{"id":26501,"date":"2021-03-29T13:00:05","date_gmt":"2021-03-29T13:00:05","guid":{"rendered":"http:\/\/toposuranos.com\/material\/?p=26501"},"modified":"2024-05-21T09:43:27","modified_gmt":"2024-05-21T09:43:27","slug":"useful-theorems-for-probability-calculation","status":"publish","type":"post","link":"https:\/\/toposuranos.com\/material\/en\/useful-theorems-for-probability-calculation\/","title":{"rendered":"Useful Theorems for Probability Calculation"},"content":{"rendered":"<div style=\"background-color:#F3F3F3; padding:20px;\">\n<center><\/p>\n<h1>Useful Theorems for Probability Calculation<\/h1>\n<p><\/p>\n<p style=\"text-align:center;\"><strong>Summary<\/strong><br \/><em>This class presents solved exercises demonstrating some useful theorems for probability calculation, including proofs and deductions. The exercises cover topics such as complementary probability, set inclusion, and event convergence. Completing these exercises will provide you with a solid foundation for further studying probability theory.<\/em><\/p>\n<p><\/center><br \/>\n<\/p>\n<p style=\"text-align:center;\"><strong>LEARNING OBJECTIVES:<\/strong><br \/>\nUpon completing this class, the student will be able to:\n<\/p>\n<ol>\n<li><strong>Demonstrate<\/strong> basic properties of probabilities<\/li>\n<\/ol>\n<p><center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/SSkPFP5FpKM\" 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 style=\"text-align: justify; color: #000000;\">What follows is a guide of solved exercises where the objective is to demonstrate some useful theorems for probability theory. Try to solve them and then compare your results &gt;:D<\/p>\n<ol style=\"text-align: justify; color: #000000;\">\n<li><strong><a href=\"https:\/\/www.youtube.com\/watch?v=SSkPFP5FpKM&amp;t=111s\" target=\"_blank\" rel=\"noopener\"><span style=\"color: #ff0000;\">Prove that <span class=\"katex-eq\" data-katex-display=\"false\">P(A^c) = 1 -P(A)<\/span> and,<\/span><\/a> based on this, make a deduction to argue that <span class=\"katex-eq\" data-katex-display=\"false\">P(\\emptyset) = 0<\/span><\/strong><br \/>\n<span class=\"collapseomatic \" id=\"id69d4d82898fd1\"  tabindex=\"0\" title=\"SHOW SOLUTION\"    >SHOW SOLUTION<\/span><div id=\"target-id69d4d82898fd1\" class=\"collapseomatic_content \">\n<p style=\"text-align: justify; color: #000000;\">From the definition of <a href=\"https:\/\/toposuranos.com\/el-espacio-de-probabilidades-medida-de-probabilidad\/\" target=\"_blank\" rel=\"noopener\">probability measure<\/a> it follows that, if <span class=\"katex-eq\" data-katex-display=\"false\">A<\/span> and <span class=\"katex-eq\" data-katex-display=\"false\">B<\/span> are any measurable events, then it will hold that<\/p>\n<table>\n<tbody>\n<tr>\n<td>[a]<\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">0\\leq P(A) \\leq 1<\/span><\/td>\n<\/tr>\n<tr>\n<td>[b]<\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">A\\cap B = \\emptyset \\rightarrow P(A\\cup B) = P(A) + P(B)<\/span><\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">P(\\Omega) = 1<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: justify; color: #000000;\">Now, since <span class=\"katex-eq\" data-katex-display=\"false\">A\\cap A^c = \\emptyset,<\/span> from part [b], it will hold that:<\/p>\n<table>\n<tbody>\n<tr>\n<td>[d]<\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">A\\cap A^c = \\emptyset \\rightarrow P(A\\cup A^c) = P(A) + P(A^c)<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: justify; color: #000000;\">Since <span class=\"katex-eq\" data-katex-display=\"false\">A\\cap A^c = \\emptyset<\/span> is always true and <span class=\"katex-eq\" data-katex-display=\"false\">A\\cup A^c = \\Omega<\/span>, it will then hold that<\/p>\n<p style=\"text-align: center; color: #000000;\"><span class=\"katex-eq\" data-katex-display=\"false\">1=P(\\Omega) = P(A\\cup A^c) = P(A) + P(A^c)<\/span>\n<p style=\"text-align: justify; color: #000000;\">And therefore<\/p>\n<p style=\"text-align: center; color: #000000;\"><span class=\"katex-eq\" data-katex-display=\"false\">P(A^c) = 1-P(A)<\/span>\n<p style=\"text-align: justify; color: #000000;\">Which is what was to be proven.