{"id":26785,"date":"2021-05-27T13:00:49","date_gmt":"2021-05-27T13:00:49","guid":{"rendered":"http:\/\/toposuranos.com\/material\/?p=26785"},"modified":"2024-05-22T02:12:57","modified_gmt":"2024-05-22T02:12:57","slug":"processus-de-poisson-approche-du-processus-binomial","status":"publish","type":"post","link":"http:\/\/toposuranos.com\/material\/fr\/processus-de-poisson-approche-du-processus-binomial\/","title":{"rendered":"Processus de Poisson : Approche du Processus Binomial"},"content":{"rendered":"<div style=\"background-color:#F3F3F3; padding:20px;\">\n<center><\/p>\n<h1>Processus de Poisson : Approche du Processus Binomial<\/h1>\n<p><\/p>\n<p style=\"text-align:center;\"><strong>R\u00e9sum\u00e9<\/strong><br \/><em>Ce cours se concentre sur le processus de Poisson comme une approximation du processus binomial, en commen\u00e7ant par la d\u00e9finition des coefficients et de la distribution de Poisson, qui d\u00e9rive d&#8217;un \u00e9v\u00e9nement de Bernoulli avec un grand nombre de tentatives et une probabilit\u00e9 individuelle tr\u00e8s faible. La partie centrale de ce cours aborde les processus approximatifs de Poisson, tant spatiaux que temporels, en utilisant des exemples de particules minuscules dans un liquide et l&#8217;\u00e9mission de particules par une substance radioactive, respectivement. Enfin, il se termine par des exemples pratiques de l&#8217;application de la distribution de Poisson dans diff\u00e9rents contextes, tels que le service \u00e0 la client\u00e8le dans un supermarch\u00e9 et la densit\u00e9 de population dans une localit\u00e9.<\/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 ce cours, l&#8217;\u00e9tudiant sera capable de :\n<\/p>\n<ol>\n<li><strong>Comprendre<\/strong> la d\u00e9finition et les coefficients de la distribution de Poisson.<\/li>\n<li><strong>Comprendre<\/strong> le processus de Poisson comme une approximation du processus binomial.<\/li>\n<li><strong>Comprendre<\/strong> l&#8217;\u00e9quivalence formelle entre les processus spatiaux et temporels de Poisson.<\/li>\n<li><strong>Utiliser<\/strong> la distribution de Poisson pour r\u00e9soudre des probl\u00e8mes pratiques.<\/li>\n<\/ol>\n<p><center><br \/>\n<strong><u>TABLE DES MATI\u00c8RES<\/u>:<\/strong><br \/>\n<a href=\"#1\">Les Coefficients et la Distribution de Poisson<\/a><br \/>\n<a href=\"#2\">Processus approximatifs de Poisson<\/a><br \/>\n<a href=\"#3\">Processus Spatial de Poisson<\/a><br \/>\n<a href=\"#4\">Processus Temporel de Poisson<\/a><br \/>\n<a href=\"#5\">Temporel et Spatial<\/a><br \/>\n<a href=\"#6\">Exemples pratiques d&#8217;utilisation de la distribution de Poisson<\/a><br \/>\n<\/center><br \/>\n<center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/mQ0j3FE8p2U\" 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>Les Coefficients et la Distribution de Poisson<\/h2>\n<p style=\"text-align: justify; color: #000000;\"><a href=\"https:\/\/www.youtube.com\/watch?v=mQ0j3FE8p2U&amp;t=154s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">Consid\u00e9rons maintenant une approximation<\/span><\/strong><\/a> de la <a href=\"https:\/\/toposuranos.com\/el-ensayo-de-bernoulli-para-n-intentos-independientes\/\" rel=\"noopener\" target=\"_blank\">distribution binomiale<\/a>, o\u00f9 l&#8217;on consid\u00e8re un nombre de tentatives <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n<\/span><\/span> tr\u00e8s grand et toutes avec une probabilit\u00e9 individuelle <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">p<\/span><\/span> tr\u00e8s faible. En faisant cela, nous passons du processus binomial typique \u00e0 un processus de Poisson. Pour visualiser cela, imaginons une suite de la forme <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\{Bi(n;k;p_n)\\}_n,<\/span><\/span> o\u00f9 <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n\\to\\infty<\/span><\/span> et <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">p_n<\/span><\/span> satisfait la relation <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">np_n=\\lambda \\gt 0<\/span><\/span>. \u00c0 partir de cela, nous verrons que<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\\lim_{n\\to\\infty}P\\left(Bi(n;k;P_n) \\right) = \\frac{\\lambda^k}{k!