{"id":26764,"date":"2021-05-12T13:00:03","date_gmt":"2021-05-12T13:00:03","guid":{"rendered":"http:\/\/toposuranos.com\/material\/?p=26764"},"modified":"2024-05-22T01:04:05","modified_gmt":"2024-05-22T01:04:05","slug":"lessai-de-bernoulli-et-la-distribution-binomiale","status":"publish","type":"post","link":"http:\/\/toposuranos.com\/material\/fr\/lessai-de-bernoulli-et-la-distribution-binomiale\/","title":{"rendered":"L&#8217;Essai de Bernoulli et la Distribution Binomiale"},"content":{"rendered":"<div style=\"background-color:#F3F3F3; padding:20px;\">\n<center><\/p>\n<h1>L&#8217;Essai de Bernoulli et la Distribution Binomiale<\/h1>\n<p><\/p>\n<p style=\"text-align:center;\"><strong>R\u00e9sum\u00e9<\/strong><br \/><em>Dans cette classe, nous \u00e9tudierons le concept des essais de Bernoulli et leurs implications dans la th\u00e9orie des probabilit\u00e9s. Nous commen\u00e7ons par une d\u00e9finition d\u00e9taill\u00e9e des essais de Bernoulli pour ensuite aborder le concept d&#8217;ind\u00e9pendance entre les \u00e9v\u00e9nements. Apr\u00e8s avoir clarifi\u00e9 ces id\u00e9es, nous appliquons le th\u00e9or\u00e8me binomial pour comprendre comment la r\u00e9p\u00e9tition d&#8217;un essai de Bernoulli produit des r\u00e9sultats avec une distribution binomiale. Enfin, des exercices pratiques sont propos\u00e9s pour appliquer et renforcer ces concepts.<\/em><\/p>\n<p><\/center><br \/>\n<\/p>\n<p style=\"text-align:center;\"><strong>OBJECTIFS D&#8217;APPRENTISSAGE :<\/strong><br \/>\n\u00c0 la fin de cette classe, l&#8217;\u00e9tudiant sera capable de :\n<\/p>\n<ol>\n<li><strong>Identifier<\/strong> les principales caract\u00e9ristiques des essais de Bernoulli, y compris l&#8217;ind\u00e9pendance entre les tentatives.<\/li>\n<li><strong>Appliquer<\/strong> correctement la notation pour les \u00e9v\u00e9nements binomiaux d\u00e9riv\u00e9s des essais de Bernoulli.<\/li>\n<li><strong>Distinguer<\/strong> entre diff\u00e9rentes formes d&#8217;ind\u00e9pendance (2-ind\u00e9pendance, 3-ind\u00e9pendance, n-ind\u00e9pendance) et comprendre leur relation et application dans les essais de Bernoulli.<\/li>\n<li><strong>Comprendre<\/strong> la relation entre l&#8217;essai de Bernoulli et le th\u00e9or\u00e8me binomial, et comment cette relation peut \u00eatre utilis\u00e9e pour calculer la probabilit\u00e9 d&#8217;une s\u00e9rie de succ\u00e8s et d&#8217;\u00e9checs.<\/li>\n<li><strong>Appliquer<\/strong> la distribution binomiale (ou de Bernoulli) pour calculer la probabilit\u00e9 d&#8217;un certain nombre de succ\u00e8s dans une s\u00e9rie de tentatives.<\/li>\n<\/ol>\n<p><center><br \/>\n<strong><u>TABLE DES MATI\u00c8RES<\/u> :<\/strong><br \/>\n<a href=\"#1\">L&#8217;essai de Bernoulli<\/a><br \/>\n<a href=\"#2\">Diff\u00e9rentes formes d&#8217;ind\u00e9pendance<\/a><br \/>\n<a href=\"#3\">L&#8217;essai de Bernoulli et le th\u00e9or\u00e8me binomial<\/a><br \/>\n<a href=\"#4\">La distribution binomiale (ou de Bernoulli) et les distributions de probabilit\u00e9<\/a><br \/>\n<a href=\"#5\">Exercices :<\/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>L&#8217;essai 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;\">Un essai de Bernoulli est une exp\u00e9rience al\u00e9atoire<\/span><\/strong><\/a> dichotomique avec une certaine probabilit\u00e9 de succ\u00e8s <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">p.<\/span><\/span> Si un essai de Bernoulli est r\u00e9p\u00e9t\u00e9 <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n<\/span><\/span> fois de mani\u00e8re identique et ind\u00e9pendante, on obtient les <strong>\u00e9v\u00e9nements de Bernoulli :<\/strong> Un certain nombre <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k<\/span><\/span> de succ\u00e8s parmi <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n<\/span><\/span> tentatives. Ceux-ci sont \u00e9galement appel\u00e9s <strong>\u00e9v\u00e9nements binomiaux<\/strong> et nous les repr\u00e9sentons par la notation<\/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;\">Une autre caract\u00e9ristique importante des essais de Bernoulli est que toutes les tentatives sont ind\u00e9pendantes les unes des autres.