{"id":26392,"date":"2021-03-29T13:00:49","date_gmt":"2021-03-29T13:00:49","guid":{"rendered":"http:\/\/toposuranos.com\/material\/?p=26392"},"modified":"2024-05-21T09:43:25","modified_gmt":"2024-05-21T09:43:25","slug":"teoremas-utiles-para-el-calculo-de-probabilidades","status":"publish","type":"post","link":"https:\/\/toposuranos.com\/material\/es\/teoremas-utiles-para-el-calculo-de-probabilidades\/","title":{"rendered":"Teoremas \u00datiles para el C\u00e1lculo de Probabilidades"},"content":{"rendered":"<div style=\"background-color:#F3F3F3; padding:20px;\">\n<center><\/p>\n<h1>Teoremas \u00datiles para el C\u00e1lculo de Probabilidades<\/h1>\n<p><\/p>\n<p style=\"text-align:center;\"><strong>Resumen<\/strong><br \/><em>En esta clase se presentan ejercicios resueltos en que se demuestran algunos teoremas \u00fatiles para el c\u00e1lculo de probabilidades, incluyendo demostraciones y deducciones. Los ejercicios abordan temas como la probabilidad complementaria, la inclusi\u00f3n de conjuntos y la convergencia de eventos. Completar estos ejercicios te proporcionar\u00e1 una base s\u00f3lida para profundizar en el estudio de la teor\u00eda de probabilidades.<\/em><\/p>\n<p><\/center><br \/>\n<\/p>\n<p style=\"text-align:center;\"><strong>OBJETIVOS DE APRENDIZAJE:<\/strong><br \/>\nAl completar esta clase el estudiante ser\u00e1 capaz de:\n<\/p>\n<ol>\n<li><strong>Demostrar<\/strong> propiedades b\u00e1sicas de las probabilidades<\/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;\">Lo que veremos a continuaci\u00f3n es una gu\u00eda de ejercicios (resueltos) donde el objetivo es el de demostrar algunos teoremas \u00fatiles para la teor\u00eda de las probabilidades. Prueba intentar resolverlos y luego compara tus resultados &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;\">Demuestre que <span class=\"katex-eq\" data-katex-display=\"false\">P(A^c) = 1 -P(A)<\/span> y,<\/span><\/a> a partir de esto, haga una deducci\u00f3n que permita argumentar que <span class=\"katex-eq\" data-katex-display=\"false\">P(\\emptyset) = 0<\/span><\/strong><br \/>\n<span class=\"collapseomatic \" id=\"id6aa1fe64c55d4\"  tabindex=\"0\" title=\"MOSTRAR SOLUCI\u00d3N\"    >MOSTRAR SOLUCI\u00d3N<\/span><div id=\"target-id6aa1fe64c55d4\" class=\"collapseomatic_content \">\n<p style=\"text-align: justify; color: #000000;\">De la definici\u00f3n de <a href=\"https:\/\/toposuranos.com\/el-espacio-de-probabilidades-medida-de-probabilidad\/\" target=\"_blank\" rel=\"noopener\">medida de probabilidad<\/a> se tiene que, si <span class=\"katex-eq\" data-katex-display=\"false\">A<\/span> y <span class=\"katex-eq\" data-katex-display=\"false\">B<\/span> son eventos medibles cualesquiera, entonces se cumplir\u00e1 que<\/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;\">Ahora, como <span class=\"katex-eq\" data-katex-display=\"false\">A\\cap A^c = \\emptyset,<\/span> de la parte [b], se tendr\u00e1 que:<\/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;\">Como <span class=\"katex-eq\" data-katex-display=\"false\">A\\cap A^c = \\emptyset<\/span> siempre es verdad y <span class=\"katex-eq\" data-katex-display=\"false\">A\\cup A^c = \\Omega<\/span>, entonces se tendr\u00e1 que<\/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;\">Y, por lo tanto<\/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;\">Que es lo que se quer\u00eda demostrar.<\/p>\n<p style=\"text-align: justify; color: #000000;\">Para demostrar que <span class=\"katex-eq\" data-katex-display=\"false\">P(\\emptyset)=0<\/span>, basta con tomar <span class=\"katex-eq\" data-katex-display=\"false\">A\\Omega<\/span> en la relaci\u00f3n que se acaba de probar y se tendr\u00e1 que<\/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) Demuestre que <span class=\"katex-eq\" data-katex-display=\"false\">A\\subseteq B \\rightarrow P(B\\setminus A) = P(B) - P(A)<\/span><\/span><\/a>,\u00bfLa igualdad es cierta en general?<\/strong><strong>b) Demostrar que <span class=\"katex-eq\" data-katex-display=\"false\">A\\subseteq B \\rightarrow P(B)\\leq P(A)<\/span><\/strong><span class=\"collapseomatic \" id=\"id6aa1fe64c5892\"  tabindex=\"0\" title=\"MOSTRAR SOLUCI\u00d3N a)\"    >MOSTRAR SOLUCI\u00d3N a)<\/span><div id=\"target-id6aa1fe64c5892\" 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>; Premisa<\/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>; Defincion de Teor. de Conjuntos<\/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>; Propiedad de los conjuntos<\/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>; Propiedad de los conjuntos<\/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>; De (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>; De (4) + Def, Medida de Probabilidad<\/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>; De (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>; De (1)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: center; color: #000000;\">Por lo tanto <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;\">Notemos que esta igualdad no es cierta en general, ya que depende de que se cumpla <span class=\"katex-eq\" data-katex-display=\"false\">A\\subseteq B<\/span> para ser obtenida. De