}e^{-\\lambda}<\/span><\/span><\/p>\n<p style=\"text-align: justify; color: #000000;\">Cela n&#8217;est en fait pas difficile \u00e0 d\u00e9montrer, si nous prenons la probabilit\u00e9 d&#8217;un \u00e9v\u00e9nement de Bernoulli <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">Bi(n;k;p_n)<\/span><\/span> et la multiplions et divisons par <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n^k<\/span><\/span>, nous obtenons le raisonnement suivant :<\/p>\n<table style=\"text-align: justify; color: #000000;\">\n<tbody>\n<tr>\n<td width=\"80px\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">P(B(n;k;p_n))<\/span><\/span><\/td>\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 style=\"color: #f00000;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\cdot \\displaystyle \\frac{n^k}{n^k}<\/span><\/span><\/span><\/td>\n<\/tr>\n<tr>\n<td width=\"80px\"><\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle=\\frac{n!}{(n-k)!k!}p^k(1-p)^{n-k} \\cdot \\frac{n^k}{n^k}<\/span><\/span><\/td>\n<\/tr>\n<tr>\n<td width=\"80px\"><\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle=\\frac{n(n-1)\\cdots[n-(k-1)]}{n^k} \\cdot \\frac{(np_n)^k}{k!} (1-p_n)^{-k}(1-p_n)^n<\/span><\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: justify; color: #000000;\">Ainsi, si nous calculons la limite lorsque <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n\\to\\infty<\/span><\/span>, nous aurons :<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><br \/>\n<span class=\"katex-eq\" data-katex-display=\"false\">\n\\begin{array}\n\n\\displaystyle \\lim_{n\\to\\infty} {{n}\\choose{k}}p_n^k(1-p_n)^{n-k} &amp;= \\lim_{n\\to\\infty} \\underbrace{\\frac{n(n-1)\\cdots[n-(k-1)]}{n^k}}_{\\to 1} \\cdot \\frac{\\overbrace{(np_n)^k}^{\\to\\lambda^k}}{k!} \\overbrace{(1-p_n)^{-k}}^{\\to 1} {(1-p_n)^n} \\\\ \\\\\n\n&amp;\\displaystyle = \\frac{\\lambda^k}{k!} \\lim_{n\\to\\infty}\\left(1 - \\frac{\\lambda}{n} \\right)^n \\\\ \\\\\n\n&amp; \\displaystyle = \\frac{\\lambda^k}{k!}e^{-\\lambda}\n\n\\end{array}\n\n<\/span>\n<\/span><\/p>\n<p style=\"text-align: justify; color: #000000;\">\u00c0 partir de cela, les coefficients de Poisson sont d\u00e9finis, <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">Po(k;\\lambda)<\/span><\/span>, par<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle Po(k;\\lambda) := \\lim_{n\\to\\infty} {{n}\\choose{k}}p^k(1-p_n)^{n-k} = \\frac{\\lambda^k}{k!}e^{-\\lambda} <\/span><\/span><\/p>\n<p style=\"text-align: justify; color: #000000;\">Et on dit qu&#8217;une variable al\u00e9atoire <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">X<\/span><\/span> suit une distribution de Poisson, <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">X\\sim Po(k,\\lambda),<\/span><\/span> si :<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\"> P(X=k) = Po(k;\\lambda) <\/span><\/span><\/p>\n<p><a name=\"2\"><\/a><\/br><\/br><\/p>\n<h2>Processus approximatifs de Poisson<\/h2>\n<p><a name=\"3\"><\/a><\/br><\/br><\/p>\n<h3>Processus Spatial de Poisson<\/h3>\n<p style=\"text-align: justify; color: #000000;\"><a href=\"https:\/\/www.youtube.com\/watch?v=mQ0j3FE8p2U&amp;t=665s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">Supposons que nous avons un r\u00e9cipient de