{"id":26752,"date":"2021-05-12T13:00:40","date_gmt":"2021-05-12T13:00:40","guid":{"rendered":"http:\/\/toposuranos.com\/material\/?p=26752"},"modified":"2024-05-22T00:10:31","modified_gmt":"2024-05-22T00:10:31","slug":"o-ensaio-de-bernoulli-e-a-distribuicao-binomial","status":"publish","type":"post","link":"https:\/\/toposuranos.com\/material\/pt\/o-ensaio-de-bernoulli-e-a-distribuicao-binomial\/","title":{"rendered":"O Ensaio de Bernoulli e a Distribui\u00e7\u00e3o Binomial"},"content":{"rendered":"<div style=\"background-color:#F3F3F3; padding:20px;\">\n<center><\/p>\n<h1>O Ensaio de Bernoulli e a Distribui\u00e7\u00e3o Binomial<\/h1>\n<p><\/p>\n<p style=\"text-align:center;\"><strong>Resumo<\/strong><br \/><em>Nesta aula, estudaremos o conceito dos ensaios de Bernoulli e suas implica\u00e7\u00f5es na teoria das probabilidades. Come\u00e7amos com uma defini\u00e7\u00e3o detalhada dos ensaios de Bernoulli para depois abordar o conceito de independ\u00eancia entre eventos. Ap\u00f3s esclarecer essas ideias, aplica-se o teorema do bin\u00f4mio para entender como a repeti\u00e7\u00e3o de um ensaio de Bernoulli produz resultados com distribui\u00e7\u00e3o binomial. Finalmente, prop\u00f5em-se exerc\u00edcios pr\u00e1ticos para aplicar e refor\u00e7ar esses conceitos.<\/em><\/p>\n<p><\/center><br \/>\n<\/p>\n<p style=\"text-align:center;\"><strong>OBJETIVOS DE APRENDIZAGEM:<\/strong><br \/>\nAo final desta aula, o aluno ser\u00e1 capaz de:\n<\/p>\n<ol>\n<li><strong>Identificar<\/strong> as principais caracter\u00edsticas dos ensaios de Bernoulli, incluindo a independ\u00eancia entre as tentativas.<\/li>\n<li><strong>Aplicar<\/strong> corretamente a nota\u00e7\u00e3o para os eventos binomiais derivados dos ensaios de Bernoulli.<\/li>\n<li><strong>Distinguir<\/strong> entre diferentes formas de independ\u00eancia (2-independ\u00eancia, 3-independ\u00eancia, n-independ\u00eancia) e compreender sua rela\u00e7\u00e3o e aplica\u00e7\u00e3o nos ensaios de Bernoulli.<\/li>\n<li><strong>Compreender<\/strong> a rela\u00e7\u00e3o entre o ensaio de Bernoulli e o teorema do bin\u00f4mio, e como essa rela\u00e7\u00e3o pode ser utilizada para calcular a probabilidade de uma s\u00e9rie de sucessos e fracassos.<\/li>\n<li><strong>Aplicar<\/strong> a distribui\u00e7\u00e3o binomial (ou de Bernoulli) para calcular a probabilidade de um certo n\u00famero de sucessos em uma s\u00e9rie de tentativas.<\/li>\n<\/ol>\n<p><center><br \/>\n<strong><u>\u00cdNDICE DE CONTE\u00daDOS<\/u>:<\/strong><br \/>\n<a href=\"#1\">O ensaio de Bernoulli<\/a><br \/>\n<a href=\"#2\">Diferentes formas de independ\u00eancia<\/a><br \/>\n<a href=\"#3\">O ensaio de Bernoulli e o teorema do bin\u00f4mio<\/a><br \/>\n<a href=\"#4\">A distribui\u00e7\u00e3o binomial (ou de Bernoulli) e as distribui\u00e7\u00f5es de Probabilidade<\/a><br \/>\n<a href=\"#5\">Exerc\u00edcios:<\/a><br \/>\n<\/center><\/p>\n<p><center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/uyljDy0bcvU\" 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><\/br><\/p>\n<h2>O ensaio de Bernoulli<\/h2>\n<p style=\"text-align: justify; color: #000000;\"><a href=\"https:\/\/www.youtube.com\/watch?v=uyljDy0bcvU&amp;t=103s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">Um ensaio Bernoulli \u00e9 um experimento aleat\u00f3rio<\/span><\/strong><\/a> dicot\u00f4mico com uma certa probabilidade de sucesso <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">p.<\/span><\/span> Se um ensaio Bernoulli se repete <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n<\/span><\/span> vezes de forma id\u00eantica e independente, ent\u00e3o se obt\u00eam os <strong>eventos de Bernoulli:<\/strong> Certo n\u00famero <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k<\/span><\/span> de sucessos entre <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n<\/span><\/span> tentativas. Esses tamb\u00e9m recebem o nome de <strong>eventos binomiais<\/strong> e os representamos atrav\u00e9s da nota\u00e7\u00e3o<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\Large \\displaystyle Bi(n;k;p)<\/span><\/span><\/p>\n<p style=\"text-align: justify; color: #000000;\">Outra caracter\u00edstica importante dos ensaios de Bernoulli \u00e9 que todas as tentativas s\u00e3o independentes entre si.