{"id":26697,"date":"2021-04-19T13:00:52","date_gmt":"2021-04-19T13:00:52","guid":{"rendered":"http:\/\/toposuranos.com\/material\/?p=26697"},"modified":"2024-05-21T12:20:01","modified_gmt":"2024-05-21T12:20:01","slug":"probabilidade-condicional-e-independencia-entre-eventos","status":"publish","type":"post","link":"https:\/\/toposuranos.com\/material\/pt\/probabilidade-condicional-e-independencia-entre-eventos\/","title":{"rendered":"Probabilidade Condicional e Independ\u00eancia entre Eventos"},"content":{"rendered":"<div style=\"background-color:#F3F3F3; padding:20px;\">\n<center><\/p>\n<h1>Probabilidade Condicional e Independ\u00eancia entre Eventos<\/h1>\n<p><\/p>\n<p style=\"text-align:center;\"><strong>Resumo<\/strong><br \/><em>Nesta sess\u00e3o, exploraremos o conceito de probabilidade condicional e a intera\u00e7\u00e3o entre eventos. Vamos adquirir as habilidades para calcular probabilidades condicionais e determinar a depend\u00eancia ou independ\u00eancia entre eventos. Aplicaremos exemplos pr\u00e1ticos, como o estudo da preval\u00eancia de c\u00e1ries em consumidores de doces, para ilustrar esses conceitos. Ao concluir, voc\u00ea ter\u00e1 uma clara compreens\u00e3o de como aplicar a probabilidade condicional e analisar eventos dependentes e independentes.<\/em><\/p>\n<p><\/center><br \/>\n<\/p>\n<p style=\"text-align:center;\"><strong>OBJETIVOS DE APRENDIZAGEM:<\/strong><br \/>\nAo finalizar esta aula, voc\u00ea ser\u00e1 capaz de:\n<\/p>\n<ol>\n<li><strong>Compreender<\/strong> a defini\u00e7\u00e3o de probabilidade condicional e sua rela\u00e7\u00e3o com a interse\u00e7\u00e3o de eventos e as probabilidades individuais.<\/li>\n<li><strong>Identificar<\/strong> associa\u00e7\u00f5es positivas e negativas entre eventos a partir da compara\u00e7\u00e3o de probabilidades condicionais.<\/li>\n<li><strong>Testar<\/strong> a independ\u00eancia entre diferentes eventos.<\/li>\n<\/ol>\n<p style=\"text-align:center;\"><strong><u>\u00cdNDICE DE CONTE\u00daDOS<\/u><\/strong><br \/>\n<a href=\"#1\">A PROBABILIDADE CONDICIONAL<\/a><br \/>\n<a href=\"#2\">DEFINI\u00c7\u00c3O FORMAL DE PROBABILIDADE CONDICIONAL<\/a><br \/>\n<a href=\"#3\">RELA\u00c7\u00c3O ENTRE EVENTOS<\/a><br \/>\n<a href=\"#4\">INDEPEND\u00caNCIA ENTRE EVENTOS E COMPLEMENTOS DE EVENTOS<\/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>A Probabilidade Condicional<\/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;\">Qual \u00e9 a probabilidade de que um evento A<\/span><\/strong><\/a> ocorra dado que j\u00e1 ocorreu B? O c\u00e1lculo desse tipo de probabilidades envolve o conceito de <strong>probabilidade condicional.<\/strong> A seguir, estudaremos a probabilidade condicional, sua defini\u00e7\u00e3o e como a partir disso se inferem as rela\u00e7\u00f5es de depend\u00eancia e independ\u00eancia entre eventos.<\/p>\n<p style=\"text-align: justify; color: #000000;\">Suponhamos que queremos medir a preval\u00eancia de c\u00e1ries entre os consumidores habituais de doces. Se examinarmos um espa\u00e7o amostral formado por <span class=\"katex-eq\" data-katex-display=\"false\">N<\/span> pessoas <span class=\"katex-eq\" data-katex-display=\"false\">\\Omega_N,<\/span> veremos que ele pode ser dividido em 4 subconjuntos:<\/p>\n<ul style=\"text-align: justify; color: #000000;\">\n<li><span class=\"katex-eq\" data-katex-display=\"false\">A:=\\left\\{  {Pessoas\\;que\\;t\u00eam\\;c\u00e1ries}\\right\\}<\/span><\/li>\n<li><span class=\"katex-eq\" data-katex-display=\"false\">A^c:=\\left\\{{Pessoas\\;que\\;N\u00c3O\\;t\u00eam\\;c\u00e1ries}\\right\\}<\/span><\/li>\n<li><span class=\"katex-eq\" data-katex-display=\"false\">B:=\\left\\{  {Pessoas\\;que\\;comem\\;doces\\;regularmente}\\right\\}<\/span><\/li>\n<li><span class=\"katex-eq\" data-katex-display=\"false\">B^c:=\\left\\{{Pessoas\\;que\\;N\u00c3O\\;comem\\;doces\\;regularmente}\\right\\}<\/span><\/li>\n<\/ul>\n<p style=\"text-align: justify; color: #000000;\">A partir disso, \u00e9 claro que <span class=\"katex-eq\" data-katex-display=\"false\">A\\cup A^c = \\Omega_N<\/span> e <span class=\"katex-eq\" data-katex-display=\"false\">B\\cup B^c = \\Omega_N,<\/span> mas <span class=\"katex-eq\" data-katex-display=\"false\">A\\cap B<\/span> n\u00e3o \u00e9 necessariamente vazio. O cen\u00e1rio geral \u00e9 representado pela figura a seguir:<\/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=\"Probabilidade Condicional\" 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=\"Probabilidade Condicional\" 