volume<\/span><\/strong><\/a> <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">V<\/span><\/span> contenant un liquide o\u00f9 se trouvent <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n<\/span><\/span> particules minuscules uniform\u00e9ment r\u00e9parties. Ici, nous supposons que le liquide est bien m\u00e9lang\u00e9 et que les particules n&#8217;interagissent pas entre elles, ne s&#8217;attirent ni ne se repoussent. Ces hypoth\u00e8ses peuvent \u00eatre formalis\u00e9es par les affirmations suivantes :<\/p>\n<ul style=\"text-align: justify; color: #000000;\">\n<li><strong>Hypoth\u00e8se d&#8217;Homog\u00e9n\u00e9it\u00e9 Spatiale :<\/strong> La probabilit\u00e9 de trouver une particule dans une r\u00e9gion <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">D<\/span><\/span> du liquide d\u00e9pend uniquement du volume de cette r\u00e9gion.<\/li>\n<li><strong>Non-Interaction :<\/strong> Les \u00e9v\u00e9nements \u00abla j-\u00e8me particule est dans la r\u00e9gion D\u00bb, avec j=1,2,&#8230;,n sont tous n-ind\u00e9pendants.<\/li>\n<li><strong>Non-Superposition :<\/strong> Deux particules ne peuvent pas occuper le m\u00eame espace.<\/li>\n<\/ul>\n<p style=\"text-align: justify; color: #000000;\">Si nous consid\u00e9rons une r\u00e9gion <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">D<\/span><\/span> de volume <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">v<\/span><\/span>, la probabilit\u00e9 de l&#8217;\u00e9v\u00e9nement \u00abil y a <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k<\/span><\/span> particules dans <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">D<\/span>\u00ab<\/span> d\u00e9pend exclusivement de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">v<\/span><\/span> ; appelons cet \u00e9v\u00e9nement <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">g_k(v)<\/span><\/span>. Soit <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">h(v)<\/span><\/span> la probabilit\u00e9 qu&#8217;une particule soit \u00e0 l&#8217;int\u00e9rieur d&#8217;une r\u00e9gion de volume <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">v<\/span><\/span>. Si <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">D_1<\/span><\/span> et <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">D_2<\/span><\/span> sont deux r\u00e9gions disjointes de volumes <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">v_1<\/span><\/span> et <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">v_2<\/span><\/span> respectivement, alors si <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">D=D_1\\cup D_2,<\/span><\/span> ayant un volume <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">v,<\/span><\/span> alors <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">v=v_1+v_2.<\/span><\/span> Et comme <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">D_1<\/span><\/span> et <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">D_2<\/span><\/span> sont disjointes (<span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">D_1\\cap D_2 = \\emptyset <\/span><\/span>), il en r\u00e9sultera que<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\"> h(v) = h(v_1) + h(v_2) <\/span><\/span><\/p>\n<p style=\"text-align: justify; color: #000000;\">Si <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">V<\/span><\/span> est le volume total du liquide, alors :<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\"> h(V) = 1 <\/span><\/span><\/p>\n<p style=\"text-align: justify; color: #000000;\">Et par cons\u00e9quent :<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\"> h(v) =\\displaystyle \\frac{v}{V} <\/span><\/span><\/p>\n<p style=\"text-align: justify; color: #000000;\">De l\u00e0, nous avons que l&#8217;\u00e9v\u00e9nement <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">g_k(v)<\/span><\/span> est en r\u00e9alit\u00e9 un \u00e9v\u00e9nement de type Bernoulli avec <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">p=v\/V<\/span><\/span> et est donn\u00e9 par :<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\"> g_k(v) =B(n;k;p=v\/V) <\/span><\/span><\/p>\n<p style=\"text-align: justify; color: #000000;\">Cependant, la plupart des situations pratiques de ce type impliquent un grand nombre de particules <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n<\/span><\/span> et les r\u00e9gions consid\u00e9r\u00e9es tendent \u00e0 \u00eatre petites par rapport \u00e0 la taille du syst\u00e8me, de sorte que les conditions sont remplies pour appliquer l&#8217;approximation de Poisson et nous avons :<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle P(g_k(v)) = \\lim_{\\begin{matrix}n\\to\\infty\\\\ v\/V=c \\end{matrix}}P(B(n;k;p=v\/V)) =\\displaystyle \\frac{(cv)^k}{k!}e^{-cv}<\/span><\/span><\/p>\n<p><a name=\"4\"><\/a><\/br><\/br><\/p>\n<h3>Processus Temporel de Poisson<\/h3>\n<p style=\"text-align: justify; color: #000000;\"><a href=\"https:\/\/www.youtube.com\/watch?v=mQ0j3FE8p2U&amp;t=944s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">Supposons que nous enregistrons le nombre<\/span><\/strong><\/a> de particules \u00e9mises par une substance radioactive \u00e0 partir du moment t=0 et, \u00e0 partir de cela, nous calculons la probabilit\u00e9 que dans l&#8217;intervalle [0,t[ soient \u00e9mises exactement k particules sous les hypoth\u00e8ses suivantes :<\/p>\n<ul style=\"text-align: justify; color: #000000;\">\n<li><strong>Invariance :<\/strong> Les conditions de l&#8217;exp\u00e9rience ne changent pas dans le temps.<\/li>\n<li><strong>Absence de M\u00e9moire :<\/strong> Ce qui s&#8217;est pass\u00e9 dans [0,t[ n&#8217;affecte pas ce qui se passe dans [t,t'[.<\/li>\n<li><strong>\u00c9v\u00e9nements Isol\u00e9s :<\/strong> Les particules sont \u00e9mises une \u00e0 une.<\/li>\n<\/ul>\n<p style=\"text-align: justify; color: #000000;\">Si nous comparons les hypoth\u00e8ses du processus temporel \u00e0 celles du processus spatial, nous remarquerons qu&#8217;elles sont formellement \u00e9quivalentes. Tout comme la probabilit\u00e9 de trouver une particule dans une r\u00e9gion ne d\u00e9pend pas de l&#8217;endroit o\u00f9 la r\u00e9gion est choisie, mais seulement de sa taille, la probabilit\u00e9 d&#8217;observer l&#8217;\u00e9mission d&#8217;une particule ne d\u00e9pend pas du moment choisi pour mesurer, mais seulement de l&#8217;intervalle d&#8217;observation. L&#8217;absence de m\u00e9moire est analogue \u00e0 la non-interaction des processus spatiaux : ce qui s&#8217;est pass\u00e9 \u00e0 un autre moment n&#8217;affecte pas ce qui se passe aux autres instants. Enfin, les \u00e9v\u00e9nements isol\u00e9s impliquent qu&#8217;\u00e0 un instant donn\u00e9, une seule particule peut \u00eatre \u00e9mise, de la m\u00eame mani\u00e8re qu&#8217;un endroit dans l&#8217;espace ne peut \u00eatre occup\u00e9 que par un seul corps \u00e0 la fois.<\/p>\n<p style=\"text-align: justify; color: #000000;\">Ainsi, si nous d\u00e9finissons l&#8217;\u00e9v\u00e9nement \u00abk particules sont \u00e9mises dans un intervalle de temps <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">t<\/span><\/span>\u00ab, sa probabilit\u00e9 d&#8217;occurrence sera un \u00e9v\u00e9nement de la forme <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">g_k(t)<\/span><\/span>, c&#8217;est-\u00e0-dire :<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">P(g_k(t)) =\\displaystyle \\frac{(ct)^k}{k!