estos desarrollos se puede ver que, si tal cosa no se cumple, entonces se tendr\u00e1 que <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=\"id6aa1fe64c5a7f\"  tabindex=\"0\" title=\"MOSTRAR SOLUCI\u00d3N b)\"    >MOSTRAR SOLUCI\u00d3N b)<\/span><div id=\"target-id6aa1fe64c5a7f\" class=\"collapseomatic_content \">\n<p style=\"text-align: justify; color: #000000;\">A partir de lo razonado en a) se tiene que:<\/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;\">Finalmente, como <span class=\"katex-eq\" data-katex-display=\"false\">P<\/span> es una medida de probabilidad, se tiene que <span class=\"katex-eq\" data-katex-display=\"false\">\\forall X (P(X)\\geq 0),<\/span> de modo que, por lo tanto:<\/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) Demuestre que <span class=\"katex-eq\" data-katex-display=\"false\">P(A\\cup B) = P(A) + P(B) - P(A\\cup B).<\/span><\/span><\/a> Acompa\u00f1e la demostraci\u00f3n con un diagrama.<\/strong><strong>b) Usando el resultado anterior demuestre que <span class=\"katex-eq\" data-katex-display=\"false\">P(A\\cup B) \\leq P(A) + P(B)<\/span><\/strong><span class=\"collapseomatic \" id=\"id6aa1fe64c5b7c\"  tabindex=\"0\" title=\"MOSTRAR SOLUCI\u00d3N PARTE a)\"    >MOSTRAR SOLUCI\u00d3N PARTE a)<\/span><div id=\"target-id6aa1fe64c5b7c\" 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\">A\\cup B = (A\\triangle B) \\cup (A\\cap B)<\/span>; Propiedad de los conjuntos<\/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\">A\\triangle B := (A\\setminus B) \\cup (B\\setminus A)<\/span>; Definici\u00f3n de diferencia sim\u00e9trica<\/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\">(A\\triangle B)\\cap ( A \\cap B) = \\emptyset<\/span>; propiedad de los conjuntos<\/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\">(A\\setminus B)\\cap (B\\setminus A) = \\emptyset<\/span>; propiedad de los conjuntos<\/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\\cup B) = P[(A\\triangle B) \\cup (A\\cap B)]<\/span>; De (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>; De (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>; De (2,4)<\/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\">P(B\\setminus A) = P(B) - P(A\\cap B)<\/span>; Aplicando resultado del ejercicio 2<\/td>\n<\/tr>\n<tr>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">(7)<\/span><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">P(A\\setminus B) = P(A) - P(A\\cap B)<\/span>; Lo mismo que (6)<\/td>\n<\/tr>\n<tr>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">(8)<\/span><\/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>; de(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=\"id6aa1fe64c5cfe\"  tabindex=\"0\" title=\"MOSTRAR SOLUCI\u00d3N PARTE b)\"    >MOSTRAR SOLUCI\u00d3N PARTE b)<\/span><div id=\"target-id6aa1fe64c5cfe\" 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>; resultado de la parte 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;\">Si <span class=\"katex-eq\" data-katex-display=\"false\">\\{E_n\\}<\/span> es una familia infinita de eventos<\/span> <\/a>tales que <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> Demostrar que se cumple<\/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=\"id6aa1fe64c5e01\"  tabindex=\"0\" title=\"MOSTRAR SOLUCI\u00d3N\"    >MOSTRAR SOLUCI\u00d3N<\/span><div id=\"target-id6aa1fe64c5e01\" 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>; Hip\u00f3tesis<\/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>; Por complementaci\u00f3n desde (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>; Propiedad de Continuidad<\/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>; Leyes de DeMorgan<\/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>; Demostrado en el ejercicio 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>; de (2,4,5) aplicado sobre (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;\">Resolviendo estos ejercicios completar\u00e1s un primer pilar que te servir\u00e1 de sustento para continuar con el estudio de la teor\u00eda de la probabilidades.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Teoremas \u00datiles para el C\u00e1lculo de Probabilidades ResumenEn esta clase se presentan ejercicios resueltos en que se demuestran algunos teoremas \u00fatiles para el c\u00e1lculo de probabilidades, incluyendo demostraciones y deducciones. Los ejercicios abordan temas como la probabilidad complementaria, la inclusi\u00f3n de conjuntos y la convergencia de eventos. Completar estos ejercicios te proporcionar\u00e1 una base s\u00f3lida [&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":36,"footnotes":""},"categories":[563,668],"tags":[],"class_list":["post-26392","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-matematica","category-probabilidades-y-estadistica"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v28.0 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>Teoremas \u00datiles para el C\u00e1lculo de Probabilidades - toposuranos.com\/material<\/title>\n<meta name=\"description\" content=\"Explora teoremas clave de probabilidad presentada con ejercicios resueltos. 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