{"id":26693,"date":"2021-04-19T13:00:30","date_gmt":"2021-04-19T13:00:30","guid":{"rendered":"http:\/\/toposuranos.com\/material\/?p=26693"},"modified":"2024-05-21T12:11:46","modified_gmt":"2024-05-21T12:11:46","slug":"conditional-probability-and-independence-between-events","status":"publish","type":"post","link":"https:\/\/toposuranos.com\/material\/en\/conditional-probability-and-independence-between-events\/","title":{"rendered":"Conditional Probability and Independence Between Events"},"content":{"rendered":"<div style=\"background-color:#F3F3F3; padding:20px;\">\n<center><\/p>\n<h1>Conditional Probability and Independence Between Events<\/h1>\n<p><\/p>\n<p style=\"text-align:center;\"><strong>Summary<\/strong><br \/><em>In this session, we will explore the concept of conditional probability and the interaction between events. We will acquire the skills to calculate conditional probabilities and determine the dependence or independence between events. We will apply practical examples, such as the study of the prevalence of cavities in candy consumers, to illustrate these concepts. By the end, you will have a clear understanding of how to apply conditional probability and analyze dependent and independent events.<\/em><\/p>\n<p><\/center><br \/>\n<\/p>\n<p style=\"text-align:center;\"><strong>LEARNING OBJECTIVES:<\/strong><br \/>\nBy the end of this class, you will be able to:\n<\/p>\n<ol>\n<li><strong>Understand<\/strong> the definition of conditional probability and its relationship with the intersection of events and individual probabilities.<\/li>\n<li><strong>Identify<\/strong> positive and negative associations between events by comparing conditional probabilities.<\/li>\n<li><strong>Test<\/strong> the independence between different events.<\/li>\n<\/ol>\n<p style=\"text-align:center;\"><strong><u>TABLE OF CONTENTS<\/u><\/strong><br \/>\n<a href=\"#1\">CONDITIONAL PROBABILITY<\/a><br \/>\n<a href=\"#2\">FORMAL DEFINITION OF CONDITIONAL PROBABILITY<\/a><br \/>\n<a href=\"#3\">RELATIONSHIP BETWEEN EVENTS<\/a><br \/>\n<a href=\"#4\">INDEPENDENCE BETWEEN EVENTS AND COMPLEMENTS OF EVENTS<\/a>\n<\/p>\n<p><center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/rWzeKPNM-Ds\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>\n<\/div>\n<p><a name=\"1\"><\/a><br \/>\n<\/br><\/br><\/p>\n<h2>Conditional Probability<\/h2>\n<p style=\"text-align: justify; color: #000000;\"><a href=\"https:\/\/www.youtube.com\/watch?v=rWzeKPNM-Ds&amp;t=132s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">What is the probability that event A<\/span><\/strong><\/a> occurs given that B has already occurred? The calculation of such probabilities involves the concept of <strong>conditional probability.<\/strong> In what follows, we will study conditional probability, its definition, and how from this we infer the relationships of dependence and independence between events.<\/p>\n<p style=\"text-align: justify; color: #000000;\">Suppose we want to measure the prevalence of cavities among regular candy consumers. If we examine a sample space formed of <span class=\"katex-eq\" data-katex-display=\"false\">N<\/span> people <span class=\"katex-eq\" data-katex-display=\"false\">\\Omega_N,<\/span> we will see that it can be divided into four subsets:<\/p>\n<ul style=\"text-align: justify; color: #000000;\">\n<li><span class=\"katex-eq\" data-katex-display=\"false\">A:=\\left\\{  {People\\;who\\;have\\;cavities}\\right\\}<\/span><\/li>\n<li><span class=\"katex-eq\" data-katex-display=\"false\">A^c:=\\left\\{{People\\;who\\;DO\\;NOT\\;have\\;cavities}\\right\\}<\/span><\/li>\n<li><span class=\"katex-eq\" data-katex-display=\"false\">B:=\\left\\{  {People\\;who\\;eat\\;candy\\;regularly}\\right\\}<\/span><\/li>\n<li><span