{"id":339,"date":"2016-10-11T18:17:06","date_gmt":"2016-10-11T18:17:06","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/math4libarts\/?post_type=chapter&#038;p=339"},"modified":"2019-05-30T17:00:57","modified_gmt":"2019-05-30T17:00:57","slug":"bayes-theorem","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/waymakermath4libarts\/chapter\/bayes-theorem\/","title":{"raw":"Bayes' Theorem","rendered":"Bayes&#8217; Theorem"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\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\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 Solution[\/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<\/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>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\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<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 Solution<\/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<\/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\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-339\">\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>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>Bayes&#039; Theorem. <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>Project<\/strong>: Math in Society. <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 disease 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><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":20,"menu_order":7,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Bayes\\' Theorem\",\"author\":\"David Lippman\",\"organization\":\"\",\"url\":\"http:\/\/www.opentextbookstore.com\/mathinsociety\/\",\"project\":\"Math in Society\",\"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 disease given a positive test: Bayes Theorem ex1\",\"author\":\"OCLPhase2\\'s channel\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/hXevfqsBino\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Probability 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