and administer the test. How many do we expect to have the disease? Since about 1\/1000 of all people are afflicted with the disease, 1\/1000 of 1000 people is 1. (Now you know why we chose 1000.) Only 1 of 1000 test subjects actually has the disease; the other 999 do not.\r\n\r\nWe also know that 5% of all people who do not have the disease will test positive. There are 999 disease-free people, so we would expect (0.05)(999)=49.95 (so, about 50) people to test positive who do not have the disease.\r\n\r\nNow back to the original question, computing <em>P<\/em>(disease|positive). There are 51 people who test positive in our example (the one unfortunate person who actually has the disease, plus the 50 people who tested positive but don't). Only one of these people has the disease, so\r\n<p style=\"text-align: center;\">P(disease | positive) [latex]\\approx\\frac{1}{51}\\approx0.0196[\/latex]<\/p>\r\nor less than 2%. Does this surprise you? This means that of all people who test positive, over 98% <em>do not have the disease<\/em>.\r\n\r\nThe answer we got was slightly approximate, since we rounded 49.95 to 50. We could redo the problem with 100,000 test subjects, 100 of whom would have the disease and (0.05)(99,900)=4995 test positive but do not have the disease, so the exact probability of having the disease if you test positive is\r\n<p style=\"text-align: center;\">P(disease | positive) [latex]\\approx\\frac{100}{5095}\\approx0.0196[\/latex]<\/p>\r\nwhich is pretty much the same answer.\r\n\r\nBut back to the surprising result.\u00a0<em>Of all people who test positive, over 98% do not have the disease.<\/em>\u00a0\u00a0If your guess for the probability a person who tests positive has the disease was wildly different from the right answer (2%), don't feel bad. The exact same problem was posed to doctors and medical students at the Harvard Medical School 25 years ago and the results revealed in a 1978 <em>New England Journal of Medicine<\/em> article. Only about 18% of the participants got the right answer. Most of the rest thought the answer was closer to 95% (perhaps they were misled by the false positive rate of 5%).\r\n\r\nSo at least you should feel a little better that a bunch of doctors didn't get the right answer either (assuming you thought the answer was much higher). But the significance of this finding and similar results from other studies in the intervening years lies not in making math students feel better but in the possibly catastrophic consequences it might have for patient care. If a doctor thinks the chances that a positive test result nearly guarantees that a patient has a disease, they might begin an unnecessary and possibly harmful treatment regimen on a healthy patient.\u00a0 Or worse, as in the early days of the AIDS crisis when being HIV-positive was often equated with a death sentence, the patient might take a drastic action and commit suicide.\r\n\r\nThis example is worked through in detail in the video here.\r\n\r\nhttps:\/\/youtu.be\/hXevfqsBino\r\n\r\n<\/div>\r\nAs we have seen in this hypothetical example, the most responsible course of action for treating a patient who tests positive would be to counsel the patient that they most likely do <em>not<\/em> have the disease and to order further, more reliable, tests to verify the diagnosis.\r\n\r\nOne of the reasons that the doctors and medical students in the study did so poorly is that such problems, when presented in the types of statistics courses that medical students often take, are solved by use of Bayes' theorem, which is stated as follows:\r\n<div class=\"textbox\">\r\n<h3>Bayes\u2019 Theorem<\/h3>\r\n<p style=\"text-align: center;\">[latex]P(A|B)=\\frac{P(A)P(B|A)}{P(A)P(B|A)+P(\\bar{A})P(B|\\bar{A})}[\/latex]<\/p>\r\n\r\n<\/div>\r\nIn our earlier example, this translates to\r\n<p style=\"text-align: center;\">[latex]P(\\text{disease}|\\text{positive})=\\frac{P(\\text{disease})P(\\text{positive}|\\text{disease})}{P(\\text{disease})P(\\text{positive}|\\text{disease})+P(\\text{nodisease})P(\\text{positive}|\\text{nodisease})}[\/latex]<\/p>\r\nPlugging in the numbers gives\r\n<p style=\"text-align: center;\">[latex]P(\\text{disease}|\\text{positive})=\\frac{(0.001)(1)}{(0.001)(1)+(0.999)(0.05)}\\approx0.0196[\/latex]<\/p>\r\nwhich is exactly the same answer as our original solution.\r\n\r\nThe problem is that you (or the typical medical student, or even the typical math professor) are much more likely to be able to remember the original solution than to remember Bayes' theorem. Psychologists, such as Gerd Gigerenzer, author of <em>Calculated Risks: How to Know When Numbers Deceive You<\/em>, have advocated that the method involved in the original solution (which Gigerenzer calls the method of \"natural frequencies\") be employed in place of Bayes' Theorem. Gigerenzer performed a study and found that those educated in the natural frequency method were able to recall it far longer than those who were taught Bayes' theorem. When one considers the possible life-and-death consequences associated with such calculations it seems wise to heed his advice.\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\nA certain disease has an incidence rate of 2%. If the false negative rate is 10% and the false positive rate is 1%, compute the probability that a person who tests positive actually has the disease.\r\n[reveal-answer q=\"507646\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"507646\"]\r\n\r\nImagine 10,000 people who are tested. Of these 10,000, 200 will have the disease; 10% of them, or 20, will test negative and the remaining 180 will test positive. Of the 9800 who do not have the disease, 98 will test positive. So of the 278 total people who test positive, 180 will have the disease. Thus\r\n<p style=\"text-align: center;\">[latex]P(\\text{disease}|\\text{positive})=\\frac{180}{278}\\approx0.647[\/latex]<\/p>\r\nso about 65% of the people who test positive will have the disease.\r\n\r\nUsing Bayes theorem directly would give the same result:\r\n<p style=\"text-align: center;\">[latex]P(\\text{disease}|\\text{positive})=\\frac{(0.02)(0.90)}{(0.02)(0.90)+(0.98)(0.01)}=\\frac{0.018}{0.0278}\\approx0.647[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\nView the following for more about this example.\r\n\r\nhttps:\/\/youtu.be\/_c3xZvHto3k\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try it now<\/h3>\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=17494&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"350\"><\/iframe>\r\n\r\n<\/div>\r\n<h2>Counting<\/h2>\r\nCounting? You already know how to count or you wouldn't be taking a college-level math class, right? Well yes, but what we'll really be investigating here are ways of counting <em>efficiently<\/em>. When we get to the probability situations a bit later in this chapter we will need to count some <em>very<\/em> large numbers, like the number of possible winning lottery tickets. One way to do this would be to write down every possible set of numbers that might show up on a lottery ticket, but believe me: you don't want to do this.\r\n<h3>Basic Counting<\/h3>\r\nWe will start, however, with some more reasonable sorts of counting problems in order to develop the ideas that we will soon need.\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\nSuppose at a particular restaurant you have three choices for an appetizer (soup, salad or breadsticks) and five choices for a main course (hamburger, sandwich, quiche, fajita or pizza). If you are allowed to choose exactly one item from each category for your meal, how many different meal options do you have?\r\n\r\n<strong>Solution 1<\/strong>: One way to solve this problem would be to systematically list each possible meal:\r\n\r\nsoup + hamburger \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 soup + sandwich \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 soup + quiche\r\n\r\nsoup + fajita \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 soup + pizza \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 salad + hamburger\r\n\r\nsalad + sandwich \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 salad + quiche \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 salad + fajita\r\n\r\nsalad + pizza \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 breadsticks + hamburger \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 breadsticks + sandwich\r\n\r\nbreadsticks + quiche \u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 breadsticks + fajita \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 breadsticks + pizza\r\n\r\nAssuming that we did this systematically and that we neither missed any possibilities nor listed any possibility more than once, the answer would be 15. Thus you could go to the restaurant 15 nights in a row and have a different meal each night.\r\n\r\n<strong>Solution 2<\/strong>: Another way to solve this problem would be to list all the possibilities in a table:\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td><\/td>\r\n<td>hamburger<\/td>\r\n<td>sandwich<\/td>\r\n<td>quiche<\/td>\r\n<td>fajita<\/td>\r\n<td>pizza<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>soup<\/td>\r\n<td>soup+burger<\/td>\r\n<td><\/td>\r\n<td><\/td>\r\n<td><\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>salad<\/td>\r\n<td>salad+burger<\/td>\r\n<td><\/td>\r\n<td><\/td>\r\n<td><\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>bread<\/td>\r\n<td>etc<\/td>\r\n<td><\/td>\r\n<td><\/td>\r\n<td><\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nIn each of the cells in the table we could list the corresponding meal: soup + hamburger in the upper left corner, salad + hamburger below it, etc. But if we didn't really care <em>what<\/em> the possible meals are, only <em>how many<\/em> possible meals there are, we could just count the number of cells and arrive at an answer of 15, which matches our answer from the first solution. (It's always good when you solve a problem two different ways and get the same answer!)\r\n\r\n<strong>Solution 3<\/strong>: We already have two perfectly good solutions. Why do we need a third?\u00a0 The first method was not very systematic, and we might easily have made an omission. The second method was better, but suppose that in addition to the appetizer and the main course we further complicated the problem by adding desserts to the menu: we've used the rows of the table for the appetizers and the columns for the main courses\u2014where will the desserts go? We would need a third dimension, and since drawing 3-D tables on a 2-D page or computer screen isn't terribly easy, we need a better way in case we have three categories to choose form instead of just two.\r\n\r\nSo, back to the problem in the example.\u00a0 What else can we do?\u00a0 Let's draw a <strong>tree diagram<\/strong>:\r\n\r\n<img class=\"aligncenter size-full wp-image-341\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/11181317\/treediagram.png\" alt=\"Tree diagram of above table\" width=\"375\" height=\"225\" \/>\r\n\r\nThis is called a \"tree\" diagram because at each stage we branch out, like the branches on a tree.\u00a0 In this case, we first drew five branches (one for each main course) and then for each of those branches we drew three more branches (one for each appetizer).\u00a0 We count the number of branches at the final level and get (surprise, surprise!) 15.\r\n\r\nIf we wanted, we could instead draw three branches at the first stage for the three appetizers and then five branches (one for each main course) branching out of each of those three branches.\r\n\r\n<\/div>\r\nOK, so now we know how to count possibilities using tables and tree diagrams. These methods will continue to be useful in certain cases, but imagine a game where you have two decks of cards (with 52 cards in each deck) and you select one card from each deck. Would you really want to draw a table or tree diagram to determine the number of outcomes of this game?\r\n\r\nLet's go back to the previous example that involved selecting a meal from three appetizers and five main courses, and look at the second solution that used a table. Notice that one way to count the number of possible meals is simply to number each of the appropriate cells in the table, as we have done above. But another way to count the number of cells in the table would be multiply the number of rows (3) by the number of columns (5) to get 15. Notice that we could have arrived at the same result without making a table at all by simply multiplying the number of choices for the appetizer (3) by the number of choices for the main course (5). We generalize this technique as the <em>basic counting rule<\/em>:\r\n<div class=\"textbox\">\r\n<h3>Basic Counting Rule<\/h3>\r\nIf we are asked to choose one item from each of two separate categories where there are <em>m<\/em> items in the first category and <em>n<\/em> items in the second category, then the total number of available choices is <em>m <\/em><strong>\u00b7 <\/strong><em>n<\/em>.\r\n\r\nThis is sometimes called the multiplication rule for probabilities.\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\nThere are 21 novels and 18 volumes of poetry on a reading list for a college English course. How many different ways can a student select one novel and one volume of poetry to read during the quarter?\r\n[reveal-answer q=\"16976\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"16976\"]There are 21 choices from the first category and 18 for the second, so there are 21\u00a0<strong>\u00b7<\/strong>\u00a018\u00a0=\u00a0378 possibilities.[\/hidden-answer]\r\n\r\n<\/div>\r\nThe Basic Counting Rule can be extended when there are more than two categories by applying it repeatedly, as we see in the next example.\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\nSuppose at a particular restaurant you have three choices for an appetizer (soup, salad or breadsticks), five choices for a main course (hamburger, sandwich, quiche, fajita or pasta) and two choices for dessert (pie or ice cream). If you are allowed to choose exactly one item from each category for your meal, how many different meal options do you have?\r\n[reveal-answer q=\"13816\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"13816\"]\r\n\r\nThere are 3 choices for an appetizer, 5 for the main course and 2 for dessert, so there are 3 <strong>\u00b7<\/strong> 5 <strong>\u00b7<\/strong> 2 = 30 possibilities.