<\/p>\n<p style=\"text-align: justify; color: #000000;\">To prove that <span class=\"katex-eq\" data-katex-display=\"false\">P(\\emptyset)=0<\/span>, it is enough to take <span class=\"katex-eq\" data-katex-display=\"false\">A=\\Omega<\/span> in the relation just proven and it will hold that<\/p>\n<p style=\"text-align: center; color: #000000;\"><span class=\"katex-eq\" data-katex-display=\"false\">P(\\emptyset) = P(\\Omega^c) =1 - P(\\Omega) = 1-1 = 0<\/span>\n<\/div><\/li>\n<li><strong><a href=\"https:\/\/www.youtube.com\/watch?v=SSkPFP5FpKM&amp;t=534s\" target=\"_blank\" rel=\"noopener\"><span style=\"color: #ff0000;\">a) Prove that <span class=\"katex-eq\" data-katex-display=\"false\">A\\subseteq B \\rightarrow P(B\\setminus A) = P(B) - P(A)<\/span><\/span><\/a>, is the equality generally true?<\/strong><strong>b) Prove that <span class=\"katex-eq\" data-katex-display=\"false\">A\\subseteq B \\rightarrow P(B)\\leq P(A)<\/span><\/strong><span class=\"collapseomatic \" id=\"id69d4d828992bd\"  tabindex=\"0\" title=\"SHOW SOLUTION a)\"    >SHOW SOLUTION a)<\/span><div id=\"target-id69d4d828992bd\" class=\"collapseomatic_content \">\n<table>\n<tbody>\n<tr>\n<td>(1)<\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">A\\subseteq B<\/span>; Premise<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">\\equiv A\\cap B = A<\/span><\/td>\n<\/tr>\n<tr>\n<td>(2)<\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">B\\setminus A = B\\cap A^c<\/span>; Definition of Set Theory<\/td>\n<\/tr>\n<tr>\n<td>(3)<\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">B= (B\\cap A) \\cup (B\\cap A^c)<\/span>; Property of sets<\/td>\n<\/tr>\n<tr>\n<td>(4)<\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">(B\\cap A)\\cap (B\\cap A^c)=\\emptyset<\/span>; Property of sets<\/td>\n<\/tr>\n<tr>\n<td>(5)<\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">P(B)= P[(B\\cap A) \\cup (B\\cap A^c)]<\/span>; From (3)<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">P(B)= P (B\\cap A) + P(B\\cap A^c)<\/span>; From (4) + Definition, Probability Measure<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">P(B)= P (B\\cap A) + P(B\\setminus A)<\/span>; From (2)<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\"> P(B\\setminus A) =P(B) - P (B\\cap A) <\/span><\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">{P(B\\setminus A) =P(B) - P (A) }<\/span>; From (1)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: center; color: #000000;\">Therefore <span class=\"katex-eq\" data-katex-display=\"false\">{\\{A\\subseteq B\\}\\vdash P(B\\setminus A) = P(B) - P(A)}.<\/span>\n<p style=\"text-align: justify; color: #000000;\">Note that this equality is not generally true, as it depends on <span class=\"katex-eq\" data-katex-display=\"false\">A\\subseteq B<\/span> being satisfied. From these developments, it can be seen that if this is not satisfied, then it will hold that <span class=\"katex-eq\" data-katex-display=\"false\">P(B\\setminus A) = P(B) - P(A\\cap B).