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;\">Ent\u00e3o, se lembrarmos a defini\u00e7\u00e3o de probabilidade<\/span><\/strong><\/a> como o limite das frequ\u00eancias relativas, poderemos dizer que a probabilidade de que uma pessoa tenha c\u00e1ries dado que se comprovou que consome doces, <span class=\"katex-eq\" data-katex-display=\"false\">P(A|B)<\/span> ser\u00e1:<\/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;\">Por outro lado, tem-se que:<\/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;\">De modo que, se substituirmos essas duas \u00faltimas express\u00f5es em <span class=\"katex-eq\" data-katex-display=\"false\">P(A|B)<\/span>, teremos:<\/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;\">Com isso, temos a seguinte defini\u00e7\u00e3o:<\/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>Defini\u00e7\u00e3o formal de probabilidade condicional<\/h2>\n<p><strong>DEFINI\u00c7\u00c3O:<\/strong><\/span> Define-se a probabilidade de <span class=\"katex-eq\" data-katex-display=\"false\">A<\/span>, dado que ocorreu <span class=\"katex-eq\" data-katex-display=\"false\">B,<\/span> <span class=\"katex-eq\" data-katex-display=\"false\">P(A|B),<\/span> atrav\u00e9s da rela\u00e7\u00e3o:<\/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;\">No pensamento cotidiano, tende-se<\/span><\/strong><\/a> a haver certa confus\u00e3o entre <span class=\"katex-eq\" data-katex-display=\"false\">P(A|B)<\/span> e <span class=\"katex-eq\" data-katex-display=\"false\">P(B|A).<\/span> Para esclarecer essa diferen\u00e7a, vamos revisar um exemplo baseado em um caso extremo: Observe que, enquanto todos os futebolistas t\u00eam duas pernas, apenas uma pequena parte das pessoas com duas pernas s\u00e3o futebolistas.<\/p>\n<p><a name=\"3\"><\/a><br \/>\n<\/br><\/br><\/p>\n<h2>Rela\u00e7\u00e3o entre eventos<\/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;\">Continuando com o exemplo da preval\u00eancia<\/span><\/strong><\/a> de c\u00e1ries entre pessoas que consomem doces habitualmente. Se o consumo de doces faz com que as pessoas sejam mais propensas a ter c\u00e1ries, ent\u00e3o deveria ocorrer que:<\/p>\n<p style=\"text-align: justify; color: #000000;\"><span class=\"katex-eq\" data-katex-display=\"false\">P(A|B) \\gt P(A).<\/span>Aqui temos que <span class=\"katex-eq\" data-katex-display=\"false\">B<\/span> potencializa <span class=\"katex-eq\" data-katex-display=\"false\">A<\/span> e, portanto, dizemos que h\u00e1 uma <strong>associa\u00e7\u00e3o positiva<\/strong> entre os eventos.<\/p>\n<p style=\"text-align: justify; color: #000000;\">Se, por outro lado, o consumo de doces previne as c\u00e1ries, ent\u00e3o dever\u00edamos ter:<\/p>\n<p style=\"text-align: justify; color: #000000;\"><span class=\"katex-eq\" data-katex-display=\"false\">P(A|B) \\lt P(A).<\/span>Neste caso, <span class=\"katex-eq\" data-katex-display=\"false\">B<\/span> inibe <span class=\"katex-eq\" data-katex-display=\"false\">A<\/span> e, portanto, dizemos que h\u00e1 uma <strong>associa\u00e7\u00e3o negativa<\/strong> entre os eventos.<\/p>\n<p style=\"text-align: justify; color: #000000;\">E se n\u00e3o houvesse nenhuma rela\u00e7\u00e3o entre esses dois eventos, nem positiva nem negativa, ent\u00e3o dever\u00edamos ter que:<\/p>\n<p style=\"text-align: justify; color: #000000;\"><span class=\"katex-eq\" data-katex-display=\"false\">P(A|B) = P(A).<\/span>Daqui, faz-se a infer\u00eancia que, na maioria dos textos de probabilidade, \u00e9 apresentada como defini\u00e7\u00e3o:<\/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;\">Esse racioc\u00ednio nos mostra a rela\u00e7\u00e3o entre a probabilidade condicional e a independ\u00eancia dos eventos.<\/p>\n<p style=\"text-align: justify; color: #000000;\"><span style=\"color: #800000;\"><strong>DEFINI\u00c7\u00c3O:<\/strong><\/span>Dado dois eventos <span class=\"katex-eq\" data-katex-display=\"false\">A<\/span> e <span class=\"katex-eq\" data-katex-display=\"false\">B<\/span>, eles s\u00e3o ditos <strong>independentes<\/strong> se satisfazem a rela\u00e7\u00e3o:<\/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>Independ\u00eancia entre eventos e complementos de eventos<\/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;\">A independ\u00eancia entre dois eventos<\/span><\/strong><\/a> <span class=\"katex-eq\" data-katex-display=\"false\">A<\/span> e <span class=\"katex-eq\" data-katex-display=\"false\">B<\/span> \u00e9 provada equivalente \u00e0 independ\u00eancia de <span class=\"katex-eq\" data-katex-display=\"false\">A<\/span> com <span class=\"katex-eq\" data-katex-display=\"false\">B^c,<\/span> a de <span class=\"katex-eq\" data-katex-display=\"false\">A^c<\/span> com <span class=\"katex-eq\" data-katex-display=\"false\">B,<\/span> e a de <span class=\"katex-eq\" data-katex-display=\"false\">A^c<\/span> com <span class=\"katex-eq\" data-katex-display=\"false\">B^c.