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try it now<\/h3>\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=7159&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"300\"><\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nA quiz consists of 3 true-or-false questions.\u00a0 In how many ways can a student answer the quiz?\r\n[reveal-answer q=\"169025\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"169025\"]There are 3 questions. Each question has 2 possible answers (true or false), so the quiz may be answered in 2 <strong>\u00b7<\/strong> 2 <strong>\u00b7<\/strong> 2 = 8 different ways.\u00a0 Recall that another way to write 2 <strong>\u00b7<\/strong> 2 <strong>\u00b7<\/strong> 2 is 2<sup>3<\/sup>, which is much more compact.[\/hidden-answer]\r\n\r\n<\/div>\r\nBasic counting examples from this section are described in the following video.\r\n\r\nhttps:\/\/youtu.be\/fROqcu-ekkw\r\n<h3>Permutations<\/h3>\r\nIn this section we will develop an even faster way to solve some of the problems we have already learned to solve by other means.\u00a0 Let's start with a couple examples.\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\nHow many different ways can the letters of the word MATH be rearranged to form a four-letter code word?\r\n[reveal-answer q=\"614648\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"614648\"]This problem is a bit different.\u00a0 Instead of choosing one item from each of several different categories, we are repeatedly choosing items from the <em>same<\/em> category (the category is: the letters of the word MATH) and each time we choose an item we <em>do not replace<\/em> it, so there is one fewer choice at the next stage: we have 4 choices for the first letter (say we choose A), then 3 choices for the second (M, T and H; say we choose H), then 2 choices for the next letter (M and T; say we choose M) and only one choice at the last stage (T).\u00a0 Thus there are 4 <strong>\u00b7 <\/strong>3 <strong>\u00b7<\/strong> 2 <strong>\u00b7<\/strong> 1 = 24 ways to spell a code worth with the letters MATH.[\/hidden-answer]\r\n\r\n<\/div>\r\nIn this example, we needed to calculate <em>n<\/em>\u00a0\u00b7 (<em>n<\/em> \u2013 1) \u00b7 (<em>n<\/em> \u2013 2) \u00b7\u00b7\u00b7 3 \u00b7 2 \u00b7 1. This calculation shows up often in mathematics, and is called the <strong>factorial<\/strong>, and is notated <em>n<\/em>!\r\n<div class=\"textbox\">\r\n<h3>Factorial<\/h3>\r\n<p style=\"text-align: center;\"><em>n<\/em>! = <em>n<\/em>\u00a0\u00b7 (<em>n<\/em> \u2013 1) \u00b7 (<em>n<\/em> \u2013 2) \u00b7\u00b7\u00b7 3 \u00b7 2 \u00b7 1<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try it now<\/h3>\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=7188&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"350\"><\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\nHow many ways can five different door prizes be distributed among five people?\r\n[reveal-answer q=\"530386\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"530386\"]There are 5 choices of prize for the first person, 4 choices for the second, and so on. The number of ways the prizes can be distributed will be 5! = 5 <strong>\u00b7 <\/strong>4 <strong>\u00b7 <\/strong>3 <strong>\u00b7<\/strong> 2 <strong>\u00b7<\/strong> 1 = 120 ways.[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try it now<\/h3>\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=7191&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"300\"><\/iframe>\r\n\r\n<\/div>\r\nNow we will consider some slightly different examples.\r\n<div class=\"textbox exercises\">\r\n<h3>examples<\/h3>\r\nA charity benefit is attended by 25 people and three gift certificates are given away as door prizes: one gift certificate is in the amount of $100, the second is worth $25 and the third is worth $10.\u00a0 Assuming that no person receives more than one prize, how many different ways can the three gift certificates be awarded?\r\n[reveal-answer q=\"75787\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"75787\"]Using the Basic Counting Rule, there are 25 choices for the person who receives the $100 certificate, 24 remaining choices for the $25 certificate and 23 choices for the $10 certificate, so there are 25 <strong>\u00b7<\/strong> 24 <strong>\u00b7<\/strong> 23 = 13,800 ways in which the prizes can be awarded.[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nEight sprinters have made it to the Olympic finals in the 100-meter race. In how many different ways can the gold, silver and bronze medals be awarded?\r\n[reveal-answer q=\"68166\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"68166\"]\r\n\r\nUsing the Basic Counting Rule, there are 8 choices for the gold medal winner, 7 remaining choices for the silver, and 6 for the bronze, so there are 8 <strong>\u00b7<\/strong> 7 <strong>\u00b7<\/strong> 6 = 336 ways the three medals can be awarded to the 8 runners.\r\n\r\nNote that in these preceding examples, the gift certificates and the Olympic medals were awarded <em>without replacement<\/em>; that is, once we have chosen a winner of the first door prize or the gold medal, they are not eligible for the other prizes. Thus, at each succeeding stage of the solution there is one fewer choice (25, then 24, then 23 in the first example; 8, then 7, then 6 in the second).\u00a0 Contrast this with the situation of a multiple choice test, where there might be five possible answers\u00a0\u2014 A, B, C, D or E \u2014 for each question on the test.\r\n\r\nNote also that <em>the order of selection was important<\/em> in each example: for the three door prizes, being chosen first means that you receive substantially more money; in the Olympics example, coming in first means that you get the gold medal instead of the silver or bronze. In each case, if we had chosen the same three people in a different order there might have been a different person who received the $100 prize, or a different goldmedalist. (Contrast this with the situation where we might draw three names out of a hat to each receive a $10 gift certificate; in this case the order of selection is <em>not<\/em> important since each of the three people receive the same prize.\u00a0 Situations where the order is <em>not<\/em> important will be discussed in the next section.)\r\n\r\n[\/hidden-answer]\r\n\r\nFactorial examples are worked in this video.\r\n\r\nhttps:\/\/youtu.be\/9maoGi5fd_M\r\n\r\n<\/div>\r\nWe can generalize the situation in the two examples above to any problem <em>without replacement<\/em> where the <em>order of selection is important<\/em>. If we are arranging in order <em>r<\/em> items out of <em>n<\/em> possibilities (instead of 3 out of 25 or 3 out of 8 as in the previous examples), the number of possible arrangements will be given by\r\n<p style=\"text-align: center;\"><em>n<\/em>\u00a0\u00b7 (<em>n<\/em> \u2013 1) \u00b7 (<em>n<\/em> \u2013 2) \u00b7\u00b7\u00b7 (<em>n<\/em> \u2013 <em>r<\/em> + 1)<\/p>\r\nIf you don't see why (<em>n <\/em>\u2014 <em>r <\/em>+ 1) is the right number to use for the last factor, just think back to the first example in this section, where we calculated 25 <strong>\u00b7<\/strong> 24 <strong>\u00b7<\/strong> 23 to get 13,800. In this case <em>n<\/em> = 25 and <em>r<\/em> = 3, so <em>n <\/em>\u2014 <em>r <\/em>+ 1 = 25 \u2014 3 + 1 = 23, which is exactly the right number for the final factor.\r\n\r\nNow, why would we want to use this complicated formula when it's actually easier to use the Basic Counting Rule, as we did in the first two examples? Well, we won't actually use this formula all that often; we only developed it so that we could attach a special notation and a special definition to this situation where we are choosing <em>r<\/em> items out of <em>n<\/em> possibilities <em>without replacement<\/em> and where the <em>order of selection is important<\/em>. In this situation we write:\r\n<div class=\"textbox\">\r\n<h3>Permutations<\/h3>\r\n<p style=\"text-align: center;\"><em>nPr<\/em> = <em>n<\/em>\u00a0\u00b7 (<em>n<\/em> \u2013 1) \u00b7 (<em>n<\/em> \u2013 2) \u00b7\u00b7\u00b7 (<em>n<\/em> \u2013 <em>r<\/em> + 1)<\/p>\r\nWe say that there are <em>nPr<\/em> <strong>permutations<\/strong> of size <em>r<\/em> that may be selected from among <em>n<\/em> choices <em>without replacement<\/em> when <em>order matters<\/em>.\r\n\r\nIt turns out that we can express this result more simply using factorials.\r\n<p style=\"text-align: center;\">[latex]{}_{n}{{P}_{r}}=\\frac{n!}{(n-r)!}[\/latex]<\/p>\r\n\r\n<\/div>\r\nIn practicality, we usually use technology rather than factorials or repeated multiplication to compute permutations.\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\nI have nine paintings and have room to display only four of them at a time on my wall. How many different ways could I do this?\r\n[reveal-answer q=\"366612\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"366612\"]Since we are choosing 4 paintings out of 9 <em>without replacement<\/em> where the <em>order of selection is important<\/em> there are 9<em>P<\/em>4 = 9 \u00b7 8 \u00b7 7 \u00b7 6 = 3,024 permutations.[\/hidden-answer]\r\n\r\n<hr \/>\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nHow many ways can a four-person executive committee (president, vice-president, secretary, treasurer) be selected from a 16-member board of directors of a non-profit organization?\r\n[reveal-answer q=\"889810\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"889810\"]We want to choose 4 people out of 16 without replacement and where the order of selection is important. So the answer is 16<em>P<\/em>4 = 16 \u00b7 15 \u00b7 14 \u00b7 13 = 43,680.[\/hidden-answer]\r\n\r\nView this video to see more about the permutations examples.\r\n\r\nhttps:\/\/youtu.be\/xlyX2UJMJQI\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It Now<\/h3>\r\nHow many 5 character passwords can be made using the letters A through Z\r\n<ul>\r\n \t<li>if repeats are allowed<\/li>\r\n \t<li>if no repeats are allowed<\/li>\r\n<\/ul>\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=7193&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"300\"><\/iframe>\r\n\r\n<\/div>\r\n&nbsp;\r\n<h3>Combinations<\/h3>\r\nIn the previous section we considered the situation where we chose <em>r<\/em> items out of <em>n<\/em> possibilities <em>without replacement<\/em> and where the <em>order of selection was important<\/em>. We now consider a similar situation in which the order of selection is <em>not<\/em> important.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nA charity benefit is attended by 25 people at which three $50 gift certificates are given away as door prizes. Assuming no person receives more than one prize, how many different ways can the gift certificates be awarded?\r\n[reveal-answer q=\"942898\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"942898\"]\r\n\r\nUsing the Basic Counting Rule, there are 25 choices for the first person, 24 remaining choices for the second person and 23 for the third, so there are 25 \u00b7 24 \u00b7 23 = 13,800 ways to choose three people. Suppose for a moment that Abe is chosen first, Bea second and Cindy third; this is one of the 13,800 possible outcomes. Another way to award the prizes would be to choose Abe first, Cindy second and Bea third; this is another of the 13,800 possible outcomes. But either way Abe, Bea and Cindy each get $50, so it doesn't really matter the order in which we select them. In how many different orders can Abe, Bea and Cindy be selected? It turns out there are 6:\r\n\r\nABC\u00a0\u00a0\u00a0\u00a0 ACB\u00a0\u00a0\u00a0\u00a0 BAC\u00a0\u00a0\u00a0\u00a0 BCA\u00a0\u00a0\u00a0\u00a0 CAB\u00a0\u00a0\u00a0\u00a0 CBA\r\n\r\nHow can we be sure that we have counted them all? We are really just choosing 3 people out of 3, so there are 3 \u00b7 2 \u00b7 1 = 6 ways to do this; we didn't really need to list them all. We can just use permutations!\r\n\r\nSo, out of the 13,800 ways to select 3 people out of 25, six of them involve Abe, Bea and Cindy. The same argument works for any other group of three people (say Abe, Bea and David or Frank, Gloria and Hildy) so each three-person group is counted <em>six times<\/em>. Thus the 13,800 figure is six times too big. The number of distinct three-person groups will be 13,800\/6 = 2300.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\nWe can generalize the situation in this example above to any problem of choosing a collection of items <em>without replacement<\/em> where the <em>order of selection is <strong>not<\/strong> important<\/em>. If we are choosing <em>r<\/em> items out of <em>n<\/em> possibilities (instead of 3 out of 25 as in the previous examples), the number of possible choices will be given by [latex]\\frac{{}_{n}{{P}_{r}}}{{}_{r}{{P}_{r}}}[\/latex], and we could use this formula for computation. However this situation arises so frequently that we attach a special notation and a special definition to this situation where we are choosing <em>r<\/em> items out of <em>n<\/em> possibilities <em>without replacement<\/em> where the <em>order of selection is <strong>not<\/strong> important<\/em>.\r\n<div class=\"textbox\">\r\n<h3>Combinations<\/h3>\r\n<p style=\"text-align: center;\">[latex]{}_{n}{{C}_{r}}=\\frac{{}_{n}{{P}_{r}}}{{}_{r}{{P}_{r}}}[\/latex]<\/p>\r\nWe say that there are <em>nCr<\/em> <strong>combinations<\/strong> of size <em>r<\/em> that may be selected from among <em>n<\/em> choices <em>without replacement<\/em> where <em>order doesn\u2019t matter<\/em>.\r\n\r\nWe can also write the combinations formula in terms of factorials:\r\n<p style=\"text-align: center;\">[latex]{}_{n}{{C}_{r}}=\\frac{n!}{(n-r)!r!}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nA group of four students is to be chosen from a 35-member class to represent the class on the student council. How many ways can this be done?\r\n[reveal-answer q=\"125383\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"125383\"]Since we are choosing 4 people out of 35 <em>without replacement<\/em> where the <em>order of selection is <strong>not<\/strong> important<\/em> there are [latex]{}_{35}{{C}_{4}}=\\frac{35\\cdot34\\cdot33\\cdot32}{4\\cdot3\\cdot2\\cdot1}[\/latex] = 52,360 combinations.