<\/p>\n<p style=\"text-align: justify; color: #000000;\"><span style=\"color: #000080;\">EXEMPLO:<\/span> Lan\u00e7a-se repetidamente um dado de 6 faces. Exemplos de eventos do tipo Bernoulli para este experimento s\u00e3o:<\/p>\n<ul style=\"text-align: justify; color: #000000;\">\n<li><strong>Obter 3 ases entre 5 tentativas:<\/strong> representado por <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">Bi(5;3;1\/6)<\/span><\/span><\/li>\n<li><strong>Obter 7 n\u00fameros pares entre 12 tentativas:<\/strong> representado por <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">Bi(12;7;1\/3)<\/span><\/span><\/li>\n<\/ul>\n<p><a name=\"2\"><\/a><\/br><\/br><\/p>\n<h3>Diferentes formas de independ\u00eancia<\/h3>\n<p style=\"text-align: justify; color: #000000;\"><a href=\"https:\/\/www.youtube.com\/watch?v=uyljDy0bcvU&amp;t=283s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">A independ\u00eancia entre as tentativas desenvolvidas<\/span><\/strong><\/a> no ensaio Bernoulli n\u00e3o \u00e9 precisamente a mesma independ\u00eancia que j\u00e1 revisamos, trata-se de uma vers\u00e3o muito mais restrita. Para explicar essa diferen\u00e7a, examinemos os tipos de independ\u00eancia entre eventos<\/p>\n<h4>2-independ\u00eancia<\/h4>\n<p style=\"text-align: justify; color: #000000;\"><a href=\"https:\/\/toposuranos.com\/probabilidad-condicional-e-independencia-entre-eventos\/\" rel=\"noopener\" target=\"_blank\">A independ\u00eancia que j\u00e1 conhecemos<\/a> \u00e9 a que se d\u00e1 entre dois eventos, a chamamos de \u00ab2-independ\u00eancia\u00bb. Em esses termos, dizemos que os eventos <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">A<\/span><\/span> e <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">B<\/span><\/span> s\u00e3o 2-independentes se<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">P(A\\cap B) = P(A)P(B)<\/span><\/span><\/p>\n<h4>3-independ\u00eancia<\/h4>\n<p style=\"text-align: justify; color: #000000;\">De forma an\u00e1loga, se define a 3-independ\u00eancia entre tr\u00eas eventos <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">A,<\/span><\/span> <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">B<\/span><\/span> e <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">C<\/span><\/span> atrav\u00e9s da rela\u00e7\u00e3o<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">P(A\\cap B\\cap C) = P(A)P(B)P(C)<\/span><\/span><\/p>\n<p style=\"text-align: justify; color: #000000;\">\u00c9 importante destacar que a 2-independ\u00eancia entre <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">A,<\/span><\/span> <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">B<\/span><\/span> e <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">C<\/span><\/span> n\u00e3o implica necessariamente a 3-independ\u00eancia, embora no caso inverso a implica\u00e7\u00e3o seja certa.<\/p>\n<h4>A n-independ\u00eancia entre os ensaios de Bernoulli<\/h4>\n<p style=\"text-align: justify; color: #000000;\">Procedendo de forma an\u00e1loga \u00e0s defini\u00e7\u00f5es anteriores, define-se a n-independ\u00eancia entre uma cole\u00e7\u00e3o de eventos <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">A_1, \\cdots, A_n<\/span><\/span> atrav\u00e9s da rela\u00e7\u00e3o<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\Large \\displaystyle P\\left(\\bigcap_{i=1}^n A_i\\right) = \\prod_{i=1}^n P(A_i) <\/span><\/span><\/p>\n<p style=\"text-align: justify; color: #000000;\">E de forma an\u00e1loga, temos que:<\/p>\n<table style=\"color: #000000;\">\n<tbody>\n<tr>\n<td style=\"text-align: center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(n-1)<\/span><\/span>-independ\u00eancia n\u00e3o implica necessariamente <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n<\/span><\/span>-independ\u00eancia<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n<\/span><\/span>-independ\u00eancia <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\Longrightarrow<\/span> <span class=\"katex-eq\" data-katex-display=\"false\">(n-1)<\/span><\/span>-independ\u00eancia<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: justify; color: #000000;\">As <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n<\/span><\/span> repeti\u00e7\u00f5es realizadas no ensaio de Bernoulli s\u00e3o <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n<\/span><\/span>-independentes.