class=\"katex-eq\" data-katex-display=\"false\">B^c:=\\left\\{{People\\;who\\;DO\\;NOT\\;eat\\;candy\\;regularly}\\right\\}<\/span><\/li>\n<\/ul>\n<p style=\"text-align: justify; color: #000000;\">From this, it is clear that <span class=\"katex-eq\" data-katex-display=\"false\">A\\cup A^c = \\Omega_N<\/span> and <span class=\"katex-eq\" data-katex-display=\"false\">B\\cup B^c = \\Omega_N,<\/span> but <span class=\"katex-eq\" data-katex-display=\"false\">A\\cap B<\/span> is not necessarily empty. The general scenario is represented by the following figure:<\/p>\n<p><img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"https:\/\/1.bp.blogspot.com\/-YZLH9zUJd2g\/YHzyLbaupDI\/AAAAAAAAE6c\/lpdThvvjvvsMYQqqIXelpU-Kcsd-uFWEwCLcBGAsYHQ\/s0\/probabilidad%2Bcondicional%2B1.PNG\" alt=\"Conditional Probability\" class=\" aligncenter lazyload\" width=\"339\" height=\"273\" \/><noscript><img decoding=\"async\" src=\"https:\/\/1.bp.blogspot.com\/-YZLH9zUJd2g\/YHzyLbaupDI\/AAAAAAAAE6c\/lpdThvvjvvsMYQqqIXelpU-Kcsd-uFWEwCLcBGAsYHQ\/s0\/probabilidad%2Bcondicional%2B1.PNG\" alt=\"Conditional Probability\" class=\" aligncenter lazyload\" width=\"339\" height=\"273\" \/><\/noscript><\/p>\n<p style=\"text-align: justify; color: #000000;\"><a href=\"https:\/\/www.youtube.com\/watch?v=rWzeKPNM-Ds&amp;t=270s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">So, if we remember the definition of probability<\/span><\/strong><\/a> as the limit of relative frequencies, we can say that the probability that a person has cavities given that they consume candy, <span class=\"katex-eq\" data-katex-display=\"false\">P(A|B)<\/span> will be:<\/p>\n<p style=\"text-align: justify; color: #000000;\"><span class=\"katex-eq\" data-katex-display=\"false\">P(A|B) =\\displaystyle \\frac{\\#(A\\cap B)}{\\#B}<\/span>\n<p style=\"text-align: justify; color: #000000;\">On the other hand, it holds that:<\/p>\n<p style=\"text-align: justify; color: #000000;\"><span class=\"katex-eq\" data-katex-display=\"false\">P(A\\cap B) = \\displaystyle \\frac{\\#(A\\cap B)}{\\#\\Omega_N}<\/span>\n<p style=\"text-align: justify; color: #000000;\">\u21b3 <span class=\"katex-eq\" data-katex-display=\"false\">\\#(A\\cap B) = \\#\\Omega_N P(A\\cap B)<\/span>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify; color: #000000;\"><span class=\"katex-eq\" data-katex-display=\"false\">P(B) =\\displaystyle \\frac{\\#B}{\\#\\Omega_N} <\/span>\n<p style=\"text-align: justify; color: #000000;\">\u21b3 <span class=\"katex-eq\" data-katex-display=\"false\">\\#B = \\#\\Omega_N P(B)<\/span>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify; color: #000000;\">So if we replace these last two expressions in <span class=\"katex-eq\" data-katex-display=\"false\">P(A|B)<\/span>, we get:<\/p>\n<p style=\"text-align: justify; color: #000000;\"><span class=\"katex-eq\" data-katex-display=\"false\">P(A|B) = \\displaystyle \\frac{\\#\\Omega_N P(A\\cap B)}{\\#\\Omega_N P(B)} = \\frac{P(A\\cap B)}{P(B)} <\/span>\n<p style=\"text-align: justify; color: #000000;\">With this, we have the following definition:<\/p>\n<p style=\"text-align: justify; color: #000000;\"><span style=\"color: #800000;\"><\/p>\n<p><a name=\"2\"><\/a><br \/>\n<\/br><\/br><\/p>\n<h2>Formal Definition of Conditional Probability<\/h2>\n<p><strong>DEFINITION:<\/strong><\/span> The probability of <span class=\"katex-eq\" data-katex-display=\"false\">A<\/span>, given that <span class=\"katex-eq\" data-katex-display=\"false\">B<\/span> has occurred, <span class=\"katex-eq\" data-katex-display=\"false\">P(A|B),<\/span> is defined as:<\/p>\n<p style=\"text-align: center; color: #000000; background-color: #b0ffb0;\"><span class=\"katex-eq\" data-katex-display=\"false\">P(A|B) = \\displaystyle \\frac{P(A\\cap B)}{P(B)}<\/span>\u25a0<\/p>\n<p style=\"text-align: justify; color: #000000;\"><a href=\"https:\/\/www.youtube.com\/watch?v=rWzeKPNM-Ds&amp;t=418s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">In everyday thinking, there is often confusion<\/span><\/strong><\/a> between <span class=\"katex-eq\" data-katex-display=\"false\">P(A|B)<\/span> and <span class=\"katex-eq\" data-katex-display=\"false\">P(B|A).