[\/hidden-answer]\r\n\r\nView the following for more explanation of the combinations examples.\r\n\r\nhttps:\/\/youtu.be\/W8kd4YosbzE\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It Now<\/h3>\r\nThe United States Senate Appropriations Committee consists of 29 members; the Defense Subcommittee of the Appropriations Committee consists of 19 members. Disregarding party affiliation or any special seats on the Subcommittee, how many different 19-member subcommittees may be chosen from among the 29 Senators on the Appropriations Committee?\r\n\r\n<\/div>\r\nIn the preceding Try It Now problem we assumed that the 19 members of the Defense Subcommittee were chosen without regard to party affiliation. In reality this would never happen: if Republicans are in the majority they would never let a majority of Democrats sit on (and thus control) any subcommittee.\u00a0 (The same of course would be true if the Democrats were in control.) So let's consider the problem again, in a slightly more complicated form:\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nThe United States Senate Appropriations Committee consists of 29 members, 15 Republicans and 14 Democrats. The Defense Subcommittee consists of 19 members, 10 Republicans and 9 Democrats. How many different ways can the members of the Defense Subcommittee be chosen from among the 29 Senators on the Appropriations Committee?\r\n[reveal-answer q=\"205320\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"205320\"]\r\n\r\nIn this case we need to choose 10 of the 15 Republicans and 9 of the 14 Democrats.\u00a0 There are 15<em>C<\/em>10 = 3003 ways to choose the 10 Republicans and 14<em>C<\/em>9 = 2002 ways to choose the 9 Democrats. But now what?\u00a0 How do we finish the problem?\r\n\r\nSuppose we listed all of the possible 10-member Republican groups on 3003 slips of red paper and \u00a0all of the possible 9-member Democratic groups on 2002 slips of blue paper.\u00a0 How many ways can we choose one red slip and one blue slip?\u00a0 This is a job for the Basic Counting Rule!\u00a0 We are simply making one choice from the first category and one choice from the second category, just like in the restaurant menu problems from earlier.\r\n\r\nThere must be 3003 \u00b7 2002 = 6,012,006 possible ways of selecting the members of the Defense Subcommittee.\r\n\r\n[\/hidden-answer]\r\n\r\nThis example is worked through below.\r\n\r\nhttps:\/\/youtu.be\/Xqc2sdYN7xo\r\n\r\n<\/div>\r\n<h2>Probability Using Permutations and Combinations<\/h2>\r\nWe can use permutations and combinations to help us answer more complex probability questions.\r\n<div class=\"textbox exercises\">\r\n<h3>examples<\/h3>\r\nA 4 digit PIN number is selected. What is the probability that there are no repeated digits?\r\n[reveal-answer q=\"108643\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"108643\"]\r\n\r\nThere are 10 possible values for each digit of the PIN (namely: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9), so there are 10 \u00b7 10 \u00b7 10 \u00b7 10 = 10<sup>4<\/sup> = 10000 total possible PIN numbers.\r\n\r\nTo have no repeated digits, all four digits would have to be different, which is selecting without replacement. We could either compute 10 \u00b7 9 \u00b7 8 \u00b7 7, or notice that this is the same as the permutation 10<em>P<\/em>4 = 5040.\r\n\r\nThe probability of no repeated digits is the number of 4 digit PIN numbers with no repeated digits divided by the total number of 4 digit PIN numbers. This probability is [latex]\\frac{{}_{10}{{P}_{4}}}{{{10}^{4}}}=\\frac{5040}{10000}=0.504[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It Now<\/h3>\r\n<iframe id=\"mom10\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=7158&amp;theme=oea&amp;iframe_resize_id=mom10\" width=\"100%\" height=\"400\"><\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nIn a certain state's lottery, 48 balls numbered 1 through 48 are placed in a machine and six of them are drawn at random. If the six numbers drawn match the numbers that a player had chosen, the player wins $1,000,000. \u00a0\u00a0\u00a0In this lottery, the order the numbers are drawn in doesn\u2019t matter. Compute the probability that you win the million-dollar prize if you purchase a single lottery ticket.\r\n[reveal-answer q=\"767485\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"767485\"]\r\n\r\nIn order to compute the probability, we need to count the total number of ways six numbers can be drawn, and the number of ways the six numbers on the player\u2019s ticket could match the six numbers drawn from the machine. Since there is no stipulation that the numbers be in any particular order, the number of possible outcomes of the lottery drawing is 48<em>C<\/em>6 = 12,271,512. Of these possible outcomes, only one would match all six numbers on the player\u2019s ticket, so the probability of winning the grand prize is:\r\n\r\n[latex]\\frac{{}_{6}{{C}_{6}}}{{}_{48}{{C}_{6}}}=\\frac{1}{12271512}\\approx0.0000000815[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nIn the state lottery from the previous example, if five of the six numbers drawn match the numbers that a player has chosen, the player wins a second prize of $1,000. Compute the probability that you win the second prize if you purchase a single lottery ticket.\r\n[reveal-answer q=\"651173\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"651173\"]\r\n\r\nAs above, the number of possible outcomes of the lottery drawing is 48<em>C<\/em>6 = 12,271,512. In order to win the second prize, five of the six numbers on the ticket must match five of the six winning numbers; in other words, we must have chosen five of the six winning numbers and one of the 42 losing numbers. The number of ways to choose 5 out of the 6 winning numbers is given by 6<em>C<\/em>5 = 6 and the number of ways to choose 1 out of the 42 losing numbers is given by 42<em>C<\/em>1 = 42. Thus the number of favorable outcomes is then given by the Basic Counting Rule: 6<em>C<\/em>5 \u00b7 42<em>C<\/em>1 = 6 \u00b7 42 = 252. So the probability of winning the second prize is.\r\n\r\n[latex]\\frac{\\left({}_{6}{{C}_{5}}\\right)\\left({}_{42}{{C}_{1}}\\right)}{{}_{48}{{C}_{6}}}=\\frac{252}{12271512}\\approx0.0000205[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe previous examples are worked in the following video.\r\n\r\nhttps:\/\/youtu.be\/b9LFbB_aNAo\r\n<div class=\"textbox exercises\">\r\n<h3>examples<\/h3>\r\nCompute the probability of randomly drawing five cards from a deck and getting exactly one Ace.\r\n\r\n[reveal-answer q=\"542442\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"542442\"]\r\n\r\nIn many card games (such as poker) the order in which the cards are drawn is not important (since the player may rearrange the cards in his hand any way he chooses); in the problems that follow, we will assume that this is the case unless otherwise stated. Thus we use combinations to compute the possible number of 5-card hands, 52<em>C<\/em>5. This number will go in the denominator of\u00a0our probability formula, since it is the number of possible outcomes.\r\n\r\nFor the numerator, we need the number of ways to draw one Ace and four other cards (none of them Aces) from the deck.\u00a0 Since there are four Aces and we want exactly one of them, there will be 4<em>C<\/em>1 ways to select one Ace; since there are 48 non-Aces and we want 4 of them, there will be 48<em>C<\/em>4 ways to select the four non-Aces.\u00a0 Now we use the Basic Counting Rule to calculate that there will be 4<em>C<\/em>1 \u00b7 48<em>C<\/em>4 ways to choose one ace and four non-Aces.\r\n\r\nPutting this all together, we have\r\n<p style=\"text-align: center;\">[latex]P(\\text{oneAce})=\\frac{\\left({}_{4}{{C}_{1}}\\right)\\left({}_{48}{{C}_{4}}\\right)}{{}_{52}{{C}_{5}}}=\\frac{778320}{2598960}\\approx0.299[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nCompute the probability of randomly drawing five cards from a deck and getting exactly two Aces.\r\n[reveal-answer q=\"49360\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"49360\"]\r\n\r\nThe solution is similar to the previous example, except now we are choosing 2 Aces out of 4 and 3 non-Aces out of 48; the denominator remains the same:\r\n\r\n[latex]P(\\text{twoAces})=\\frac{\\left({}_{4}{{C}_{2}}\\right)\\left({}_{48}{{C}_{3}}\\right)}{{}_{52}{{C}_{5}}}=\\frac{103776}{2598960}\\approx0.0399[\/latex]\r\n\r\nIt is useful to note that these card problems are remarkably similar to the lottery problems discussed earlier.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nView the following for further demonstration of these examples.\r\n\r\nhttps:\/\/youtu.be\/RU3e3KTkjoA\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try it now<\/h3>\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=7157&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"400\"><\/iframe>\r\n\r\n<\/div>\r\n<h3>Birthday Problem<\/h3>\r\nLet's take a pause to consider a famous problem in probability theory:\r\n<div class=\"textbox shaded\">Suppose you have a room full of 30 people. What is the probability that there is at least one shared birthday?<\/div>\r\nTake a guess at the answer to the above problem. Was your guess fairly low, like around 10%? That seems to be the intuitive answer (30\/365, perhaps?). Let's see if we should listen to our intuition. Let's start with a simpler problem, however.\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\nSuppose three people are in a room.\u00a0 What is the probability that there is at least one shared birthday among these three people?\r\n[reveal-answer q=\"226841\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"226841\"]\r\n\r\nThere are a lot of ways there could be at least one shared birthday. Fortunately there is an easier way. We ask ourselves \u201cWhat is the alternative to having at least one shared birthday?\u201d In this case, the alternative is that there are <strong>no<\/strong> shared birthdays. In other words, the alternative to \u201cat least one\u201d is having <strong>none<\/strong>. In other words, since this is a complementary event,\r\n<p style=\"text-align: center;\">P(at least one) = 1 \u2013 P(none)<\/p>\r\nWe will start, then, by computing the probability that there is no shared birthday.\u00a0 Let's imagine that you are one of these three people.\u00a0 Your birthday can be anything without conflict, so there are 365 choices out of 365 for your birthday. What is the probability that the second person does not share your birthday?\u00a0 There are 365 days in the year (let's ignore leap years) and removing your birthday from contention, there are 364 choices that will guarantee that you do not share a birthday with this person, so the probability that the second person does not share your birthday is 364\/365.\u00a0 Now we move to the third person.\u00a0 What is the probability that this third person does not have the same birthday as either you or the second person?\u00a0 There are 363 days that will not duplicate your birthday or the second person's, so the probability that the third person does not share a birthday with the first two is 363\/365.\r\n\r\nWe want the second person not to share a birthday with you <em>and<\/em> the third person not to share a birthday with the first two people, so we use the multiplication rule:\r\n<p style=\"text-align: center;\">[latex]P(\\text{nosharedbirthday})=\\frac{365}{365}\\cdot\\frac{364}{365}\\cdot\\frac{363}{365}\\approx0.9918[\/latex]<\/p>\r\nand then subtract from 1 to get\r\n<p style=\"text-align: center;\">P(shared birthday) = 1 \u2013 P(no shared birthday) = 1 \u2013 0.9918 = 0.0082.<\/p>\r\nThis is a pretty small number, so maybe it makes sense that the answer to our original problem will be small.\u00a0 Let's make our group a bit bigger.\r\n\r\n[\/hidden-answer]\r\n\r\n<hr \/>\r\n\r\nSuppose five people are in a room.\u00a0 What is the probability that there is at least one shared birthday among these five people?\r\n[reveal-answer q=\"6410\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"6410\"]\r\n\r\nContinuing the pattern of the previous example, the answer should be\r\n<p style=\"text-align: center;\">[latex]P(\\text{sharedbirthday})=1-\\frac{365}{365}\\cdot\\frac{364}{365}\\cdot\\frac{363}{365}\\cdot\\frac{362}{365}\\cdot\\frac{361}{365}\\approx0.0271[\/latex]<\/p>\r\nNote that we could rewrite this more compactly as\r\n<p style=\"text-align: center;\">[latex]P(\\text{sharedbirthday})=1-\\frac{{}_{365}{{P}_{5}}}{{{365}^{5}}}\\approx0.0271[\/latex]<\/p>\r\nwhich makes it a bit easier to type into a calculator or computer, and which suggests a nice formula as we continue to expand the population of our group.\r\n\r\n[\/hidden-answer]\r\n\r\n<hr \/>\r\n\r\nSuppose 30 people are in a room.\u00a0 What is the probability that there is at least one shared birthday among these 30 people?\r\n[reveal-answer q=\"741907\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"741907\"]\r\n\r\nHere we can calculate\r\n<p style=\"text-align: center;\">[latex]P(\\text{sharedbirthday})=1-\\frac{{}_{365}{{P}_{30}}}{{{365}^{30}}}\\approx0.706[\/latex]<\/p>\r\nwhich gives us the surprising result that when you are in a room with 30 people there is a 70% chance that there will be at least one shared birthday!\r\n\r\n[\/hidden-answer]\r\n\r\nThe birthday problem\u00a0is examined in detail in the following.\r\n\r\nhttps:\/\/youtu.be\/UUmTfiJ_0k4\r\n\r\n<\/div>\r\nIf you like to bet, and if you can convince 30 people to reveal their birthdays, you might be able to win some money by betting a friend that there will be at least two people with the same birthday in the room anytime you are in a room of 30 or more people. (Of course, you would need to make sure your friend hasn't studied probability!) You wouldn't be guaranteed to win, but you should win more than half the time.\r\n\r\nThis is one of many results in probability theory that is counterintuitive; that is, it goes against our gut instincts.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It Now<\/h3>\r\nSuppose 10 people are in a room. What is the probability that there is at least one shared birthday among these 10 people?\r\n\r\n<\/div>\r\n<h2>Expected Value<\/h2>\r\n<h3>Repeating Procedures Over Time<\/h3>\r\nExpected value is perhaps the most useful probability concept we will discuss.