<\/span>\n<\/div>\n<span class=\"collapseomatic \" id=\"id69d4d828994ce\"  tabindex=\"0\" title=\"SHOW SOLUTION b)\"    >SHOW SOLUTION b)<\/span><div id=\"target-id69d4d828994ce\" class=\"collapseomatic_content \">\n<p style=\"text-align: justify; color: #000000;\">Based on the reasoning in a) it follows that:<\/p>\n<table>\n<tbody>\n<tr>\n<td><span class=\"katex-eq\" data-katex-display=\"false\"> \\{A\\subseteq B\\}\\vdash P(B\\setminus A) = P(B) - P(A)<\/span><\/td>\n<\/tr>\n<tr>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">\\equiv \\{A\\subseteq B\\}\\vdash P(A) + P(B\\setminus A) = P(B)<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: justify; color: #000000;\">Finally, since <span class=\"katex-eq\" data-katex-display=\"false\">P<\/span> is a probability measure, it holds that <span class=\"katex-eq\" data-katex-display=\"false\">\\forall X (P(X)\\geq 0),<\/span> so, therefore:<\/p>\n<p style=\"text-align: center; color: #000000;\"><span class=\"katex-eq\" data-katex-display=\"false\">{\\{A\\subseteq B\\}\\vdash P(A) \\leq P(B)}.<\/span>\n<\/div>\n<p>&nbsp;<\/li>\n<li><strong><a href=\"https:\/\/www.youtube.com\/watch?v=SSkPFP5FpKM&amp;t=892s\" target=\"_blank\" rel=\"noopener\"><span style=\"color: #ff0000;\">a) Prove that <span class=\"katex-eq\" data-katex-display=\"false\">P(A\\cup B) = P(A) + P(B) - P(A\\cup B).<\/span><\/span><\/a> Accompany the proof with a diagram.<\/strong><strong>b) Using the previous result, prove that <span class=\"katex-eq\" data-katex-display=\"false\">P(A\\cup B) \\leq P(A) + P(B)<\/span><\/strong><span class=\"collapseomatic \" id=\"id69d4d828995f2\"  tabindex=\"0\" title=\"SHOW SOLUTION PART a)\"    >SHOW SOLUTION PART a)<\/span><div id=\"target-id69d4d828995f2\" class=\"collapseomatic_content \">\n<table>\n<tbody>\n<tr>\n<td> [\/latex]<\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">A\\cup B = (A\\triangle B) \\cup (A\\cap B)<\/span>; Property of sets<\/td>\n<\/tr>\n<tr>\n<td> [\/latex]<\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">A\\triangle B := (A\\setminus B) \\cup (B\\setminus A)<\/span>; Definition of symmetric difference<\/td>\n<\/tr>\n<tr>\n<td> [\/latex]<\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">(A\\triangle B)\\cap ( A \\cap B) = \\emptyset<\/span>; property of sets<\/td>\n<\/tr>\n<tr>\n<td> [\/latex]<\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">(A\\setminus B)\\cap (B\\setminus A) = \\emptyset<\/span>; property of sets<\/td>\n<\/tr>\n<tr>\n<td> [\/latex]<\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">P(A\\cup B) = P[(A\\triangle B) \\cup (A\\cap B)]<\/span>; From (1)<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">P(A\\cup B) = P(A\\triangle B) + P(A\\cap B)]<\/span>; From (3)<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">P(A\\cup B) = P(A\\setminus B) + P(B\\setminus A) + P(A\\cap B)]<\/span>; From (2,4)<\/td>\n<\/tr>\n<tr>\n<td> [\/latex]<\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">P(B\\setminus A) = P(B) - P(A\\cap B)<\/span>; Applying result from exercise 2<\/td>\n<\/tr>\n<tr>\n<td> [\/latex]<\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">P(A\\setminus B) = P(A) - P(A\\cap B)<\/span>; Same as (6)<\/td>\n<\/tr>\n<tr>\n<td> [\/latex]<\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">P(A\\cup B) = P(A) - P(A\\cap B) + P(B) - P(A\\cap B) + P(A\\cap B)<\/span>; from (5,6,7)<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">{P(A\\cup B) = P(A) + P(B) - P(A\\cap B)}. <\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<span class=\"collapseomatic \" id=\"id69d4d82899750\"  tabindex=\"0\" title=\"SHOW SOLUTION PART b)\"    >SHOW SOLUTION PART b)<\/span><div id=\"target-id69d4d82899750\" class=\"collapseomatic_content \">\n<table>\n<tbody>\n<tr>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">\\vdash P(A\\cup B) = P(A) + P(B) - P(A\\cap B)<\/span>; result from part a)<\/td>\n<\/tr>\n<tr>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">\\equiv\\; \\vdash P(A\\cup B) + P(A\\cap B) = P(A) + P(B) <\/span><\/td>\n<\/tr>\n<tr>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">\\equiv\\; \\vdash {P(A\\cup B) \\leq P(A) + P(B)} <\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<p>&nbsp;<\/li>\n<li><strong><a href=\"https:\/\/www.youtube.com\/watch?v=SSkPFP5FpKM&amp;t=1625s\" target=\"_blank\" rel=\"noopener\"><span style=\"color: #ff0000;\">If <span class=\"katex-eq\" data-katex-display=\"false\">\\{E_n\\}<\/span> is an infinite family of events<\/span> <\/a>such that <span class=\"katex-eq\" data-katex-display=\"false\">E_1\\supseteq E_2 \\supseteq E_2 \\supseteq \\cdots \\supseteq E_n \\supseteq E_{n+1}\\supseteq \\cdots.