<\/span> To shed light on this difference, let us review an example based on an extreme case: Note that while all footballers have two legs, only a tiny fraction of people with two legs are footballers.<\/p>\n<p><a name=\"3\"><\/a><br \/>\n<\/br><\/br><\/p>\n<h2>Relationship Between Events<\/h2>\n<p style=\"text-align: justify; color: #000000;\"><a href=\"https:\/\/www.youtube.com\/watch?v=rWzeKPNM-Ds&amp;t=534s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">Continuing with the example of the prevalence<\/span><\/strong><\/a> of cavities among people who regularly consume candy. If consuming candy makes people more prone to getting cavities, then it should occur that:<\/p>\n<p style=\"text-align: justify; color: #000000;\"><span class=\"katex-eq\" data-katex-display=\"false\">P(A|B) \\gt P(A).<\/span>Here we have that <span class=\"katex-eq\" data-katex-display=\"false\">B<\/span> boosts <span class=\"katex-eq\" data-katex-display=\"false\">A<\/span> and therefore we say that there is a <strong>positive association<\/strong> between the events.<\/p>\n<p style=\"text-align: justify; color: #000000;\">If, on the contrary, consuming candy prevents cavities, then it should occur that:<\/p>\n<p style=\"text-align: justify; color: #000000;\"><span class=\"katex-eq\" data-katex-display=\"false\">P(A|B) \\lt P(A).<\/span>In this case, <span class=\"katex-eq\" data-katex-display=\"false\">B<\/span> inhibits <span class=\"katex-eq\" data-katex-display=\"false\">A<\/span> and therefore we say that there is a <strong>negative association<\/strong> between the events.<\/p>\n<p style=\"text-align: justify; color: #000000;\">And if there is no relationship between these two events, neither positive nor negative, then it should occur that:<\/p>\n<p style=\"text-align: justify; color: #000000;\"><span class=\"katex-eq\" data-katex-display=\"false\">P(A|B) = P(A).<\/span>From here, the inference is made that in most probability texts is presented as a definition:<\/p>\n<table style=\"text-align: justify; color: #000000;\">\n<tbody>\n<tr>\n<td><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">P(A|B) = P(A)<\/span><\/td>\n<\/tr>\n<tr>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">\\equiv<\/span><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{P(A\\cap B)}{P(B)} = P(A)<\/span><\/td>\n<\/tr>\n<tr>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">\\equiv<\/span><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">P(A\\cap B)= P(A) P(B)<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: justify; color: #000000;\">This reasoning shows us the relationship between conditional probability and the independence of events.<\/p>\n<p style=\"text-align: justify; color: #000000;\"><span style=\"color: #800000;\"><strong>DEFINITION:<\/strong><\/span> Given two events <span class=\"katex-eq\" data-katex-display=\"false\">A<\/span> and <span class=\"katex-eq\" data-katex-display=\"false\">B<\/span>, they are said to be <strong>independent<\/strong> if they satisfy the relationship:<\/p>\n<p style=\"text-align: center; color: #000000; background-color: #b0ffb0;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\color{black}{P(A\\cap B)= P(A) P(B)}<\/span>\u25a0<\/p>\n<p><a name=\"4\"><\/a><br \/>\n<\/br><\/br><\/p>\n<h2>Independence Between Events and Complements of Events<\/h2>\n<p style=\"text-align: justify; color: #000000;\"><a href=\"https:\/\/www.youtube.com\/watch?v=rWzeKPNM-Ds&amp;t=662s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">The