<\/span>\n<p style=\"text-align: justify; color: #000000;\"><span style=\"color: #000080;\"><strong>DEMONSTRA\u00c7\u00c3O<\/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>; Pressuposi\u00e7\u00e3o<\/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>; porque <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\">Ver desenvolvimento do exerc\u00edcio 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>; De (2) e (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>; De (1) e (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>; Fatorando por <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\">ver desenvolvimento do exerc\u00edcio 1<\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: justify; color: #000000;\">Essa \u00faltima express\u00e3o se l\u00ea como: \u00abDo fato de que <span class=\"katex-eq\" data-katex-display=\"false\">A<\/span> e <span class=\"katex-eq\" data-katex-display=\"false\">B<\/span> s\u00e3o independentes, infere-se que <span class=\"katex-eq\" data-katex-display=\"false\">A<\/span> e <span class=\"katex-eq\" data-katex-display=\"false\">B^c<\/span> tamb\u00e9m s\u00e3o independentes.<\/p>\n<p style=\"text-align: justify; color: #000000;\">A demonstra\u00e7\u00e3o no sentido inverso \u00e9 feita de forma semelhante.<\/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>; Pressuposi\u00e7\u00e3o<\/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\">ver desenvolvimento do exerc\u00edcio 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>; Realizando o produto do par\u00eantese no lado direito.<\/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>; Porque <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\">Ver desenvolvimento do exerc\u00edcio 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>; De (1), (2) e (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>; Eliminando termos semelhantes<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: justify; color: #000000;\">E essa express\u00e3o se l\u00ea como: \u00abDo fato de que <span class=\"katex-eq\" data-katex-display=\"false\">A<\/span> e <span class=\"katex-eq\" data-katex-display=\"false\">B^c<\/span> s\u00e3o independentes, infere-se que <span class=\"katex-eq\" data-katex-display=\"false\">A<\/span> e <span class=\"katex-eq\" data-katex-display=\"false\">B<\/span> tamb\u00e9m s\u00e3o independentes.<\/p>\n<p style=\"text-align: justify; color: #000000;\">Finalmente, desses dois racioc\u00ednios, tem-se a equival\u00eancia provada entre as independ\u00eancias de <span class=\"katex-eq\" data-katex-display=\"false\">A<\/span> com <span class=\"katex-eq\" data-katex-display=\"false\">B<\/span> e a de <span class=\"katex-eq\" data-katex-display=\"false\">A<\/span> com <span class=\"katex-eq\" data-katex-display=\"false\">B^c.<\/span>\n<p style=\"text-align: justify; color: #000000;\">As outras equival\u00eancias provadas podem ser obtidas de forma semelhante. Estas ser\u00e3o deixadas como um desafio para o leitor &gt;:D<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Probabilidade Condicional e Independ\u00eancia entre Eventos ResumoNesta sess\u00e3o, exploraremos o conceito de probabilidade condicional e a intera\u00e7\u00e3o entre eventos. Vamos adquirir as habilidades para calcular probabilidades condicionais e determinar a depend\u00eancia ou independ\u00eancia entre eventos. Aplicaremos exemplos pr\u00e1ticos, como o estudo da preval\u00eancia de c\u00e1ries em consumidores de doces, para ilustrar esses conceitos. Ao concluir, [&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":15,"footnotes":""},"categories":[571,672],"tags":[],"class_list":["post-26697","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-matematica-pt","category-probabilidades-e-estatistica"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.4 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>Probabilidade Condicional e Independ\u00eancia entre Eventos - toposuranos.com\/material<\/title>\n<meta name=\"description\" content=\"Quer saber o que \u00e9 probabilidade condicional? 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