\u00a0 It has many applications, from insurance policies to making financial decisions, and it's one thing that the casinos and government agencies that run gambling operations and lotteries hope most people never learn about.<img class=\"aligncenter wp-image-343\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/11182717\/roulette.jpg\" alt=\"A roulette wheel\" width=\"600\" height=\"400\" \/>\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\nIn the casino game roulette, a wheel with 38 spaces (18 red, 18 black, and 2 green) is spun. In one possible bet, the player bets $1 on a single number. If that number is spun on the wheel, then they receive $36 (their original $1 + $35). Otherwise, they lose their $1. On average, how much money should a player expect to win or lose if they play this game repeatedly?\r\n[reveal-answer q=\"408866\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"408866\"]\r\n\r\nSuppose you bet $1 on each of the 38 spaces on the wheel, for a total of $38 bet. When the winning number is spun, you are paid $36 on that number. While you won on that one number, overall you\u2019ve lost $2. On a per-space basis, you have \u201cwon\u201d -$2\/$38 \u2248 -$0.053. In other words, on average you lose 5.3 cents per space you bet on.\r\n\r\nWe call this average gain or loss the expected value of playing roulette. Notice that no one ever loses exactly 5.3 cents: most people (in fact, about 37 out of every 38) lose $1 and a very few people (about 1 person out of every 38) gain $35 (the $36 they win minus the $1 they spent to play the game).\r\n\r\nThere is another way to compute expected value without imagining what would happen if we play every possible space.\u00a0 There are 38 possible outcomes when the wheel spins, so the probability of winning is [latex]\\frac{1}{38}[\/latex]. The complement, the probability of losing, is [latex]\\frac{37}{38}[\/latex].\r\n\r\nSummarizing these along with the values, we get this table:\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td>Outcome<\/td>\r\n<td>Probability of outcome<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>$35<\/td>\r\n<td>[latex]\\frac{1}{38}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>-$1<\/td>\r\n<td>[latex]\\frac{37}{38}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nNotice that if we multiply each outcome by its corresponding probability we get [latex]\\$35\\cdot \\frac{1}{38}=0.9211[\/latex] and [latex]-\\$1\\cdot \\frac{37}{38}=-0.9737[\/latex], and if we add these numbers we get\r\n\r\n0.9211 + (-0.9737) \u2248 -0.053, which is the expected value we computed above.[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox\">\r\n<h3>Expected Value<\/h3>\r\n<ul>\r\n \t<li><strong>Expected Value<\/strong> is the average gain or loss of an event if the procedure is repeated many times.<\/li>\r\n<\/ul>\r\nWe can compute the expected value by multiplying each outcome by the probability of that outcome, then adding up the products.\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It Now<\/h3>\r\nYou purchase a raffle ticket to help out a charity. The raffle ticket costs $5. The charity is selling 2000 tickets. One of them will be drawn and the person holding the ticket will be given a prize worth $4000. Compute the expected value for this raffle.\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nIn a certain state's lottery, 48 balls numbered 1 through 48 are placed in a machine and six of them are drawn at random. If the six numbers drawn match the numbers that a player had chosen, the player wins $1,000,000. If they match 5 numbers, then win $1,000.\u00a0\u00a0 It costs $1 to buy a ticket. Find the expected value.\r\n[reveal-answer q=\"737029\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"737029\"]\r\n\r\nEarlier, we calculated the probability of matching all 6 numbers and the probability of matching 5 numbers:\r\n\r\n[latex]\\frac{{}_{6}{{C}_{6}}}{{}_{48}{{C}_{6}}}=\\frac{1}{12271512}\\approx0.0000000815[\/latex] for all 6 numbers,\r\n\r\n[latex]\\frac{\\left({}_{6}{{C}_{5}}\\right)\\left({}_{42}{{C}_{1}}\\right)}{{}_{48}{{C}_{6}}}=\\frac{252}{12271512}\\approx0.0000205[\/latex] for 5 numbers.\r\n\r\nOur probabilities and outcome values are:\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td>Outcome<\/td>\r\n<td>Probability of outcome<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>$999,999<\/td>\r\n<td>[latex]\\frac{1}{12271512}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>$999<\/td>\r\n<td>[latex]\\frac{252}{12271512}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>-$1<\/td>\r\n<td>[latex]1-\\frac{253}{12271512}=\\frac{12271259}{12271512}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThe expected value, then is:\r\n\r\n[latex]\\left(\\$999,999 \\right)\\cdot \\frac{1}{12271512}+\\left( \\$999\\right)\\cdot\\frac{252}{12271512}+\\left(-\\$1\\right)\\cdot\\frac{12271259}{12271512}\\approx-\\$0.898[\/latex]\r\n\r\nOn average, one can expect to lose about 90 cents on a lottery ticket. Of course, most players will lose $1.\r\n\r\n[\/hidden-answer]\r\n\r\nView more about the expected value examples in the following video.\r\n\r\nhttps:\/\/youtu.be\/pFzgxGVltS8\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It Now<\/h3>\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=17430&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"300\"><\/iframe>\r\n\r\n<\/div>\r\nIn general, if the expected value of a game is negative, it is not a good idea to play the game, since on average you will lose money.\u00a0 It would be better to play a game with a positive expected value (good luck trying to find one!), although keep in mind that even if the <em>average<\/em> winnings are positive it could be the case that most people lose money and one very fortunate individual wins a great deal of money.\u00a0 If the expected value of a game is 0, we call it a <strong>fair game<\/strong>, since neither side has an advantage.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It Now<\/h3>\r\nA friend offers to play a game, in which you roll 3 standard 6-sided dice. If all the dice roll different values, you give him $1. If any two dice match values, you get $2. What is the expected value of this game? Would you play?\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=17431&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"350\"><\/iframe>\r\n\r\n<\/div>\r\nExpected value also has applications outside of gambling. Expected value is very common in making insurance decisions.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nA 40-year-old man in the U.S. has a 0.242% risk of dying during the next year.[footnote]According to the estimator at <a href=\"http:\/\/www.numericalexample.com\/index.php?view=article&amp;id=91\" target=\"_blank\">http:\/\/www.numericalexample.com\/index.php?view=article&amp;id=91<\/a>[\/footnote] An insurance company charges $275 for a life-insurance policy that pays a $100,000 death benefit. What is the expected value for the person buying the insurance?\r\n[reveal-answer q=\"90556\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"90556\"]\r\n\r\nThe probabilities and outcomes are\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td>Outcome<\/td>\r\n<td>Probability of outcome<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>$100,000 - $275 = $99,725<\/td>\r\n<td>0.00242<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>-$275<\/td>\r\n<td>1 \u2013 0.00242 = 0.99758<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThe expected value is ($99,725)(0.00242) + (-$275)(0.99758) = -$33.\r\n\r\n[\/hidden-answer]\r\n\r\nThe insurance applications of expected value are detailed in the following video.\r\n\r\nhttps:\/\/youtu.be\/Bnai8apt8vw\r\n\r\n<\/div>\r\nNot surprisingly, the expected value is negative; the insurance company can only afford to offer policies if they, on average, make money on each policy. They can afford to pay out the occasional benefit because they offer enough policies that those benefit payouts are balanced by the rest of the insured people.\r\n\r\nFor people buying the insurance, there is a negative expected value, but there is a security that comes from insurance that is worth that cost.","rendered":"<p>In the next section, we will explore more complex conditional probabilities and ways to compute them. Conditional probabilities can give us information such as the likelihood of getting a positive test result for a disease without actually having the disease. If a doctor thinks the chances that a positive test result nearly guarantees that a patient has a disease,they might begin an unnecessary and possibly harmful treatment regimen on a healthy patient. If you were to get a positive test result, knowing the likelihood of getting a false positive can guide you to get a second opinion.<\/p>\n<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<p>Applications With Probability<\/p>\n<ul>\n<li>Compute a conditional probability for an event<\/li>\n<li>Use Baye\u2019s theorem to compute a conditional probability<\/li>\n<li>Calculate the expected value of an event<\/li>\n<\/ul>\n<\/div>\n<h2>Bayes&#8217; Theorem<\/h2>\n<p>In this section we concentrate on the more complex conditional probability problems we began looking at in the last section.<\/p>\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/28211759\/800px-Bayes_Theorem_MMB_01.jpg\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-996\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/28211759\/800px-Bayes_Theorem_MMB_01.jpg\" alt=\"Blue neon sign of Bayes' Theorem equation\" width=\"800\" height=\"513\" \/><\/a><\/p>\n<p>For example, suppose a certain disease has an incidence rate of 0.1% (that is, it afflicts 0.1% of the population). A test has been devised to detect this disease. The test does not produce false negatives (that is, anyone who has the disease will test positive for it), but the false positive rate is 5% (that is, about 5% of people who take the test will test positive, even though they do not have the disease). Suppose a randomly selected person takes the test and tests positive.\u00a0 What is the probability that this person actually has the disease?<\/p>\n<p>There are two ways to approach the solution to this problem. One involves an important result in probability theory called Bayes&#8217; theorem. We will discuss this theorem a bit later, but for now we will use an alternative and, we hope, much more intuitive approach.<\/p>\n<p>Let&#8217;s break down the information in the problem piece by piece as an example.<\/p>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p><strong>Suppose a certain disease has an incidence rate of 0.1% (that is, it afflicts 0.1% of the population)<\/strong><em>.<\/em>\u00a0The percentage 0.1% can be converted to a decimal number by moving the decimal place two places to the left, to get 0.001. In turn, 0.001 can be rewritten as a fraction: 1\/1000. This tells us that about 1 in every 1000 people has the disease. (If we wanted we could write <em>P<\/em>(disease)=0.001.)<\/p>\n<p><strong>A test has been devised to detect this disease.\u00a0 The test does not produce false negatives (that is, anyone who has the disease will test positive for it)<\/strong><em>.<\/em>\u00a0This part is fairly straightforward: everyone who has the disease will test positive, or alternatively everyone who tests negative does not have the disease. (We could also say <em>P<\/em>(positive | disease)=1.)<\/p>\n<p><strong>The false positive rate is 5% (that is, about 5% of people who take the test will test positive, even though they do not have the disease)<\/strong><em>.<\/em>\u00a0This is even more straightforward. Another way of looking at it is that of every 100 people who are tested and do not have the disease, 5 will test positive even though they do not have the disease. (We could also say that <em>P<\/em>(positive | no\u00a0disease)=0.05.)<\/p>\n<p><strong>Suppose a randomly selected person takes the test and tests positive.\u00a0 What is the probability that this person actually has the disease?<\/strong>\u00a0Here we want to compute <em>P<\/em>(disease|positive). We already know that <em>P<\/em>(positive|disease)=1, but remember that conditional probabilities are not equal if the conditions are switched.<\/p>\n<p>Rather than thinking in terms of all these probabilities we have developed, let&#8217;s create a hypothetical situation and apply the facts as set out above. First, suppose we randomly select 1000 people and administer the test. How many do we expect to have the disease? Since about 1\/1000 of all people are afflicted with the disease, 1\/1000 of 1000 people is 1. (Now you know why we chose 1000.) Only 1 of 1000 test subjects actually has the disease; the other 999 do not.<\/p>\n<p>We also know that 5% of all people who do not have the disease will test positive. There are 999 disease-free people, so we would expect (0.05)(999)=49.95 (so, about 50) people to test positive who do not have the disease.<\/p>\n<p>Now back to the original question, computing <em>P<\/em>(disease|positive). There are 51 people who test positive in our example (the one unfortunate person who actually has the disease, plus the 50 people who tested positive but don&#8217;t). Only one of these people has the disease, so<\/p>\n<p style=\"text-align: center;\">P(disease | positive) [latex]\\approx\\frac{1}{51}\\approx0.0196[\/latex]<\/p>\n<p>or less than 2%. Does this surprise you? This means that of all people who test positive, over 98% <em>do not have the disease<\/em>.<\/p>\n<p>The answer we got was slightly approximate, since we rounded 49.95 to 50. We could redo the problem with 100,000 test subjects, 100 of whom would have the disease and (0.05)(99,900)=4995 test positive but do not have the disease, so the exact probability of having the disease if you test positive is<\/p>\n<p style=\"text-align: center;\">P(disease | positive) [latex]\\approx\\frac{100}{5095}\\approx0.0196[\/latex]<\/p>\n<p>which is pretty much the same answer.<\/p>\n<p>But back to the surprising result.\u00a0<em>Of all people who test positive, over 98% do not have the disease.