<\/p>\n<p><a name=\"3\"><\/a><\/br><\/br><\/p>\n<h2>O ensaio de Bernoulli e o teorema do bin\u00f4mio<\/h2>\n<p style=\"text-align: justify; color: #000000;\"><a href=\"https:\/\/www.youtube.com\/watch?v=uyljDy0bcvU&amp;t=478s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">Consideremos um experimento de sucesso e fracasso<\/span><\/strong><\/a> com probabilidade de sucesso de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">p<\/span><\/span>; em cada tentativa ter\u00e1, consequentemente, uma probabilidade <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">1-p<\/span><\/span> de fracasso. \u00c9 claro que a probabilidade de ocorrer um sucesso ou um fracasso entre cada tentativa \u00e9 1; e como todas as tentativas s\u00e3o independentes, a probabilidade de ocorrer sucesso ou fracasso nas <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n<\/span><\/span> tentativas ser\u00e1 <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">1^n.<\/span><\/span> A partir disso, teremos que:<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\Large \\displaystyle 1 = 1^n = [p + (1-p)]^n = \\sum_{k=0}^n {{n}\\choose{k}} p^k(1-p)^{n-k} <\/span><\/span><\/p>\n<p style=\"text-align: justify; color: #000000;\">Na \u00faltima igualdade, aplicou-se o <a href=\"https:\/\/toposuranos.com\/ejercicios-de-induccion-matematica-teorema-del-binomio-de-newton\/\" rel=\"noopener\" target=\"_blank\">Teorema do Bin\u00f4mio de Newton,<\/a> e os termos dentro da soma podem ser interpretados da seguinte maneira:<\/p>\n<ul style=\"text-align: justify; color: #000000;\">\n<li><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle {{n}\\choose{k}}<\/span><\/span>: o n\u00famero de formas em que podem ocorrer <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k<\/span><\/span> sucessos ao realizar <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n<\/span><\/span> tentativas<\/li>\n<li><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">p^k<\/span><\/span>: A probabilidade de ocorrerem <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k<\/span><\/span> sucessos independentes<\/li>\n<li><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(1-p)^{n-k}<\/span><\/span>: A probabilidade de ocorrerem <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n-k<\/span><\/span> fracassos independentes<\/li>\n<\/ul>\n<p style=\"text-align: justify; color: #000000;\">Ao juntar esses elementos da forma em que aparecem na soma, obtemos: a probabilidade de obter <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k<\/span><\/span> sucessos entre <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n<\/span><\/span> tentativas; ou, equivalentemente, a probabilidade de obter <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n-k<\/span><\/span> fracassos entre <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n<\/span><\/span> tentativas.<\/p>\n<p style=\"text-align: justify; color: #000000;\">Se separarmos cada termo da soma, teremos as probabilidades de obter:<\/p>\n<table style=\"text-align: justify; color: #000000;\">\n<tbody>\n<tr>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle {{n}\\choose{0}} p^0(1-p)^{n-0} = (1-p)^n<\/span><\/span><\/td>\n<td>0 sucessos entre n tentativas<\/td>\n<\/tr>\n<tr>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle {{n}\\choose{1}} p^1(1-p)^{n-1} = n p(1-p)^{n-1}<\/span><\/span><\/td>\n<td>1 sucesso entre n tentativas<\/td>\n<\/tr>\n<tr>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle {{n}\\choose{2}} p^2(1-p)^{n-2}<\/span><\/span><\/td>\n<td>2 sucessos entre n tentativas<\/td>\n<\/tr>\n<tr>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\vdots<\/span><\/span><\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\vdots<\/span><\/span><\/td>\n<\/tr>\n<tr>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle {{n}\\choose{k}} p^k(1-p)^{n-k}<\/span><\/span><\/td>\n<td>k sucessos entre n tentativas<\/td>\n<\/tr>\n<tr>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\vdots<\/span><\/span><\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\vdots<\/span><\/span><\/td>\n<\/tr>\n<tr>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle {{n}\\choose{n-1}} p^{n-1}(1-p)^{n-(n-1)} = n p^{n-1}(1-p)<\/span><\/span><\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n-1<\/span><\/span> sucessos entre <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n<\/span><\/span> tentativas<\/td>\n<\/tr>\n<tr>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle {{n}\\choose{n}} p^{n}(1-p)^{0} = p^{n}<\/span><\/span><\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n<\/span><\/span> sucessos entre <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n<\/span><\/span> tentativas<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: justify; color: #000000;\">E a soma de todos estes, como j\u00e1 vimos, \u00e9 \u00ab1\u00bb. Mostrando que todas as possibilidades foram cobertas.