} e^{-ct}<\/span><\/span><\/p>\n<p><a name=\"5\"><\/a><\/br><\/br><\/p>\n<h3>Temporel et Spatial<\/h3>\n<p style=\"text-align: justify; color: #000000;\"><a href=\"https:\/\/www.youtube.com\/watch?v=mQ0j3FE8p2U&amp;t=1102s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">Les deux processus, spatial et temporel, sont formellement \u00e9quivalents.<\/span><\/strong><\/a> Ils ne diff\u00e8rent que par la mani\u00e8re dont ils sont interpr\u00e9t\u00e9s \u00e0 des fins pratiques. Une mani\u00e8re rapide de clarifier cette distinction est d&#8217;observer le r\u00f4le de la constante \u00abc\u00bb qui appara\u00eet dans les deux cas. Pour que la fonction exponentielle soit bien d\u00e9finie, il est n\u00e9cessaire que son argument soit adimensionnel ; cependant, celle-ci contient des unit\u00e9s de temps ou d&#8217;espace selon qu&#8217;il s&#8217;agit de processus temporels ou spatiaux. Ce probl\u00e8me est r\u00e9solu pr\u00e9cis\u00e9ment par la constante c. Nous avons :<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">Po(k;\\lambda)=\\displaystyle \\frac{\\lambda^k}{k!}e^{-\\lambda}=\\left\\{\\begin{matrix} {En prenant\\,\\lambda = \\rho v } &amp; \\longmapsto &amp;\\displaystyle \\frac{(\\rho v)^k}{k!}e^{-\\rho v} &amp; {Processus\\,Spatial} \\\\ {En prenant\\,\\lambda = \\nu t } &amp; \\longmappe &amp;\\displaystyle \\frac{(\\nu t)^k}{k!}e^{-\\nu t} &amp; {Processus\\,Temporel} \\end{matrix} \\right.<\/span><\/span><\/p>\n<ul style=\"text-align: justify; color: #000000;\">\n<li>Si <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">c=\\rho<\/span><\/span>, il s&#8217;agit d&#8217;une densit\u00e9 spatiale (nombre de choses par unit\u00e9 d&#8217;espace), d\u00e9finissant ainsi un processus spatial de Poisson.<\/li>\n<li>Si <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">c=\\nu<\/span><\/span>, il s&#8217;agit d&#8217;une densit\u00e9 temporelle (ou fr\u00e9quence, nombre d&#8217;occurrences par unit\u00e9 de temps), d\u00e9finissant ainsi un processus temporel de Poisson.<\/li>\n<\/ul>\n<p><center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/8qvHRoEckSc\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center><br \/>\n<a name=\"6\"><\/a><\/br><\/br><\/p>\n<h2>Exemples pratiques d&#8217;utilisation de la distribution de Poisson<\/h2>\n<ol style=\"text-align: justify; color: #000000;\">\n<li>La caisse d&#8217;un supermarch\u00e9 sert en moyenne 2 clients toutes les 9 minutes. \u00c9laborer un tableau montrant les probabilit\u00e9s qu&#8217;elle serve entre 1, 2, 3, et ainsi de suite, jusqu&#8217;\u00e0 5 personnes dans un intervalle de temps de 5 minutes.<\/li>\n<li>Une clinique v\u00e9t\u00e9rinaire a la capacit\u00e9 de recevoir au maximum 12 clients par jour. Si elle re\u00e7oit en moyenne 9 clients par jour, quelle est la probabilit\u00e9 qu&#8217;un jour donn\u00e9, elle d\u00e9passe la capacit\u00e9 d&#8217;accueil de la clinique ?<\/li>\n<li>Une localit\u00e9 a une densit\u00e9 de population de 10 personnes pour 1000 m\u00e8tres carr\u00e9s. Quelle est la probabilit\u00e9 que dans un site de 60 m\u00e8tres carr\u00e9s, nous trouvions moins de 15 personnes ?<\/li>\n<li>Une poule veut traverser la rue. En marchant en ligne droite, cela lui prend 58 secondes. Si la rue a un trafic de 3 v\u00e9hicules par minute, et si un v\u00e9hicule passe pendant que la poule essaie de traverser, elle sera s\u00fbrement \u00e9cras\u00e9e avec des r\u00e9sultats mortels. Quelle est la probabilit\u00e9 que la poule atteigne l&#8217;autre c\u00f4t\u00e9 vivante ?