<\/em>\u00a0\u00a0If your guess for the probability a person who tests positive has the disease was wildly different from the right answer (2%), don&#8217;t feel bad. The exact same problem was posed to doctors and medical students at the Harvard Medical School 25 years ago and the results revealed in a 1978 <em>New England Journal of Medicine<\/em> article. Only about 18% of the participants got the right answer. Most of the rest thought the answer was closer to 95% (perhaps they were misled by the false positive rate of 5%).<\/p>\n<p>So at least you should feel a little better that a bunch of doctors didn&#8217;t get the right answer either (assuming you thought the answer was much higher). But the significance of this finding and similar results from other studies in the intervening years lies not in making math students feel better but in the possibly catastrophic consequences it might have for patient care. If a doctor thinks the chances that a positive test result nearly guarantees that a patient has a disease, they might begin an unnecessary and possibly harmful treatment regimen on a healthy patient.\u00a0 Or worse, as in the early days of the AIDS crisis when being HIV-positive was often equated with a death sentence, the patient might take a drastic action and commit suicide.<\/p>\n<p>This example is worked through in detail in the video here.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Probability of a diease given a positive test: Bayes Thorem ex1\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/hXevfqsBino?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<p>As we have seen in this hypothetical example, the most responsible course of action for treating a patient who tests positive would be to counsel the patient that they most likely do <em>not<\/em> have the disease and to order further, more reliable, tests to verify the diagnosis.<\/p>\n<p>One of the reasons that the doctors and medical students in the study did so poorly is that such problems, when presented in the types of statistics courses that medical students often take, are solved by use of Bayes&#8217; theorem, which is stated as follows:<\/p>\n<div class=\"textbox\">\n<h3>Bayes\u2019 Theorem<\/h3>\n<p style=\"text-align: center;\">[latex]P(A|B)=\\frac{P(A)P(B|A)}{P(A)P(B|A)+P(\\bar{A})P(B|\\bar{A})}[\/latex]<\/p>\n<\/div>\n<p>In our earlier example, this translates to<\/p>\n<p style=\"text-align: center;\">[latex]P(\\text{disease}|\\text{positive})=\\frac{P(\\text{disease})P(\\text{positive}|\\text{disease})}{P(\\text{disease})P(\\text{positive}|\\text{disease})+P(\\text{nodisease})P(\\text{positive}|\\text{nodisease})}[\/latex]<\/p>\n<p>Plugging in the numbers gives<\/p>\n<p style=\"text-align: center;\">[latex]P(\\text{disease}|\\text{positive})=\\frac{(0.001)(1)}{(0.001)(1)+(0.999)(0.05)}\\approx0.0196[\/latex]<\/p>\n<p>which is exactly the same answer as our original solution.<\/p>\n<p>The problem is that you (or the typical medical student, or even the typical math professor) are much more likely to be able to remember the original solution than to remember Bayes&#8217; theorem. Psychologists, such as Gerd Gigerenzer, author of <em>Calculated Risks: How to Know When Numbers Deceive You<\/em>, have advocated that the method involved in the original solution (which Gigerenzer calls the method of &#8220;natural frequencies&#8221;) be employed in place of Bayes&#8217; Theorem. Gigerenzer performed a study and found that those educated in the natural frequency method were able to recall it far longer than those who were taught Bayes&#8217; theorem. When one considers the possible life-and-death consequences associated with such calculations it seems wise to heed his advice.<\/p>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>A certain disease has an incidence rate of 2%. If the false negative rate is 10% and the false positive rate is 1%, compute the probability that a person who tests positive actually has the disease.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q507646\">Show Answer<\/span><\/p>\n<div id=\"q507646\" class=\"hidden-answer\" style=\"display: none\">\n<p>Imagine 10,000 people who are tested. Of these 10,000, 200 will have the disease; 10% of them, or 20, will test negative and the remaining 180 will test positive. Of the 9800 who do not have the disease, 98 will test positive. So of the 278 total people who test positive, 180 will have the disease. Thus<\/p>\n<p style=\"text-align: center;\">[latex]P(\\text{disease}|\\text{positive})=\\frac{180}{278}\\approx0.647[\/latex]<\/p>\n<p>so about 65% of the people who test positive will have the disease.<\/p>\n<p>Using Bayes theorem directly would give the same result:<\/p>\n<p style=\"text-align: center;\">[latex]P(\\text{disease}|\\text{positive})=\\frac{(0.02)(0.90)}{(0.02)(0.90)+(0.98)(0.01)}=\\frac{0.018}{0.0278}\\approx0.647[\/latex]<\/p>\n<\/div>\n<\/div>\n<p>View the following for more about this example.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Probability of a disease given a postiive test: Bayes Theorem ex2\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/_c3xZvHto3k?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try it now<\/h3>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=17494&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<h2>Counting<\/h2>\n<p>Counting? You already know how to count or you wouldn&#8217;t be taking a college-level math class, right? Well yes, but what we&#8217;ll really be investigating here are ways of counting <em>efficiently<\/em>. When we get to the probability situations a bit later in this chapter we will need to count some <em>very<\/em> large numbers, like the number of possible winning lottery tickets. One way to do this would be to write down every possible set of numbers that might show up on a lottery ticket, but believe me: you don&#8217;t want to do this.<\/p>\n<h3>Basic Counting<\/h3>\n<p>We will start, however, with some more reasonable sorts of counting problems in order to develop the ideas that we will soon need.<\/p>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>Suppose at a particular restaurant you have three choices for an appetizer (soup, salad or breadsticks) and five choices for a main course (hamburger, sandwich, quiche, fajita or pizza). If you are allowed to choose exactly one item from each category for your meal, how many different meal options do you have?<\/p>\n<p><strong>Solution 1<\/strong>: One way to solve this problem would be to systematically list each possible meal:<\/p>\n<p>soup + hamburger \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 soup + sandwich \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 soup + quiche<\/p>\n<p>soup + fajita \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 soup + pizza \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 salad + hamburger<\/p>\n<p>salad + sandwich \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 salad + quiche \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 salad + fajita<\/p>\n<p>salad + pizza \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 breadsticks + hamburger \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 breadsticks + sandwich<\/p>\n<p>breadsticks + quiche \u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 breadsticks + fajita \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 breadsticks + pizza<\/p>\n<p>Assuming that we did this systematically and that we neither missed any possibilities nor listed any possibility more than once, the answer would be 15. Thus you could go to the restaurant 15 nights in a row and have a different meal each night.<\/p>\n<p><strong>Solution 2<\/strong>: Another way to solve this problem would be to list all the possibilities in a table:<\/p>\n<table>\n<tbody>\n<tr>\n<td><\/td>\n<td>hamburger<\/td>\n<td>sandwich<\/td>\n<td>quiche<\/td>\n<td>fajita<\/td>\n<td>pizza<\/td>\n<\/tr>\n<tr>\n<td>soup<\/td>\n<td>soup+burger<\/td>\n<td><\/td>\n<td><\/td>\n<td><\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>salad<\/td>\n<td>salad+burger<\/td>\n<td><\/td>\n<td><\/td>\n<td><\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>bread<\/td>\n<td>etc<\/td>\n<td><\/td>\n<td><\/td>\n<td><\/td>\n<td><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>In each of the cells in the table we could list the corresponding meal: soup + hamburger in the upper left corner, salad + hamburger below it, etc. But if we didn&#8217;t really care <em>what<\/em> the possible meals are, only <em>how many<\/em> possible meals there are, we could just count the number of cells and arrive at an answer of 15, which matches our answer from the first solution. (It&#8217;s always good when you solve a problem two different ways and get the same answer!)<\/p>\n<p><strong>Solution 3<\/strong>: We already have two perfectly good solutions. Why do we need a third?\u00a0 The first method was not very systematic, and we might easily have made an omission. The second method was better, but suppose that in addition to the appetizer and the main course we further complicated the problem by adding desserts to the menu: we&#8217;ve used the rows of the table for the appetizers and the columns for the main courses\u2014where will the desserts go? We would need a third dimension, and since drawing 3-D tables on a 2-D page or computer screen isn&#8217;t terribly easy, we need a better way in case we have three categories to choose form instead of just two.<\/p>\n<p>So, back to the problem in the example.\u00a0 What else can we do?\u00a0 Let&#8217;s draw a <strong>tree diagram<\/strong>:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-341\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/11181317\/treediagram.png\" alt=\"Tree diagram of above table\" width=\"375\" height=\"225\" \/><\/p>\n<p>This is called a &#8220;tree&#8221; diagram because at each stage we branch out, like the branches on a tree.\u00a0 In this case, we first drew five branches (one for each main course) and then for each of those branches we drew three more branches (one for each appetizer).\u00a0 We count the number of branches at the final level and get (surprise, surprise!) 15.<\/p>\n<p>If we wanted, we could instead draw three branches at the first stage for the three appetizers and then five branches (one for each main course) branching out of each of those three branches.<\/p>\n<\/div>\n<p>OK, so now we know how to count possibilities using tables and tree diagrams. These methods will continue to be useful in certain cases, but imagine a game where you have two decks of cards (with 52 cards in each deck) and you select one card from each deck. Would you really want to draw a table or tree diagram to determine the number of outcomes of this game?<\/p>\n<p>Let&#8217;s go back to the previous example that involved selecting a meal from three appetizers and five main courses, and look at the second solution that used a table. Notice that one way to count the number of possible meals is simply to number each of the appropriate cells in the table, as we have done above. But another way to count the number of cells in the table would be multiply the number of rows (3) by the number of columns (5) to get 15. Notice that we could have arrived at the same result without making a table at all by simply multiplying the number of choices for the appetizer (3) by the number of choices for the main course (5). We generalize this technique as the <em>basic counting rule<\/em>:<\/p>\n<div class=\"textbox\">\n<h3>Basic Counting Rule<\/h3>\n<p>If we are asked to choose one item from each of two separate categories where there are <em>m<\/em> items in the first category and <em>n<\/em> items in the second category, then the total number of available choices is <em>m <\/em><strong>\u00b7 <\/strong><em>n<\/em>.<\/p>\n<p>This is sometimes called the multiplication rule for probabilities.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>There are 21 novels and 18 volumes of poetry on a reading list for a college English course. How many different ways can a student select one novel and one volume of poetry to read during the quarter?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q16976\">Show Answer<\/span><\/p>\n<div id=\"q16976\" class=\"hidden-answer\" style=\"display: none\">There are 21 choices from the first category and 18 for the second, so there are 21\u00a0<strong>\u00b7<\/strong>\u00a018\u00a0=\u00a0378 possibilities.<\/div>\n<\/div>\n<\/div>\n<p>The Basic Counting Rule can be extended when there are more than two categories by applying it repeatedly, as we see in the next example.<\/p>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>Suppose at a particular restaurant you have three choices for an appetizer (soup, salad or breadsticks), five choices for a main course (hamburger, sandwich, quiche, fajita or pasta) and two choices for dessert (pie or ice cream). If you are allowed to choose exactly one item from each category for your meal, how many different meal options do you have?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q13816\">Show Answer<\/span><\/p>\n<div id=\"q13816\" class=\"hidden-answer\" style=\"display: none\">\n<p>There are 3 choices for an appetizer, 5 for the main course and 2 for dessert, so there are 3 <strong>\u00b7<\/strong> 5 <strong>\u00b7<\/strong> 2 = 30 possibilities.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try it now<\/h3>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=7159&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"300\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>A quiz consists of 3 true-or-false questions.\u00a0 In how many ways can a student answer the quiz?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q169025\">Show Answer<\/span><\/p>\n<div id=\"q169025\" class=\"hidden-answer\" style=\"display: none\">There are 3 questions. Each question has 2 possible answers (true or false), so the quiz may be answered in 2 <strong>\u00b7<\/strong> 2 <strong>\u00b7<\/strong> 2 = 8 different ways.\u00a0 Recall that another way to write 2 <strong>\u00b7<\/strong> 2 <strong>\u00b7<\/strong> 2 is 2<sup>3<\/sup>, which is much more compact.<\/div>\n<\/div>\n<\/div>\n<p>Basic counting examples from this section are described in the following video.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Basic counting\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/fROqcu-ekkw?