<\/p>\n<p style=\"text-align: justify; color: #000000;\"><span style=\"color: #000080;\">EXEMPLE :<\/span> On lance plusieurs fois un d\u00e9 \u00e0 6 faces. Des exemples d&#8217;\u00e9v\u00e9nements de type Bernoulli pour cette exp\u00e9rience sont :<\/p>\n<ul style=\"text-align: justify; color: #000000;\">\n<li><strong>Obtenir 3 as en 5 tentatives :<\/strong> repr\u00e9sent\u00e9 par <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">Bi(5;3;1\/6)<\/span><\/span><\/li>\n<li><strong>Obtenir 7 nombres pairs en 12 tentatives :<\/strong> repr\u00e9sent\u00e9 par <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>Diff\u00e9rentes formes d&#8217;ind\u00e9pendance<\/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;\">L&#8217;ind\u00e9pendance entre les tentatives d\u00e9velopp\u00e9es<\/span><\/strong><\/a> dans l&#8217;essai de Bernoulli n&#8217;est pas pr\u00e9cis\u00e9ment la m\u00eame ind\u00e9pendance que nous avons d\u00e9j\u00e0 examin\u00e9e, c&#8217;est une version beaucoup plus restreinte. Pour expliquer cette diff\u00e9rence, examinons les types d&#8217;ind\u00e9pendance entre les \u00e9v\u00e9nements<\/p>\n<h4>2-ind\u00e9pendance<\/h4>\n<p style=\"text-align: justify; color: #000000;\"><a href=\"https:\/\/toposuranos.com\/probabilidad-condicional-e-independencia-entre-eventos\/\" rel=\"noopener\" target=\"_blank\">L&#8217;ind\u00e9pendance que nous connaissons d\u00e9j\u00e0<\/a> est celle qui existe entre deux \u00e9v\u00e9nements. Nous l&#8217;appelons \u00ab2-ind\u00e9pendance\u00bb. En ces termes, nous disons que les \u00e9v\u00e9nements <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">A<\/span><\/span> et <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">B<\/span><\/span> sont 2-ind\u00e9pendants si<\/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-ind\u00e9pendance<\/h4>\n<p style=\"text-align: justify; color: #000000;\">De mani\u00e8re analogue, la 3-ind\u00e9pendance entre trois \u00e9v\u00e9nements <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> et <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">C<\/span><\/span> est d\u00e9finie par la relation<\/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;\">Il est important de noter que la 2-ind\u00e9pendance 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> et <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">C<\/span><\/span> n&#8217;implique pas n\u00e9cessairement la 3-ind\u00e9pendance, bien que l&#8217;inverse soit vrai.<\/p>\n<h4>La n-ind\u00e9pendance entre les essais de Bernoulli<\/h4>\n<p style=\"text-align: justify; color: #000000;\">De mani\u00e8re analogue aux d\u00e9finitions pr\u00e9c\u00e9dentes, la n-ind\u00e9pendance entre une collection d&#8217;\u00e9v\u00e9nements <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">A_1, \\cdots, A_n<\/span><\/span> est d\u00e9finie par la relation<\/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;\">Et de mani\u00e8re analogue, nous avons 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>-ind\u00e9pendance n&#8217;implique pas n\u00e9cessairement <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n<\/span><\/span>-ind\u00e9pendance<\/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>-ind\u00e9pendance <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>-ind\u00e9pendance<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: justify; color: #000000;\">Les <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n<\/span><\/span> r\u00e9p\u00e9titions r\u00e9alis\u00e9es dans l&#8217;essai de Bernoulli sont <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n<\/span><\/span>-ind\u00e9pendantes.