<\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"<p>Processus de Poisson : Approche du Processus Binomial R\u00e9sum\u00e9Ce cours se concentre sur le processus de Poisson comme une approximation du processus binomial, en commen\u00e7ant par la d\u00e9finition des coefficients et de la distribution de Poisson, qui d\u00e9rive d&#8217;un \u00e9v\u00e9nement de Bernoulli avec un grand nombre de tentatives et une probabilit\u00e9 individuelle tr\u00e8s faible. La [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":26422,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"iawp_total_views":1,"footnotes":""},"categories":[569,682],"tags":[],"class_list":["post-26785","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>Processus de Poisson : Approche du Processus Binomial - toposuranos.com\/material<\/title>\n<meta name=\"description\" content=\"Comprendre que la distribution de Poisson est un r\u00e9sultat obtenu \u00e0 partir des processus binomiaux port\u00e9s \u00e0 leur limite\" \/>\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\/processus-de-poisson-approche-du-processus-binomial\/\" \/>\n<meta property=\"og:locale\" content=\"es_ES\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Processus de Poisson : Approche du Processus Binomial\" \/>\n<meta property=\"og:description\" content=\"Comprendre que la distribution de Poisson est un r\u00e9sultat obtenu \u00e0 partir des processus binomiaux port\u00e9s \u00e0 leur limite\" \/>\n<meta property=\"og:url\" content=\"http:\/\/toposuranos.com\/material\/fr\/processus-de-poisson-approche-du-processus-binomial\/\" \/>\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=\"2021-05-27T13:00:49+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2024-05-22T02:12:57+00:00\" \/>\n<meta property=\"og:image\" content=\"http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2024\/04\/poisson-e1712962118751-1024x285.jpg\" \/>\n<meta name=\"author\" content=\"giorgio.reveco\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:title\" content=\"Processus de Poisson : Approche du Processus Binomial\" \/>\n<meta name=\"twitter:description\" content=\"Comprendre que la distribution de Poisson est un r\u00e9sultat obtenu \u00e0 partir des processus binomiaux port\u00e9s \u00e0 leur limite\" \/>\n<meta name=\"twitter:image\" content=\"http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2024\/04\/poisson-e1712962118751.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\/processus-de-poisson-approche-du-processus-binomial\/#article\",\"isPartOf\":{\"@id\":\"http:\/\/toposuranos.com\/material\/fr\/processus-de-poisson-approche-du-processus-binomial\/\"},\"author\":{\"name\":\"giorgio.reveco\",\"@id\":\"http:\/\/toposuranos.com\/material\/#\/schema\/person\/e15164361c3f9a2a02cf6c234cf7fdc1\"},\"headline\":\"Processus de Poisson : Approche du Processus Binomial\",\"datePublished\":\"2021-05-27T13:00:49+00:00\",\"dateModified\":\"2024-05-22T02:12:57+00:00\",\"mainEntityOfPage\":{\"@id\":\"http:\/\/toposuranos.com\/material\/fr\/processus-de-poisson-approche-du-processus-binomial\/\"},\"wordCount\":1751,\"commentCount\":0,\"publisher\":{\"@id\":\"http:\/\/toposuranos.com\/material\/#organization\"},\"image\":{\"@id\":\"http:\/\/toposuranos.com\/material\/fr\/processus-de-poisson-approche-du-processus-binomial\/#primaryimage\"},\"thumbnailUrl\":\"http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2024\/04\/poisson-e1712962118751.jpg\",\"articleSection\":[\"Math\u00e9matiques\",\"Probabilit\u00e9s et Statistiques\"],\"inLanguage\":\"es\",\"potentialAction\":[{\"@type\":\"CommentAction\",\"name\":\"Comment\",\"target\":[\"http:\/\/toposuranos.com\/material\/fr\/processus-de-poisson-approche-du-processus-binomial\/#respond\"]}]},{\"@type\":\"WebPage\",\"@id\":\"http:\/\/toposuranos.com\/material\/fr\/processus-de-poisson-approche-du-processus-binomial\/\",\"url\":\"http:\/\/toposuranos.com\/material\/fr\/processus-de-poisson-approche-du-processus-binomial\/\",\"name\":\"Processus de Poisson : Approche du Processus Binomial - 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