<\/span> Prove that<\/strong>\n<p style=\"text-align: center; color: #000000;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle P\\left( \\bigcap_{n=1}^\\infty E_n \\right) = \\lim_{n\\to \\infty}P(E_n)<\/span>\n<span class=\"collapseomatic \" id=\"id69d4d82899848\"  tabindex=\"0\" title=\"SHOW SOLUTION\"    >SHOW SOLUTION<\/span><div id=\"target-id69d4d82899848\" class=\"collapseomatic_content \">\n<table>\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\">E_n \\supseteq E_{n+1}<\/span>; Hypothesis<\/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\">E_n^c \\subseteq E_{n+1}^c<\/span>; By complementation from (1)<\/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\">\\displaystyle(E_n^c \\subseteq E_{n+1}^c) \\rightarrow P\\left( \\bigcup_{n=1}^\\infty E_n^c \\right)= \\lim_{n\\to\\infty}P(E_n^c)<\/span>; Continuity Property<\/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\">\\displaystyle \\bigcup_{n=1}^\\infty E_n^c = \\left( \\bigcap_{n=1}^\\infty E_n \\right)^c<\/span>; DeMorgan&#8217;s Laws<\/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\\left(E_n^c\\right) = 1 - P(E_n)<\/span>; Proven in exercise 1<\/td>\n<\/tr>\n<tr>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">(6)<\/span><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle P\\left(\\left[ \\bigcap_{n=1}^\\infty E_n\\right]^c \\right) = \\lim_{n\\to\\infty}[1-P(E_n)] = 1 - \\lim_{n\\to\\infty}P(E_n)<\/span>; from (2,4,5) applied on (3)<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle 1 - P\\left( \\bigcap_{n=1}^\\infty E_n \\right) = 1 - \\lim_{n\\to\\infty}P(E_n) <\/span><\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">{\\displaystyle P\\left( \\bigcap_{n=1}^\\infty E_n \\right) = \\lim_{n\\to\\infty}P(E_n)} <\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: center; color: #000000;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\\therefore\\{E_n \\supseteq E_{n+1}\\}\\vdash P\\left( \\bigcap_{n=1}^\\infty E_n \\right) = \\lim_{n\\to\\infty}P(E_n).<\/span>\n<p>&nbsp;<\/li>\n<\/ol>\n<\/div>\n<p style=\"text-align: justify; color: #000000;\">By solving these exercises, you will complete a foundational pillar that will support you in continuing the study of probability theory.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Useful Theorems for Probability Calculation SummaryThis class presents solved exercises demonstrating some useful theorems for probability calculation, including proofs and deductions. The exercises cover topics such as complementary probability, set inclusion, and event convergence. Completing these exercises will provide you with a solid foundation for further studying probability theory. LEARNING OBJECTIVES: Upon completing this class, [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":26394,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"iawp_total_views":4,"footnotes":""},"categories":[567,670],"tags":[],"class_list":["post-26501","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 v26.7 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Useful Theorems for Probability Calculation - toposuranos.com\/material<\/title>\n<meta name=\"description\" content=\"Explore key probability theorems presented with solved exercises. 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