<\/p>\n<p><a name=\"3\"><\/a><\/br><\/br><\/p>\n<h2>L&#8217;essai de Bernoulli et le th\u00e9or\u00e8me binomial<\/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;\">Consid\u00e9rons une exp\u00e9rience de succ\u00e8s et d&#8217;\u00e9chec<\/span><\/strong><\/a> avec une probabilit\u00e9 de succ\u00e8s de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">p<\/span><\/span> ; \u00e0 chaque tentative, il y aura, en cons\u00e9quence, une probabilit\u00e9 <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">1-p<\/span><\/span> d&#8217;\u00e9chec. Il est clair que la probabilit\u00e9 qu&#8217;un succ\u00e8s ou un \u00e9chec se produise \u00e0 chaque tentative est de 1 ; et comme toutes les tentatives sont ind\u00e9pendantes, la probabilit\u00e9 qu&#8217;un succ\u00e8s ou un \u00e9chec se produise dans les <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n<\/span><\/span> tentatives sera <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">1^n.<\/span><\/span> \u00c0 partir de cela, nous aurons :<\/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;\">Dans la derni\u00e8re \u00e9galit\u00e9, le <a href=\"https:\/\/toposuranos.com\/ejercicios-de-induccion-matematica-teorema-del-binomio-de-newton\/\" rel=\"noopener\" target=\"_blank\">Th\u00e9or\u00e8me binomial de Newton<\/a> a \u00e9t\u00e9 appliqu\u00e9, et les termes dans la somme peuvent \u00eatre interpr\u00e9t\u00e9s de la mani\u00e8re suivante :<\/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> : le nombre de fa\u00e7ons dont <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k<\/span><\/span> succ\u00e8s peuvent se produire en <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n<\/span><\/span> tentatives<\/li>\n<li><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">p^k<\/span><\/span> : La probabilit\u00e9 que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k<\/span><\/span> succ\u00e8s ind\u00e9pendants se produisent<\/li>\n<li><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(1-p)^{n-k}<\/span><\/span> : La probabilit\u00e9 que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n-k<\/span><\/span> \u00e9checs ind\u00e9pendants se produisent<\/li>\n<\/ul>\n<p style=\"text-align: justify; color: #000000;\">En combinant ces \u00e9l\u00e9ments de la mani\u00e8re dont ils apparaissent dans la somme, nous obtenons : la probabilit\u00e9 d&#8217;obtenir <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k<\/span><\/span> succ\u00e8s parmi <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n<\/span><\/span> tentatives ; ou \u00e9quivalemment, la probabilit\u00e9 d&#8217;obtenir <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n-k<\/span><\/span> \u00e9checs parmi <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n<\/span><\/span> tentatives.<\/p>\n<p style=\"text-align: justify; color: #000000;\">Si nous s\u00e9parons chaque terme de la somme, nous obtenons les probabilit\u00e9s d&#8217;obtenir :<\/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 succ\u00e8s parmi n tentatives<\/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 succ\u00e8s parmi n tentatives<\/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 succ\u00e8s parmi n tentatives<\/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 succ\u00e8s parmi n tentatives<\/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> succ\u00e8s parmi <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n<\/span><\/span> tentatives<\/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> succ\u00e8s parmi <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n<\/span><\/span> tentatives<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: justify; color: #000000;\">Et la somme de tous ceux-ci, comme nous l&#8217;avons d\u00e9j\u00e0 vu, est \u00ab1\u00bb. Montrant que toutes les possibilit\u00e9s ont \u00e9t\u00e9 couvertes.