independence between two events<\/span><\/strong><\/a> <span class=\"katex-eq\" data-katex-display=\"false\">A<\/span> and <span class=\"katex-eq\" data-katex-display=\"false\">B<\/span> is proven to be equivalent to the independence of <span class=\"katex-eq\" data-katex-display=\"false\">A<\/span> with <span class=\"katex-eq\" data-katex-display=\"false\">B^c,<\/span> the independence of <span class=\"katex-eq\" data-katex-display=\"false\">A^c<\/span> with <span class=\"katex-eq\" data-katex-display=\"false\">B,<\/span> and the independence of <span class=\"katex-eq\" data-katex-display=\"false\">A^c<\/span> with <span class=\"katex-eq\" data-katex-display=\"false\">B^c.<\/span>\n<p style=\"text-align: justify; color: #000000;\"><span style=\"color: #000080;\"><strong>PROOF<\/strong><\/span><\/p>\n<table style=\"text-align: justify; color: #000000;\">\n<tbody>\n<tr>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">(1)<\/span> <\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">\\{P(A\\cap B) = P(A)P(B)\\}\\vdash P(A\\cap B) = P(A)P(B)<\/span><\/td>\n<td>; Assumption<\/td>\n<\/tr>\n<tr>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">(2)<\/span><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">\\{P(A\\cap B) = P(A)P(B)\\}\\vdash P(A\\cap B^c) = P(A\\setminus B)<\/span><\/td>\n<td>; because <span class=\"katex-eq\" data-katex-display=\"false\"> A\\cap B^c := A\\setminus B <\/span><\/td>\n<\/tr>\n<tr>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">(3)<\/span><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\"> \\{P(A\\cap B) = P(A)P(B)\\}\\vdash P(A\\setminus B)= P(A) - P(A\\cap B)<\/span><\/td>\n<td>; <a href=\"https:\/\/toposuranos.com\/probabilidades-y-estadistica-ejercicios-de-teoria\/\" rel=\"noopener\" target=\"_blank\">See development of exercise 2<\/a><\/td>\n<\/tr>\n<tr>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">(4)<\/span><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">\\{P(A\\cap B) = P(A)P(B)\\}\\vdash P(A\\cap B^c)= P(A) - P(A\\cap B)<\/span><\/td>\n<td>; From (2) and (3)<\/td>\n<\/tr>\n<tr>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">(5)<\/span><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">\\{P(A\\cap B) = P(A)P(B)\\}\\vdash P(A\\cap B^c)= P(A) - P(A)P(B)<\/span><\/td>\n<td>; From (1) and (4)<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">\\{P(A\\cap B) = P(A)P(B)\\}\\vdash P(A\\cap B^c)= P(A)(1 -P(B))<\/span><\/td>\n<td>; Factoring by <span class=\"katex-eq\" data-katex-display=\"false\">P(A)<\/span><\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">\\color{red}{\\{P(A\\cap B) = P(A)P(B)\\}\\vdash P(A\\cap B^c)= P(A)P(B^c)}<\/span><\/td>\n<td>; <a href=\"https:\/\/toposuranos.com\/probabilidades-y-estadistica-ejercicios-de-teoria\/\" rel=\"noopener\" target=\"_blank\">see development of exercise 1<\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: justify; color: #000000;\">This last expression is read as: \u00abFrom the fact that <span class=\"katex-eq\" data-katex-display=\"false\">A<\/span> and <span class=\"katex-eq\" data-katex-display=\"false\">B<\/span> are independent, it is inferred that <span class=\"katex-eq\" data-katex-display=\"false\">A<\/span> and <span class=\"katex-eq\" data-katex-display=\"false\">B^c<\/span> are also independent.<\/p>\n<p style=\"text-align: justify; color: #000000;\">The proof in the reverse sense is done similarly.