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h3>Permutations<\/h3>\n<p>In this section we will develop an even faster way to solve some of the problems we have already learned to solve by other means.\u00a0 Let&#8217;s start with a couple examples.<\/p>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>How many different ways can the letters of the word MATH be rearranged to form a four-letter code word?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q614648\">Show Answer<\/span><\/p>\n<div id=\"q614648\" class=\"hidden-answer\" style=\"display: none\">This problem is a bit different.\u00a0 Instead of choosing one item from each of several different categories, we are repeatedly choosing items from the <em>same<\/em> category (the category is: the letters of the word MATH) and each time we choose an item we <em>do not replace<\/em> it, so there is one fewer choice at the next stage: we have 4 choices for the first letter (say we choose A), then 3 choices for the second (M, T and H; say we choose H), then 2 choices for the next letter (M and T; say we choose M) and only one choice at the last stage (T).\u00a0 Thus there are 4 <strong>\u00b7 <\/strong>3 <strong>\u00b7<\/strong> 2 <strong>\u00b7<\/strong> 1 = 24 ways to spell a code worth with the letters MATH.<\/div>\n<\/div>\n<\/div>\n<p>In this example, we needed to calculate <em>n<\/em>\u00a0\u00b7 (<em>n<\/em> \u2013 1) \u00b7 (<em>n<\/em> \u2013 2) \u00b7\u00b7\u00b7 3 \u00b7 2 \u00b7 1. This calculation shows up often in mathematics, and is called the <strong>factorial<\/strong>, and is notated <em>n<\/em>!<\/p>\n<div class=\"textbox\">\n<h3>Factorial<\/h3>\n<p style=\"text-align: center;\"><em>n<\/em>! = <em>n<\/em>\u00a0\u00b7 (<em>n<\/em> \u2013 1) \u00b7 (<em>n<\/em> \u2013 2) \u00b7\u00b7\u00b7 3 \u00b7 2 \u00b7 1<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try it now<\/h3>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=7188&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>How many ways can five different door prizes be distributed among five people?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q530386\">Show Answer<\/span><\/p>\n<div id=\"q530386\" class=\"hidden-answer\" style=\"display: none\">There are 5 choices of prize for the first person, 4 choices for the second, and so on. The number of ways the prizes can be distributed will be 5! = 5 <strong>\u00b7 <\/strong>4 <strong>\u00b7 <\/strong>3 <strong>\u00b7<\/strong> 2 <strong>\u00b7<\/strong> 1 = 120 ways.<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try it now<\/h3>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=7191&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"300\"><\/iframe><\/p>\n<\/div>\n<p>Now we will consider some slightly different examples.<\/p>\n<div class=\"textbox exercises\">\n<h3>examples<\/h3>\n<p>A charity benefit is attended by 25 people and three gift certificates are given away as door prizes: one gift certificate is in the amount of $100, the second is worth $25 and the third is worth $10.\u00a0 Assuming that no person receives more than one prize, how many different ways can the three gift certificates be awarded?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q75787\">Show Answer<\/span><\/p>\n<div id=\"q75787\" class=\"hidden-answer\" style=\"display: none\">Using the Basic Counting Rule, there are 25 choices for the person who receives the $100 certificate, 24 remaining choices for the $25 certificate and 23 choices for the $10 certificate, so there are 25 <strong>\u00b7<\/strong> 24 <strong>\u00b7<\/strong> 23 = 13,800 ways in which the prizes can be awarded.<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Eight sprinters have made it to the Olympic finals in the 100-meter race. In how many different ways can the gold, silver and bronze medals be awarded?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q68166\">Show Answer<\/span><\/p>\n<div id=\"q68166\" class=\"hidden-answer\" style=\"display: none\">\n<p>Using the Basic Counting Rule, there are 8 choices for the gold medal winner, 7 remaining choices for the silver, and 6 for the bronze, so there are 8 <strong>\u00b7<\/strong> 7 <strong>\u00b7<\/strong> 6 = 336 ways the three medals can be awarded to the 8 runners.<\/p>\n<p>Note that in these preceding examples, the gift certificates and the Olympic medals were awarded <em>without replacement<\/em>; that is, once we have chosen a winner of the first door prize or the gold medal, they are not eligible for the other prizes. Thus, at each succeeding stage of the solution there is one fewer choice (25, then 24, then 23 in the first example; 8, then 7, then 6 in the second).\u00a0 Contrast this with the situation of a multiple choice test, where there might be five possible answers\u00a0\u2014 A, B, C, D or E \u2014 for each question on the test.<\/p>\n<p>Note also that <em>the order of selection was important<\/em> in each example: for the three door prizes, being chosen first means that you receive substantially more money; in the Olympics example, coming in first means that you get the gold medal instead of the silver or bronze. In each case, if we had chosen the same three people in a different order there might have been a different person who received the $100 prize, or a different goldmedalist. (Contrast this with the situation where we might draw three names out of a hat to each receive a $10 gift certificate; in this case the order of selection is <em>not<\/em> important since each of the three people receive the same prize.\u00a0 Situations where the order is <em>not<\/em> important will be discussed in the next section.)<\/p>\n<\/div>\n<\/div>\n<p>Factorial examples are worked in this video.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-4\" title=\"Counting using the factorial\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/9maoGi5fd_M?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<p>We can generalize the situation in the two examples above to any problem <em>without replacement<\/em> where the <em>order of selection is important<\/em>. If we are arranging in order <em>r<\/em> items out of <em>n<\/em> possibilities (instead of 3 out of 25 or 3 out of 8 as in the previous examples), the number of possible arrangements will be given by<\/p>\n<p style=\"text-align: center;\"><em>n<\/em>\u00a0\u00b7 (<em>n<\/em> \u2013 1) \u00b7 (<em>n<\/em> \u2013 2) \u00b7\u00b7\u00b7 (<em>n<\/em> \u2013 <em>r<\/em> + 1)<\/p>\n<p>If you don&#8217;t see why (<em>n <\/em>\u2014 <em>r <\/em>+ 1) is the right number to use for the last factor, just think back to the first example in this section, where we calculated 25 <strong>\u00b7<\/strong> 24 <strong>\u00b7<\/strong> 23 to get 13,800. In this case <em>n<\/em> = 25 and <em>r<\/em> = 3, so <em>n <\/em>\u2014 <em>r <\/em>+ 1 = 25 \u2014 3 + 1 = 23, which is exactly the right number for the final factor.<\/p>\n<p>Now, why would we want to use this complicated formula when it&#8217;s actually easier to use the Basic Counting Rule, as we did in the first two examples? Well, we won&#8217;t actually use this formula all that often; we only developed it so that we could attach a special notation and a special definition to this situation where we are choosing <em>r<\/em> items out of <em>n<\/em> possibilities <em>without replacement<\/em> and where the <em>order of selection is important<\/em>. In this situation we write:<\/p>\n<div class=\"textbox\">\n<h3>Permutations<\/h3>\n<p style=\"text-align: center;\"><em>nPr<\/em> = <em>n<\/em>\u00a0\u00b7 (<em>n<\/em> \u2013 1) \u00b7 (<em>n<\/em> \u2013 2) \u00b7\u00b7\u00b7 (<em>n<\/em> \u2013 <em>r<\/em> + 1)<\/p>\n<p>We say that there are <em>nPr<\/em> <strong>permutations<\/strong> of size <em>r<\/em> that may be selected from among <em>n<\/em> choices <em>without replacement<\/em> when <em>order matters<\/em>.<\/p>\n<p>It turns out that we can express this result more simply using factorials.<\/p>\n<p style=\"text-align: center;\">[latex]{}_{n}{{P}_{r}}=\\frac{n!}{(n-r)!}[\/latex]<\/p>\n<\/div>\n<p>In practicality, we usually use technology rather than factorials or repeated multiplication to compute permutations.<\/p>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>I have nine paintings and have room to display only four of them at a time on my wall. How many different ways could I do this?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q366612\">Show Answer<\/span><\/p>\n<div id=\"q366612\" class=\"hidden-answer\" style=\"display: none\">Since we are choosing 4 paintings out of 9 <em>without replacement<\/em> where the <em>order of selection is important<\/em> there are 9<em>P<\/em>4 = 9 \u00b7 8 \u00b7 7 \u00b7 6 = 3,024 permutations.<\/div>\n<\/div>\n<hr \/>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>How many ways can a four-person executive committee (president, vice-president, secretary, treasurer) be selected from a 16-member board of directors of a non-profit organization?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q889810\">Show Answer<\/span><\/p>\n<div id=\"q889810\" class=\"hidden-answer\" style=\"display: none\">We want to choose 4 people out of 16 without replacement and where the order of selection is important. So the answer is 16<em>P<\/em>4 = 16 \u00b7 15 \u00b7 14 \u00b7 13 = 43,680.<\/div>\n<\/div>\n<p>View this video to see more about the permutations examples.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-5\" title=\"Permutations\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/xlyX2UJMJQI?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It Now<\/h3>\n<p>How many 5 character passwords can be made using the letters A through Z<\/p>\n<ul>\n<li>if repeats are allowed<\/li>\n<li>if no repeats are allowed<\/li>\n<\/ul>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=7193&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"300\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<h3>Combinations<\/h3>\n<p>In the previous section we considered the situation where we chose <em>r<\/em> items out of <em>n<\/em> possibilities <em>without replacement<\/em> and where the <em>order of selection was important<\/em>. We now consider a similar situation in which the order of selection is <em>not<\/em> important.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>A charity benefit is attended by 25 people at which three $50 gift certificates are given away as door prizes. Assuming no person receives more than one prize, how many different ways can the gift certificates be awarded?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q942898\">Show Answer<\/span><\/p>\n<div id=\"q942898\" class=\"hidden-answer\" style=\"display: none\">\n<p>Using the Basic Counting Rule, there are 25 choices for the first person, 24 remaining choices for the second person and 23 for the third, so there are 25 \u00b7 24 \u00b7 23 = 13,800 ways to choose three people. Suppose for a moment that Abe is chosen first, Bea second and Cindy third; this is one of the 13,800 possible outcomes. Another way to award the prizes would be to choose Abe first, Cindy second and Bea third; this is another of the 13,800 possible outcomes. But either way Abe, Bea and Cindy each get $50, so it doesn&#8217;t really matter the order in which we select them. In how many different orders can Abe, Bea and Cindy be selected? It turns out there are 6:<\/p>\n<p>ABC\u00a0\u00a0\u00a0\u00a0 ACB\u00a0\u00a0\u00a0\u00a0 BAC\u00a0\u00a0\u00a0\u00a0 BCA\u00a0\u00a0\u00a0\u00a0 CAB\u00a0\u00a0\u00a0\u00a0 CBA<\/p>\n<p>How can we be sure that we have counted them all? We are really just choosing 3 people out of 3, so there are 3 \u00b7 2 \u00b7 1 = 6 ways to do this; we didn&#8217;t really need to list them all. We can just use permutations!<\/p>\n<p>So, out of the 13,800 ways to select 3 people out of 25, six of them involve Abe, Bea and Cindy. The same argument works for any other group of three people (say Abe, Bea and David or Frank, Gloria and Hildy) so each three-person group is counted <em>six times<\/em>. Thus the 13,800 figure is six times too big. The number of distinct three-person groups will be 13,800\/6 = 2300.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p>We can generalize the situation in this example above to any problem of choosing a collection of items <em>without replacement<\/em> where the <em>order of selection is <strong>not<\/strong> important<\/em>. If we are choosing <em>r<\/em> items out of <em>n<\/em> possibilities (instead of 3 out of 25 as in the previous examples), the number of possible choices will be given by [latex]\\frac{{}_{n}{{P}_{r}}}{{}_{r}{{P}_{r}}}[\/latex], and we could use this formula for computation. However this situation arises so frequently that we attach a special notation and a special definition to this situation where we are choosing <em>r<\/em> items out of <em>n<\/em> possibilities <em>without replacement<\/em> where the <em>order of selection is <strong>not<\/strong> important<\/em>.<\/p>\n<div class=\"textbox\">\n<h3>Combinations<\/h3>\n<p style=\"text-align: center;\">[latex]{}_{n}{{C}_{r}}=\\frac{{}_{n}{{P}_{r}}}{{}_{r}{{P}_{r}}}[\/latex]<\/p>\n<p>We say that there are <em>nCr<\/em> <strong>combinations<\/strong> of size <em>r<\/em> that may be selected from among <em>n<\/em> choices <em>without replacement<\/em> where <em>order doesn\u2019t matter<\/em>.<\/p>\n<p>We can also write the combinations formula in terms of factorials:<\/p>\n<p style=\"text-align: center;\">[latex]{}_{n}{{C}_{r}}=\\frac{n!}{(n-r)!r!}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>A group of four students is to be chosen from a 35-member class to represent the class on the student council. How many ways can this be done?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q125383\">Show Answer<\/span><\/p>\n<div id=\"q125383\" class=\"hidden-answer\" style=\"display: none\">Since we are choosing 4 people out of 35 <em>without replacement<\/em> where the <em>order of selection is <strong>not<\/strong> important<\/em> there are [latex]{}_{35}{{C}_{4}}=\\frac{35\\cdot34\\cdot33\\cdot32}{4\\cdot3\\cdot2\\cdot1}[\/latex] = 52,360 combinations.<\/div>\n<\/div>\n<p>View the following for more explanation of the combinations examples.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-6\" title=\"Combinations\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/W8kd4YosbzE?