<\/p>\n<p style=\"text-align: justify; color: #000000;\">\u00c0 partir de cela, la <strong>probabilit\u00e9 de l&#8217;\u00e9v\u00e9nement de Bernoulli<\/strong> est d\u00e9finie :<\/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 nous disons aussi que <strong>le nombre de succ\u00e8s <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">X<\/span><\/span> suit une distribution binomiale :<\/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>La distribution binomiale (ou de Bernoulli) et les distributions de probabilit\u00e9<\/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;\">\u00c0 travers la distribution binomiale<\/span><\/strong><\/a> nous commen\u00e7ons \u00e0 avoir les premi\u00e8res notions des distributions de probabilit\u00e9 et de variable al\u00e9atoire. Dans ce cas, la variable al\u00e9atoire (discr\u00e8te) est associ\u00e9e au nombre de succ\u00e8s et sa distribution de probabilit\u00e9 est donn\u00e9e par les termes du th\u00e9or\u00e8me binomial<\/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>Exercices :<\/h2>\n<ol style=\"text-align: justify; color: #000000;\">\n<li>On lance un d\u00e9 \u00e9quilibr\u00e9 \u00e0 6 faces 5 fois. Calculez la probabilit\u00e9 d&#8217;obtenir 3 fois un nombre pair comme r\u00e9sultat.<\/li>\n<li>On lance une pi\u00e8ce de monnaie 10 fois. Calculez la probabilit\u00e9 d&#8217;obtenir de 0 \u00e0 10 faces et faites un graphique montrant la probabilit\u00e9 pour chaque r\u00e9sultat. \u00c0 quoi ressemblera le graphique si on augmente le nombre de lancers et qu&#8217;on examine la probabilit\u00e9 d&#8217;obtenir un nombre de faces allant de 0 \u00e0 ce nombre de lancers ? Un tableur Excel peut \u00eatre utile ici.<\/li>\n<li>On a une loterie avec une quantit\u00e9 <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">s<\/span><\/span> de boules, o\u00f9 <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r<\/span><\/span> sont dor\u00e9es et le reste blanches. On m\u00e9lange toutes les boules et on en tire une au hasard, et on gagne lorsqu&#8217;on tire la boule dor\u00e9e. Si cet exp\u00e9rience est r\u00e9p\u00e9t\u00e9e de mani\u00e8re identique <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">20<\/span><\/span> fois, estimez le nombre de victoires les plus probables pour chaque valeur possible de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0\\leq r\\leq s.<\/span><\/span> Un tableur Excel peut \u00e9galement \u00eatre utile ici.<\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"<p>L&#8217;Essai de Bernoulli et la Distribution Binomiale R\u00e9sum\u00e9Dans cette classe, nous \u00e9tudierons le concept des essais de Bernoulli et leurs implications dans la th\u00e9orie des probabilit\u00e9s. Nous commen\u00e7ons par une d\u00e9finition d\u00e9taill\u00e9e des essais de Bernoulli pour ensuite aborder le concept d&#8217;ind\u00e9pendance entre les \u00e9v\u00e9nements. Apr\u00e8s avoir clarifi\u00e9 ces id\u00e9es, nous appliquons le th\u00e9or\u00e8me binomial [&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":3,"footnotes":""},"categories":[569,682],"tags":[],"class_list":["post-26764","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-mathematiques","category-probabilites-et-statistiques"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.7 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>L&#039;Essai de Bernoulli et la Distribution Binomiale - toposuranos.com\/material<\/title>\n<meta name=\"description\" content=\"D\u00e9couvrez comment la r\u00e9p\u00e9tition d&#039;un essai de Bernoulli produit des r\u00e9sultats avec une Distribution Binomiale de Probabilit\u00e9s.