<\/p>\n<table style=\"text-align: justify; color: #000000;\">\n<tbody>\n<tr>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">(1)<\/span><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">\\{P(A\\cap B^c) = P(A)P(B^c)\\}\\vdash P(A\\cap B^c) = P(A)P(B^c)<\/span><\/td>\n<td>; Assumption<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">\\{P(A\\cap B^c) = P(A)P(B^c)\\}\\vdash P(A\\cap B^c) = P(A)(1 - P(B))<\/span><\/td>\n<td>; <a href=\"https:\/\/toposuranos.com\/probabilidades-y-estadistica-ejercicios-de-teoria\/\" rel=\"noopener\" target=\"_blank\">see development of exercise 1<\/a><\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">\\{P(A\\cap B^c) = P(A)P(B^c)\\}\\vdash P(A\\cap B^c) = P(A) - P(A)P(B)<\/span><\/td>\n<td>; Performing the product of the parenthesis on the right side.<\/td>\n<\/tr>\n<tr>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">(2)<\/span><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">\\{P(A\\cap B^c) = P(A)P(B^c)\\}\\vdash P(A\\cap B^c) = P(A\\setminus B)<\/span><\/td>\n<td>; Because <span class=\"katex-eq\" data-katex-display=\"false\">A\\setminus B := A\\cap B^c<\/span>.<\/td>\n<\/tr>\n<tr>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">(3)<\/span><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">\\{P(A\\cap B^c) = P(A)P(B^c)\\}\\vdash P(A\\setminus B) = P(A) - P(A\\cap B) <\/span><\/td>\n<td>; <a href=\"https:\/\/toposuranos.com\/probabilidades-y-estadistica-ejercicios-de-teoria\/\" rel=\"noopener\" target=\"_blank\">See development of exercise 2<\/a><\/td>\n<\/tr>\n<tr>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">(4)<\/span><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">\\{P(A\\cap B^c) = P(A)P(B^c)\\}\\vdash P(A) - P(A)P(B) = P(A) - P(A\\cap B) <\/span><\/td>\n<td>; From (1), (2) and (3)<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">\\color{red}{\\{P(A\\cap B^c) = P(A)P(B^c)\\}\\vdash P(A)P(B) = P(A\\cap B)} <\/span><\/td>\n<td>; Eliminating similar terms<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: justify; color: #000000;\">And this expression is read as: \u00abFrom the fact that <span class=\"katex-eq\" data-katex-display=\"false\">A<\/span> and <span class=\"katex-eq\" data-katex-display=\"false\">B^c<\/span> are independent, it is inferred that <span class=\"katex-eq\" data-katex-display=\"false\">A<\/span> and <span class=\"katex-eq\" data-katex-display=\"false\">B<\/span> are also independent.<\/p>\n<p style=\"text-align: justify; color: #000000;\">Finally, from these two reasonings, the equivalence between the independence of <span class=\"katex-eq\" data-katex-display=\"false\">A<\/span> with <span class=\"katex-eq\" data-katex-display=\"false\">B<\/span> and that of <span class=\"katex-eq\" data-katex-display=\"false\">A<\/span> with <span class=\"katex-eq\" data-katex-display=\"false\">B^c<\/span> is proven.<\/p>\n<p style=\"text-align: justify; color: #000000;\">The other proven equivalences can be obtained similarly. These will be left as a challenge for the reader &gt;:D<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Conditional Probability and Independence Between Events SummaryIn this session, we will explore the concept of conditional probability and the interaction between events. We will acquire the skills to calculate conditional probabilities and determine the dependence or independence between events. We will apply practical examples, such as the study of the prevalence of cavities in candy [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":26406,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"iawp_total_views":5,"footnotes":""},"categories":[567,670],"tags":[],"class_list":["post-26693","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-mathematics","category-probabilities-and-statistics"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.4 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>Conditional Probability and Independence Between Events - toposuranos.com\/material<\/title>\n<meta name=\"description\" content=\"Do you want to know what conditional probability is? 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