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It Now<\/h3>\n<p>The United States Senate Appropriations Committee consists of 29 members; the Defense Subcommittee of the Appropriations Committee consists of 19 members. Disregarding party affiliation or any special seats on the Subcommittee, how many different 19-member subcommittees may be chosen from among the 29 Senators on the Appropriations Committee?<\/p>\n<\/div>\n<p>In the preceding Try It Now problem we assumed that the 19 members of the Defense Subcommittee were chosen without regard to party affiliation. In reality this would never happen: if Republicans are in the majority they would never let a majority of Democrats sit on (and thus control) any subcommittee.\u00a0 (The same of course would be true if the Democrats were in control.) So let&#8217;s consider the problem again, in a slightly more complicated form:<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>The United States Senate Appropriations Committee consists of 29 members, 15 Republicans and 14 Democrats. The Defense Subcommittee consists of 19 members, 10 Republicans and 9 Democrats. How many different ways can the members of the Defense Subcommittee be chosen from among the 29 Senators on the Appropriations Committee?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q205320\">Show Answer<\/span><\/p>\n<div id=\"q205320\" class=\"hidden-answer\" style=\"display: none\">\n<p>In this case we need to choose 10 of the 15 Republicans and 9 of the 14 Democrats.\u00a0 There are 15<em>C<\/em>10 = 3003 ways to choose the 10 Republicans and 14<em>C<\/em>9 = 2002 ways to choose the 9 Democrats. But now what?\u00a0 How do we finish the problem?<\/p>\n<p>Suppose we listed all of the possible 10-member Republican groups on 3003 slips of red paper and \u00a0all of the possible 9-member Democratic groups on 2002 slips of blue paper.\u00a0 How many ways can we choose one red slip and one blue slip?\u00a0 This is a job for the Basic Counting Rule!\u00a0 We are simply making one choice from the first category and one choice from the second category, just like in the restaurant menu problems from earlier.<\/p>\n<p>There must be 3003 \u00b7 2002 = 6,012,006 possible ways of selecting the members of the Defense Subcommittee.<\/p>\n<\/div>\n<\/div>\n<p>This example is worked through below.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-7\" title=\"Combinations 2\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/Xqc2sdYN7xo?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<h2>Probability Using Permutations and Combinations<\/h2>\n<p>We can use permutations and combinations to help us answer more complex probability questions.<\/p>\n<div class=\"textbox exercises\">\n<h3>examples<\/h3>\n<p>A 4 digit PIN number is selected. What is the probability that there are no repeated digits?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q108643\">Show Answer<\/span><\/p>\n<div id=\"q108643\" class=\"hidden-answer\" style=\"display: none\">\n<p>There are 10 possible values for each digit of the PIN (namely: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9), so there are 10 \u00b7 10 \u00b7 10 \u00b7 10 = 10<sup>4<\/sup> = 10000 total possible PIN numbers.<\/p>\n<p>To have no repeated digits, all four digits would have to be different, which is selecting without replacement. We could either compute 10 \u00b7 9 \u00b7 8 \u00b7 7, or notice that this is the same as the permutation 10<em>P<\/em>4 = 5040.<\/p>\n<p>The probability of no repeated digits is the number of 4 digit PIN numbers with no repeated digits divided by the total number of 4 digit PIN numbers. This probability is [latex]\\frac{{}_{10}{{P}_{4}}}{{{10}^{4}}}=\\frac{5040}{10000}=0.504[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It Now<\/h3>\n<p><iframe loading=\"lazy\" id=\"mom10\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=7158&amp;theme=oea&amp;iframe_resize_id=mom10\" width=\"100%\" height=\"400\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>In a certain state&#8217;s lottery, 48 balls numbered 1 through 48 are placed in a machine and six of them are drawn at random. If the six numbers drawn match the numbers that a player had chosen, the player wins $1,000,000. \u00a0\u00a0\u00a0In this lottery, the order the numbers are drawn in doesn\u2019t matter. Compute the probability that you win the million-dollar prize if you purchase a single lottery ticket.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q767485\">Show Answer<\/span><\/p>\n<div id=\"q767485\" class=\"hidden-answer\" style=\"display: none\">\n<p>In order to compute the probability, we need to count the total number of ways six numbers can be drawn, and the number of ways the six numbers on the player\u2019s ticket could match the six numbers drawn from the machine. Since there is no stipulation that the numbers be in any particular order, the number of possible outcomes of the lottery drawing is 48<em>C<\/em>6 = 12,271,512. Of these possible outcomes, only one would match all six numbers on the player\u2019s ticket, so the probability of winning the grand prize is:<\/p>\n<p>[latex]\\frac{{}_{6}{{C}_{6}}}{{}_{48}{{C}_{6}}}=\\frac{1}{12271512}\\approx0.0000000815[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>In the state lottery from the previous example, if five of the six numbers drawn match the numbers that a player has chosen, the player wins a second prize of $1,000. Compute the probability that you win the second prize if you purchase a single lottery ticket.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q651173\">Show Answer<\/span><\/p>\n<div id=\"q651173\" class=\"hidden-answer\" style=\"display: none\">\n<p>As above, the number of possible outcomes of the lottery drawing is 48<em>C<\/em>6 = 12,271,512. In order to win the second prize, five of the six numbers on the ticket must match five of the six winning numbers; in other words, we must have chosen five of the six winning numbers and one of the 42 losing numbers. The number of ways to choose 5 out of the 6 winning numbers is given by 6<em>C<\/em>5 = 6 and the number of ways to choose 1 out of the 42 losing numbers is given by 42<em>C<\/em>1 = 42. Thus the number of favorable outcomes is then given by the Basic Counting Rule: 6<em>C<\/em>5 \u00b7 42<em>C<\/em>1 = 6 \u00b7 42 = 252. So the probability of winning the second prize is.<\/p>\n<p>[latex]\\frac{\\left({}_{6}{{C}_{5}}\\right)\\left({}_{42}{{C}_{1}}\\right)}{{}_{48}{{C}_{6}}}=\\frac{252}{12271512}\\approx0.0000205[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>The previous examples are worked in the following video.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-8\" title=\"Probabilities using combinations\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/b9LFbB_aNAo?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox exercises\">\n<h3>examples<\/h3>\n<p>Compute the probability of randomly drawing five cards from a deck and getting exactly one Ace.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q542442\">Show Answer<\/span><\/p>\n<div id=\"q542442\" class=\"hidden-answer\" style=\"display: none\">\n<p>In many card games (such as poker) the order in which the cards are drawn is not important (since the player may rearrange the cards in his hand any way he chooses); in the problems that follow, we will assume that this is the case unless otherwise stated. Thus we use combinations to compute the possible number of 5-card hands, 52<em>C<\/em>5. This number will go in the denominator of\u00a0our probability formula, since it is the number of possible outcomes.<\/p>\n<p>For the numerator, we need the number of ways to draw one Ace and four other cards (none of them Aces) from the deck.\u00a0 Since there are four Aces and we want exactly one of them, there will be 4<em>C<\/em>1 ways to select one Ace; since there are 48 non-Aces and we want 4 of them, there will be 48<em>C<\/em>4 ways to select the four non-Aces.\u00a0 Now we use the Basic Counting Rule to calculate that there will be 4<em>C<\/em>1 \u00b7 48<em>C<\/em>4 ways to choose one ace and four non-Aces.<\/p>\n<p>Putting this all together, we have<\/p>\n<p style=\"text-align: center;\">[latex]P(\\text{oneAce})=\\frac{\\left({}_{4}{{C}_{1}}\\right)\\left({}_{48}{{C}_{4}}\\right)}{{}_{52}{{C}_{5}}}=\\frac{778320}{2598960}\\approx0.299[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Compute the probability of randomly drawing five cards from a deck and getting exactly two Aces.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q49360\">Show Answer<\/span><\/p>\n<div id=\"q49360\" class=\"hidden-answer\" style=\"display: none\">\n<p>The solution is similar to the previous example, except now we are choosing 2 Aces out of 4 and 3 non-Aces out of 48; the denominator remains the same:<\/p>\n<p>[latex]P(\\text{twoAces})=\\frac{\\left({}_{4}{{C}_{2}}\\right)\\left({}_{48}{{C}_{3}}\\right)}{{}_{52}{{C}_{5}}}=\\frac{103776}{2598960}\\approx0.0399[\/latex]<\/p>\n<p>It is useful to note that these card problems are remarkably similar to the lottery problems discussed earlier.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>View the following for further demonstration of these examples.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-9\" title=\"Probabilities using combinations: cards\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/RU3e3KTkjoA?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try it now<\/h3>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=7157&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"400\"><\/iframe><\/p>\n<\/div>\n<h3>Birthday Problem<\/h3>\n<p>Let&#8217;s take a pause to consider a famous problem in probability theory:<\/p>\n<div class=\"textbox shaded\">Suppose you have a room full of 30 people. What is the probability that there is at least one shared birthday?<\/div>\n<p>Take a guess at the answer to the above problem. Was your guess fairly low, like around 10%? That seems to be the intuitive answer (30\/365, perhaps?). Let&#8217;s see if we should listen to our intuition. Let&#8217;s start with a simpler problem, however.<\/p>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>Suppose three people are in a room.\u00a0 What is the probability that there is at least one shared birthday among these three people?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q226841\">Show Answer<\/span><\/p>\n<div id=\"q226841\" class=\"hidden-answer\" style=\"display: none\">\n<p>There are a lot of ways there could be at least one shared birthday. Fortunately there is an easier way. We ask ourselves \u201cWhat is the alternative to having at least one shared birthday?\u201d In this case, the alternative is that there are <strong>no<\/strong> shared birthdays. In other words, the alternative to \u201cat least one\u201d is having <strong>none<\/strong>. In other words, since this is a complementary event,<\/p>\n<p style=\"text-align: center;\">P(at least one) = 1 \u2013 P(none)<\/p>\n<p>We will start, then, by computing the probability that there is no shared birthday.\u00a0 Let&#8217;s imagine that you are one of these three people.\u00a0 Your birthday can be anything without conflict, so there are 365 choices out of 365 for your birthday. What is the probability that the second person does not share your birthday?\u00a0 There are 365 days in the year (let&#8217;s ignore leap years) and removing your birthday from contention, there are 364 choices that will guarantee that you do not share a birthday with this person, so the probability that the second person does not share your birthday is 364\/365.\u00a0 Now we move to the third person.\u00a0 What is the probability that this third person does not have the same birthday as either you or the second person?\u00a0 There are 363 days that will not duplicate your birthday or the second person&#8217;s, so the probability that the third person does not share a birthday with the first two is 363\/365.<\/p>\n<p>We want the second person not to share a birthday with you <em>and<\/em> the third person not to share a birthday with the first two people, so we use the multiplication rule:<\/p>\n<p style=\"text-align: center;\">[latex]P(\\text{nosharedbirthday})=\\frac{365}{365}\\cdot\\frac{364}{365}\\cdot\\frac{363}{365}\\approx0.9918[\/latex]<\/p>\n<p>and then subtract from 1 to get<\/p>\n<p style=\"text-align: center;\">P(shared birthday) = 1 \u2013 P(no shared birthday) = 1 \u2013 0.9918 = 0.0082.<\/p>\n<p>This is a pretty small number, so maybe it makes sense that the answer to our original problem will be small.\u00a0 Let&#8217;s make our group a bit bigger.<\/p>\n<\/div>\n<\/div>\n<hr \/>\n<p>Suppose five people are in a room.\u00a0 What is the probability that there is at least one shared birthday among these five people?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q6410\">Show Answer<\/span><\/p>\n<div id=\"q6410\" class=\"hidden-answer\" style=\"display: none\">\n<p>Continuing the pattern of the previous example, the answer should be<\/p>\n<p style=\"text-align: center;\">[latex]P(\\text{sharedbirthday})=1-\\frac{365}{365}\\cdot\\frac{364}{365}\\cdot\\frac{363}{365}\\cdot\\frac{362}{365}\\cdot\\frac{361}{365}\\approx0.0271[\/latex]<\/p>\n<p>Note that we could rewrite this more compactly as<\/p>\n<p style=\"text-align: center;\">[latex]P(\\text{sharedbirthday})=1-\\frac{{}_{365}{{P}_{5}}}{{{365}^{5}}}\\approx0.0271[\/latex]<\/p>\n<p>which makes it a bit easier to type into a calculator or computer, and which suggests a nice formula as we continue to expand the population of our group.<\/p>\n<\/div>\n<\/div>\n<hr \/>\n<p>Suppose 30 people are in a room.\u00a0 What is the probability that there is at least one shared birthday among these 30 people?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q741907\">Show Answer<\/span><\/p>\n<div id=\"q741907\" class=\"hidden-answer\" style=\"display: none\">\n<p>Here we can calculate<\/p>\n<p style=\"text-align: center;\">[latex]P(\\text{sharedbirthday})=1-\\frac{{}_{365}{{P}_{30}}}{{{365}^{30}}}\\approx0.706[\/latex]<\/p>\n<p>which gives us the surprising result that when you are in a room with 30 people there is a 70% chance that there will be at least one shared birthday!<\/p>\n<\/div>\n<\/div>\n<p>The birthday problem\u00a0is examined in detail in the following.