<\/p>\n<p style=\"text-align: justify; color: #000000;\">A partir disso, define-se a <strong>probabilidade do evento Bernoulli<\/strong>:<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\\Large \\color{blue}{P(Bi(n;k;p)) = {{n}\\choose{k}}p^k(1-p)^{n-k}}<\/span><\/span><\/p>\n<p style=\"text-align: justify; color: #000000;\">Ou tamb\u00e9m dizemos que <strong>o n\u00famero de sucessos <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">X<\/span><\/span> tem distribui\u00e7\u00e3o binomial:<\/strong><\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\color{blue}{\\Large \\displaystyle X\\sim Bi(n;p) \\longmapsto P(X=x) = {{n}\\choose{x}}p^x(1-p)^{n-x}}<\/span><\/span><\/p>\n<p><a name=\"4\"><\/a><\/br><\/br><\/p>\n<h3>A distribui\u00e7\u00e3o binomial (ou de Bernoulli) e as distribui\u00e7\u00f5es de Probabilidade<\/h3>\n<p style=\"text-align: justify; color: #000000;\"><a href=\"https:\/\/www.youtube.com\/watch?v=uyljDy0bcvU&amp;t=779s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">Atrav\u00e9s da distribui\u00e7\u00e3o binomial<\/span><\/strong><\/a> \u00e9 que come\u00e7amos a ter as primeiras no\u00e7\u00f5es de distribui\u00e7\u00f5es de probabilidade e de vari\u00e1vel aleat\u00f3ria. Neste caso, a vari\u00e1vel aleat\u00f3ria (discreta) est\u00e1 associada ao n\u00famero de sucessos e sua distribui\u00e7\u00e3o de probabilidade \u00e9 dada pelos termos do teorema do bin\u00f4mio<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">{\\Large \\displaystyle P(X=x) = {{n}\\choose{x}}p^x(1-p)^{n-x}}<\/span><\/span><\/p>\n<p><center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/36s4rcfQn7M\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen><\/iframe><\/center><\/p>\n<p><a name=\"5\"><\/a><\/br><\/br><\/p>\n<h2>Exerc\u00edcios:<\/h2>\n<ol style=\"text-align: justify; color: #000000;\">\n<li>Lan\u00e7a-se um dado de 6 faces equilibrado 5 vezes. Calcule a probabilidade de obter 3 vezes um n\u00famero par como resultado.<\/li>\n<li>Lan\u00e7a-se uma moeda 10 vezes. Calcule a probabilidade de obter, de 0 a 10 caras e fa\u00e7a um gr\u00e1fico que mostre a probabilidade para cada resultado. Como ser\u00e1 o gr\u00e1fico se aumentar o n\u00famero de lan\u00e7amentos e examinar a probabilidade de obter um n\u00famero de caras que vai de 0 at\u00e9 esse n\u00famero de lan\u00e7amentos? Uma planilha de Excel pode ser \u00fatil aqui.<\/li>\n<li>Tem-se uma tombola com uma quantidade <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">s<\/span><\/span> de bolinhas, onde <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r<\/span><\/span> s\u00e3o douradas e o resto brancas. Misturam-se todas e tira-se uma ao acaso, e ganha-se quando sai a dourada. Se este experimento se repete de maneira id\u00eantica <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">20<\/span><\/span> vezes, estime o n\u00famero de vit\u00f3rias mais prov\u00e1veis para cada valor poss\u00edvel de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0\\leq r\\leq s.<\/span><\/span> Uma planilha em Excel tamb\u00e9m pode ser \u00fatil aqui.<\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"<p>O Ensaio de Bernoulli e a Distribui\u00e7\u00e3o Binomial ResumoNesta aula, estudaremos o conceito dos ensaios de Bernoulli e suas implica\u00e7\u00f5es na teoria das probabilidades. Come\u00e7amos com uma defini\u00e7\u00e3o detalhada dos ensaios de Bernoulli para depois abordar o conceito de independ\u00eancia entre eventos. Ap\u00f3s esclarecer essas ideias, aplica-se o teorema do bin\u00f4mio para entender como a [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":26419,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"iawp_total_views":19,"footnotes":""},"categories":[571,672],"tags":[],"class_list":["post-26752","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>O Ensaio de Bernoulli e a Distribui\u00e7\u00e3o Binomial - toposuranos.com\/material<\/title>\n<meta name=\"description\" content=\"Descubra como a repeti\u00e7\u00e3o de um ensaio de Bernoulli produz resultados com uma Distribui\u00e7\u00e3o Binomial de Probabilidades.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, 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