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"http:\/\/toposuranos.com\/material\/fr\/lessai-de-bernoulli-et-la-distribution-binomiale\/\" \/>\n<meta property=\"og:locale\" content=\"es_ES\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"L&#039;Essai de Bernoulli et la Distribution Binomiale\" \/>\n<meta property=\"og:description\" content=\"D\u00e9couvrez comment la r\u00e9p\u00e9tition d&#039;un essai de Bernoulli produit des r\u00e9sultats avec une Distribution Binomiale de Probabilit\u00e9s.\" \/>\n<meta property=\"og:url\" content=\"http:\/\/toposuranos.com\/material\/fr\/lessai-de-bernoulli-et-la-distribution-binomiale\/\" \/>\n<meta property=\"og:site_name\" content=\"toposuranos.com\/material\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/groups\/toposuranos\" \/>\n<meta property=\"article:published_time\" content=\"2021-05-12T13:00:03+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2024-05-22T01:04:05+00:00\" \/>\n<meta property=\"og:image\" content=\"http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2024\/04\/bernoulli-e1712960368446-1024x285.jpg\" \/>\n<meta name=\"author\" content=\"giorgio.reveco\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:title\" content=\"L&#039;Essai de Bernoulli et la Distribution Binomiale\" \/>\n<meta name=\"twitter:description\" content=\"D\u00e9couvrez comment la r\u00e9p\u00e9tition d&#039;un essai de Bernoulli produit des r\u00e9sultats avec une Distribution Binomiale de Probabilit\u00e9s.\" \/>\n<meta name=\"twitter:image\" content=\"http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2024\/04\/bernoulli-e1712960368446.jpg\" \/>\n<meta name=\"twitter:creator\" content=\"@topuranos\" \/>\n<meta name=\"twitter:site\" content=\"@topuranos\" \/>\n<meta name=\"twitter:label1\" content=\"Escrito por\" \/>\n\t<meta name=\"twitter:data1\" content=\"giorgio.reveco\" \/>\n\t<meta name=\"twitter:label2\" content=\"Tiempo de lectura\" \/>\n\t<meta name=\"twitter:data2\" content=\"1 minuto\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"http:\/\/toposuranos.com\/material\/fr\/lessai-de-bernoulli-et-la-distribution-binomiale\/#article\",\"isPartOf\":{\"@id\":\"http:\/\/toposuranos.com\/material\/fr\/lessai-de-bernoulli-et-la-distribution-binomiale\/\"},\"author\":{\"name\":\"giorgio.reveco\",\"@id\":\"http:\/\/toposuranos.com\/material\/#\/schema\/person\/e15164361c3f9a2a02cf6c234cf7fdc1\"},\"headline\":\"L&#8217;Essai de Bernoulli et la Distribution Binomiale\",\"datePublished\":\"2021-05-12T13:00:03+00:00\",\"dateModified\":\"2024-05-22T01:04:05+00:00\",\"mainEntityOfPage\":{\"@id\":\"http:\/\/toposuranos.com\/material\/fr\/lessai-de-bernoulli-et-la-distribution-binomiale\/\"},\"wordCount\":1465,\"commentCount\":0,\"publisher\":{\"@id\":\"http:\/\/toposuranos.com\/material\/#organization\"},\"image\":{\"@id\":\"http:\/\/toposuranos.com\/material\/fr\/lessai-de-bernoulli-et-la-distribution-binomiale\/#primaryimage\"},\"thumbnailUrl\":\"http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2024\/04\/bernoulli-e1712960368446.jpg\",\"articleSection\":[\"Math\u00e9matiques\",\"Probabilit\u00e9s et Statistiques\"],\"inLanguage\":\"es\",\"potentialAction\":[{\"@type\":\"CommentAction\",\"name\":\"Comment\",\"target\":[\"http:\/\/toposuranos.com\/material\/fr\/lessai-de-bernoulli-et-la-distribution-binomiale\/#respond\"]}]},{\"@type\":\"WebPage\",\"@id\":\"http:\/\/toposuranos.com\/material\/fr\/lessai-de-bernoulli-et-la-distribution-binomiale\/\",\"url\":\"http:\/\/toposuranos.com\/material\/fr\/lessai-de-bernoulli-et-la-distribution-binomiale\/\",\"name\":\"L'Essai de Bernoulli et la Distribution Binomiale - 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