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-10\" title=\"Probability: the birthday problem\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/UUmTfiJ_0k4?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<p>If you like to bet, and if you can convince 30 people to reveal their birthdays, you might be able to win some money by betting a friend that there will be at least two people with the same birthday in the room anytime you are in a room of 30 or more people. (Of course, you would need to make sure your friend hasn&#8217;t studied probability!) You wouldn&#8217;t be guaranteed to win, but you should win more than half the time.<\/p>\n<p>This is one of many results in probability theory that is counterintuitive; that is, it goes against our gut instincts.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try It Now<\/h3>\n<p>Suppose 10 people are in a room. What is the probability that there is at least one shared birthday among these 10 people?<\/p>\n<\/div>\n<h2>Expected Value<\/h2>\n<h3>Repeating Procedures Over Time<\/h3>\n<p>Expected value is perhaps the most useful probability concept we will discuss.\u00a0 It has many applications, from insurance policies to making financial decisions, and it&#8217;s one thing that the casinos and government agencies that run gambling operations and lotteries hope most people never learn about.<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-343\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/11182717\/roulette.jpg\" alt=\"A roulette wheel\" width=\"600\" height=\"400\" \/><\/p>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>In the casino game roulette, a wheel with 38 spaces (18 red, 18 black, and 2 green) is spun. In one possible bet, the player bets $1 on a single number. If that number is spun on the wheel, then they receive $36 (their original $1 + $35). Otherwise, they lose their $1. On average, how much money should a player expect to win or lose if they play this game repeatedly?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q408866\">Show Answer<\/span><\/p>\n<div id=\"q408866\" class=\"hidden-answer\" style=\"display: none\">\n<p>Suppose you bet $1 on each of the 38 spaces on the wheel, for a total of $38 bet. When the winning number is spun, you are paid $36 on that number. While you won on that one number, overall you\u2019ve lost $2. On a per-space basis, you have \u201cwon\u201d -$2\/$38 \u2248 -$0.053. In other words, on average you lose 5.3 cents per space you bet on.<\/p>\n<p>We call this average gain or loss the expected value of playing roulette. Notice that no one ever loses exactly 5.3 cents: most people (in fact, about 37 out of every 38) lose $1 and a very few people (about 1 person out of every 38) gain $35 (the $36 they win minus the $1 they spent to play the game).<\/p>\n<p>There is another way to compute expected value without imagining what would happen if we play every possible space.\u00a0 There are 38 possible outcomes when the wheel spins, so the probability of winning is [latex]\\frac{1}{38}[\/latex]. The complement, the probability of losing, is [latex]\\frac{37}{38}[\/latex].<\/p>\n<p>Summarizing these along with the values, we get this table:<\/p>\n<table>\n<tbody>\n<tr>\n<td>Outcome<\/td>\n<td>Probability of outcome<\/td>\n<\/tr>\n<tr>\n<td>$35<\/td>\n<td>[latex]\\frac{1}{38}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>-$1<\/td>\n<td>[latex]\\frac{37}{38}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Notice that if we multiply each outcome by its corresponding probability we get [latex]\\$35\\cdot \\frac{1}{38}=0.9211[\/latex] and [latex]-\\$1\\cdot \\frac{37}{38}=-0.9737[\/latex], and if we add these numbers we get<\/p>\n<p>0.9211 + (-0.9737) \u2248 -0.053, which is the expected value we computed above.<\/p><\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox\">\n<h3>Expected Value<\/h3>\n<ul>\n<li><strong>Expected Value<\/strong> is the average gain or loss of an event if the procedure is repeated many times.<\/li>\n<\/ul>\n<p>We can compute the expected value by multiplying each outcome by the probability of that outcome, then adding up the products.<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It Now<\/h3>\n<p>You purchase a raffle ticket to help out a charity. The raffle ticket costs $5. The charity is selling 2000 tickets. One of them will be drawn and the person holding the ticket will be given a prize worth $4000. Compute the expected value for this raffle.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>In a certain state&#8217;s lottery, 48 balls numbered 1 through 48 are placed in a machine and six of them are drawn at random. If the six numbers drawn match the numbers that a player had chosen, the player wins $1,000,000. If they match 5 numbers, then win $1,000.\u00a0\u00a0 It costs $1 to buy a ticket. Find the expected value.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q737029\">Show Answer<\/span><\/p>\n<div id=\"q737029\" class=\"hidden-answer\" style=\"display: none\">\n<p>Earlier, we calculated the probability of matching all 6 numbers and the probability of matching 5 numbers:<\/p>\n<p>[latex]\\frac{{}_{6}{{C}_{6}}}{{}_{48}{{C}_{6}}}=\\frac{1}{12271512}\\approx0.0000000815[\/latex] for all 6 numbers,<\/p>\n<p>[latex]\\frac{\\left({}_{6}{{C}_{5}}\\right)\\left({}_{42}{{C}_{1}}\\right)}{{}_{48}{{C}_{6}}}=\\frac{252}{12271512}\\approx0.0000205[\/latex] for 5 numbers.<\/p>\n<p>Our probabilities and outcome values are:<\/p>\n<table>\n<tbody>\n<tr>\n<td>Outcome<\/td>\n<td>Probability of outcome<\/td>\n<\/tr>\n<tr>\n<td>$999,999<\/td>\n<td>[latex]\\frac{1}{12271512}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>$999<\/td>\n<td>[latex]\\frac{252}{12271512}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>-$1<\/td>\n<td>[latex]1-\\frac{253}{12271512}=\\frac{12271259}{12271512}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>The expected value, then is:<\/p>\n<p>[latex]\\left(\\$999,999 \\right)\\cdot \\frac{1}{12271512}+\\left( \\$999\\right)\\cdot\\frac{252}{12271512}+\\left(-\\$1\\right)\\cdot\\frac{12271259}{12271512}\\approx-\\$0.898[\/latex]<\/p>\n<p>On average, one can expect to lose about 90 cents on a lottery ticket. Of course, most players will lose $1.<\/p>\n<\/div>\n<\/div>\n<p>View more about the expected value examples in the following video.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-11\" title=\"Expected value\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/pFzgxGVltS8?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It Now<\/h3>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=17430&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"300\"><\/iframe><\/p>\n<\/div>\n<p>In general, if the expected value of a game is negative, it is not a good idea to play the game, since on average you will lose money.\u00a0 It would be better to play a game with a positive expected value (good luck trying to find one!), although keep in mind that even if the <em>average<\/em> winnings are positive it could be the case that most people lose money and one very fortunate individual wins a great deal of money.\u00a0 If the expected value of a game is 0, we call it a <strong>fair game<\/strong>, since neither side has an advantage.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try It Now<\/h3>\n<p>A friend offers to play a game, in which you roll 3 standard 6-sided dice. If all the dice roll different values, you give him $1. If any two dice match values, you get $2. What is the expected value of this game? Would you play?<br \/>\n<iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=17431&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<p>Expected value also has applications outside of gambling. Expected value is very common in making insurance decisions.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>A 40-year-old man in the U.S. has a 0.242% risk of dying during the next year.<a class=\"footnote\" title=\"According to the estimator at http:\/\/www.numericalexample.com\/index.php?view=article&amp;id=91\" id=\"return-footnote-863-1\" href=\"#footnote-863-1\" aria-label=\"Footnote 1\"><sup class=\"footnote\">[1]<\/sup><\/a> An insurance company charges $275 for a life-insurance policy that pays a $100,000 death benefit. What is the expected value for the person buying the insurance?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q90556\">Show Answer<\/span><\/p>\n<div id=\"q90556\" class=\"hidden-answer\" style=\"display: none\">\n<p>The probabilities and outcomes are<\/p>\n<table>\n<tbody>\n<tr>\n<td>Outcome<\/td>\n<td>Probability of outcome<\/td>\n<\/tr>\n<tr>\n<td>$100,000 &#8211; $275 = $99,725<\/td>\n<td>0.00242<\/td>\n<\/tr>\n<tr>\n<td>-$275<\/td>\n<td>1 \u2013 0.00242 = 0.99758<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>The expected value is ($99,725)(0.00242) + (-$275)(0.99758) = -$33.<\/p>\n<\/div>\n<\/div>\n<p>The insurance applications of expected value are detailed in the following video.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-12\" title=\"Expected value of insurance\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/Bnai8apt8vw?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<p>Not surprisingly, the expected value is negative; the insurance company can only afford to offer policies if they, on average, make money on each policy. They can afford to pay out the occasional benefit because they offer enough policies that those benefit payouts are balanced by the rest of the insured people.<\/p>\n<p>For people buying the insurance, there is a negative expected value, but there is a security that comes from insurance that is worth that cost.<\/p>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-863\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Introduction and Learning Outcomes. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Math in Society. <strong>Authored by<\/strong>: David Lippman. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/www.opentextbookstore.com\/mathinsociety\/\">http:\/\/www.opentextbookstore.com\/mathinsociety\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-sa\/4.0\/\">CC BY-SA: Attribution-ShareAlike<\/a><\/em><\/li><li>Bayes&#039; Theorem MMB 01. <strong>Authored by<\/strong>: mattbuck. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/commons.wikimedia.org\/wiki\/File:Bayes%27_Theorem_MMB_01.jpg\">https:\/\/commons.wikimedia.org\/wiki\/File:Bayes%27_Theorem_MMB_01.jpg<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-sa\/4.0\/\">CC BY-SA: Attribution-ShareAlike<\/a><\/em><\/li><li>Probability of a diesase given a positive test: Bayes Theorem ex1. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/hXevfqsBino\">https:\/\/youtu.be\/hXevfqsBino<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Probability of a disease given a positive test: Bayes Theorem ex2. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/_c3xZvHto3k\">https:\/\/youtu.be\/_c3xZvHto3k<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Basic counting. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/fROqcu-ekkw\">https:\/\/youtu.be\/fROqcu-ekkw<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Counting using the factorial. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/9maoGi5fd_M\">https:\/\/youtu.be\/9maoGi5fd_M<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Permutations. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/xlyX2UJMJQI\">https:\/\/youtu.be\/xlyX2UJMJQI<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Combinations. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/W8kd4YosbzE\">https:\/\/youtu.be\/W8kd4YosbzE<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Combinations 2. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/Xqc2sdYN7xo\">https:\/\/youtu.be\/Xqc2sdYN7xo<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Probabilities using combinations. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/b9LFbB_aNAo\">https:\/\/youtu.be\/b9LFbB_aNAo<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Probabilities using combinations: cards. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/RU3e3KTkjoA\">https:\/\/youtu.be\/RU3e3KTkjoA<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Probability: the birthday problem. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/UUmTfiJ_0k4\">https:\/\/youtu.be\/UUmTfiJ_0k4<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Question ID 17479. <strong>Authored by<\/strong>: Lippman, David. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>Roulette. <strong>Authored by<\/strong>: Chris Yiu. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/www.flickr.com\/photos\/clry2\/1366937217\/\">https:\/\/www.flickr.com\/photos\/clry2\/1366937217\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-sa\/4.0\/\">CC BY-SA: Attribution-ShareAlike<\/a><\/em><\/li><li>Expected value. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/pFzgxGVltS8\">https:\/\/youtu.be\/pFzgxGVltS8<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Expected value of insurance. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/Bnai8apt8vw\">https:\/\/youtu.be\/Bnai8apt8vw<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section><hr class=\"before-footnotes clear\" \/><div class=\"footnotes\"><ol><li id=\"footnote-863-1\">According to the estimator at <a href=\"http:\/\/www.numericalexample.com\/index.php?view=article&amp;id=91\" target=\"_blank\">http:\/\/www.numericalexample.com\/index.php?view=article&amp;id=91<\/a> <a href=\"#return-footnote-863-1\" class=\"return-footnote\" aria-label=\"Return to footnote 1\">&crarr;<\/a><\/li><\/ol><\/div>","protected":false},"author":21,"menu_order":3,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Introduction and Learning Outcomes\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Math in Society\",\"author\":\"David Lippman\",\"organization\":\"\",\"url\":\"http:\/\/www.opentextbookstore.com\/mathinsociety\/\",\"project\":\"\",\"license\":\"cc-by-sa\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Bayes\\' Theorem MMB 01\",\"author\":\"mattbuck\",\"organization\":\"\",\"url\":\"https:\/\/commons.wikimedia.org\/wiki\/File:Bayes%27_Theorem_MMB_01.jpg\",\"project\":\"\",\"license\":\"cc-by-sa\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Probability of a diesase given a positive test: 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