{"id":26717,"date":"2021-04-19T13:00:45","date_gmt":"2021-04-19T13:00:45","guid":{"rendered":"http:\/\/toposuranos.com\/material\/?p=26717"},"modified":"2024-05-21T21:20:13","modified_gmt":"2024-05-21T21:20:13","slug":"probabilite-conditionnelle-et-independance-entre-evenements","status":"publish","type":"post","link":"http:\/\/toposuranos.com\/material\/fr\/probabilite-conditionnelle-et-independance-entre-evenements\/","title":{"rendered":"Probabilit\u00e9 Conditionnelle et Ind\u00e9pendance entre \u00c9v\u00e9nements"},"content":{"rendered":"<div style=\"background-color:#F3F3F3; padding:20px;\">\n<center><\/p>\n<h1>Probabilit\u00e9 Conditionnelle et Ind\u00e9pendance entre \u00c9v\u00e9nements<\/h1>\n<p><\/p>\n<p style=\"text-align:center;\"><strong>R\u00e9sum\u00e9<\/strong><br \/><em>Dans cette session, nous explorerons le concept de probabilit\u00e9 conditionnelle et l&#8217;interaction entre les \u00e9v\u00e9nements. Nous allons acqu\u00e9rir les comp\u00e9tences n\u00e9cessaires pour calculer des probabilit\u00e9s conditionnelles et d\u00e9terminer la d\u00e9pendance ou l&#8217;ind\u00e9pendance entre les \u00e9v\u00e9nements. Nous appliquerons des exemples pratiques, comme l&#8217;\u00e9tude de la pr\u00e9valence des caries chez les consommateurs de sucreries, pour illustrer ces concepts. \u00c0 la fin, vous aurez une compr\u00e9hension claire de la mani\u00e8re d&#8217;appliquer la probabilit\u00e9 conditionnelle et d&#8217;analyser les \u00e9v\u00e9nements d\u00e9pendants et ind\u00e9pendants.<\/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, vous serez capable de :\n<\/p>\n<ol>\n<li><strong>Comprendre<\/strong> la d\u00e9finition de la probabilit\u00e9 conditionnelle et sa relation avec l&#8217;intersection des \u00e9v\u00e9nements et les probabilit\u00e9s individuelles.<\/li>\n<li><strong>Identifier<\/strong> les associations positives et n\u00e9gatives entre les \u00e9v\u00e9nements \u00e0 partir de la comparaison des probabilit\u00e9s conditionnelles.<\/li>\n<li><strong>Tester<\/strong> l&#8217;ind\u00e9pendance entre diff\u00e9rents \u00e9v\u00e9nements.<\/li>\n<\/ol>\n<p style=\"text-align:center;\"><strong><u>INDEX DES CONTENUS<\/u><\/strong><br \/>\n<a href=\"#1\">LA PROBABILIT\u00c9 CONDITIONNELLE<\/a><br \/>\n<a href=\"#2\">D\u00c9FINITION FORMELLE DE LA PROBABILIT\u00c9 CONDITIONNELLE<\/a><br \/>\n<a href=\"#3\">RELATION ENTRE \u00c9V\u00c9NEMENTS<\/a><br \/>\n<a href=\"#4\">IND\u00c9PENDANCE ENTRE \u00c9V\u00c9NEMENTS ET COMPL\u00c9MENTS D&#8217;\u00c9V\u00c9NEMENTS<\/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>La Probabilit\u00e9 Conditionnelle<\/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;\">Quelle est la probabilit\u00e9 qu&#8217;un \u00e9v\u00e9nement A<\/span><\/strong><\/a> se produise \u00e9tant donn\u00e9 que B s&#8217;est d\u00e9j\u00e0 produit ? Le calcul de ce type de probabilit\u00e9s implique le concept de <strong>probabilit\u00e9 conditionnelle.<\/strong> Dans ce qui suit, nous \u00e9tudierons la probabilit\u00e9 conditionnelle, sa d\u00e9finition et comment, \u00e0 partir de cela, on peut inf\u00e9rer les relations de d\u00e9pendance et d&#8217;ind\u00e9pendance entre les \u00e9v\u00e9nements.<\/p>\n<p style=\"text-align: justify; color: #000000;\">Supposons que nous voulions mesurer la pr\u00e9valence des caries chez les consommateurs r\u00e9guliers de sucreries. Si nous examinons un espace d&#8217;\u00e9chantillonnage constitu\u00e9 de <span class=\"katex-eq\" data-katex-display=\"false\">N<\/span> personnes <span class=\"katex-eq\" data-katex-display=\"false\">\\Omega_N,<\/span> nous verrons qu&#8217;il peut \u00eatre divis\u00e9 en 4 sous-ensembles :<\/p>\n<ul style=\"text-align: justify; color: #000000;\">\n<li><span class=\"katex-eq\" data-katex-display=\"false\">A:=\\left\\{  {Personnes\\;qui\\;ont\\;des\\;caries}\\right\\}<\/span><\/li>\n<li><span class=\"katex-eq\" data-katex-display=\"false\">A^c:=\\left\\{{Personnes\\;qui\\;n&#039;ont\\;PAS\\;de\\;caries}\\right\\}<\/span><\/li>\n<li><span class=\"katex-eq\" data-katex-display=\"false\">B:=\\left\\{  {Personnes\\;qui\\;mangent\\;des\\;sucreries\\;r\u00e9guli\u00e8rement}\\right\\}<\/span><\/li>\n<li><span class=\"katex-eq\" data-katex-display=\"false\">B^c:=\\left\\{{Personnes\\;qui\\;ne\\;mangent\\;PAS\\;de\\;sucreries\\;r\u00e9guli\u00e8rement}\\right\\}<\/span><\/li>\n<\/ul>\n<p style=\"text-align: justify; color: #000000;\">\u00c0 partir de cela, il est clair que <span class=\"katex-eq\" data-katex-display=\"false\">A\\cup A^c = \\Omega_N<\/span> et <span class=\"katex-eq\" data-katex-display=\"false\">B\\cup B^c = \\Omega_N,<\/span> mais <span class=\"katex-eq\" data-katex-display=\"false\">A\\cap B<\/span> n&#8217;est pas n\u00e9cessairement vide. Le sc\u00e9nario g\u00e9n\u00e9ral est repr\u00e9sent\u00e9 par la figure suivante :<\/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=\"Probabilit\u00e9 Conditionnelle\" 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=\"Probabilit\u00e9 Conditionnelle\" 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;\">Ainsi, si nous nous rappelons la d\u00e9finition de la probabilit\u00e9<\/span><\/strong><\/a> comme la limite des fr\u00e9quences relatives, nous pourrons dire que la probabilit\u00e9 qu&#8217;une personne ait des caries \u00e9tant donn\u00e9 qu&#8217;il est confirm\u00e9 qu&#8217;elle consomme des sucreries, <span class=\"katex-eq\" data-katex-display=\"false\">P(A|B)<\/span> sera :<\/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;\">D&#8217;autre part, il est vrai 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;\">Ainsi, si nous rempla\u00e7ons ces deux derni\u00e8res expressions dans <span class=\"katex-eq\" data-katex-display=\"false\">P(A|B)<\/span>, nous aurons :<\/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;\">Ceci constitue la d\u00e9finition suivante :<\/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>D\u00e9finition formelle de la probabilit\u00e9 conditionnelle<\/h2>\n<p><strong>D\u00c9FINITION :<\/strong><\/span> La probabilit\u00e9 de <span class=\"katex-eq\" data-katex-display=\"false\">A<\/span>, \u00e9tant donn\u00e9 que <span class=\"katex-eq\" data-katex-display=\"false\">B<\/span> s&#8217;est produit, <span class=\"katex-eq\" data-katex-display=\"false\">P(A|B),<\/span> est d\u00e9finie par la relation<\/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;\">Dans la pens\u00e9e courante, il y a souvent<\/span><\/strong><\/a> une confusion entre <span class=\"katex-eq\" data-katex-display=\"false\">P(A|B)<\/span> et <span class=\"katex-eq\" data-katex-display=\"false\">P(B|A).<\/span> Pour clarifier cette diff\u00e9rence, examinons un exemple bas\u00e9 sur un cas extr\u00eame : Notons que si la totalit\u00e9 des footballeurs ont deux jambes, seule une infime partie des personnes ayant deux jambes sont des footballeurs.<\/p>\n<p><a name=\"3\"><\/a><br \/>\n<\/br><\/br><\/p>\n<h2>Relation entre \u00e9v\u00e9nements<\/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;\">Continuant avec l&#8217;exemple de la pr\u00e9valence<\/span><\/strong><\/a> des caries chez les gens qui consomment r\u00e9guli\u00e8rement des sucreries. Si la consommation de sucreries rend les personnes plus susceptibles d&#8217;avoir des caries, alors il devrait en r\u00e9sulter 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>Ici, nous avons que <span class=\"katex-eq\" data-katex-display=\"false\">B<\/span> renforce <span class=\"katex-eq\" data-katex-display=\"false\">A<\/span> et nous disons donc qu&#8217;il y a une <strong>association positive<\/strong> entre les \u00e9v\u00e9nements.<\/p>\n<p style=\"text-align: justify; color: #000000;\">En revanche, si la consommation de sucreries pr\u00e9vient les caries, alors il devrait en r\u00e9sulter que :<\/p>\n<p style=\"text-align: justify; color: #000000;\"><span class=\"katex-eq\" data-katex-display=\"false\">P(A|B) \\lt P(A).<\/span>Dans ce cas, <span class=\"katex-eq\" data-katex-display=\"false\">B<\/span> inhibe <span class=\"katex-eq\" data-katex-display=\"false\">A<\/span> et nous disons donc qu&#8217;il y a une <strong>association n\u00e9gative<\/strong> entre les \u00e9v\u00e9nements.<\/p>\n<p style=\"text-align: justify; color: #000000;\">Et s&#8217;il n&#8217;y avait aucune relation entre ces deux \u00e9v\u00e9nements, ni positive ni n\u00e9gative, alors il devrait en r\u00e9sulter que :<\/p>\n<p style=\"text-align: justify; color: #000000;\"><span class=\"katex-eq\" data-katex-display=\"false\">P(A|B) = P(A).<\/span>De l\u00e0, on fait l&#8217;inf\u00e9rence qui est pr\u00e9sent\u00e9e dans la plupart des textes de probabilit\u00e9s comme une d\u00e9finition :<\/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;\">Ce raisonnement montre la relation entre la probabilit\u00e9 conditionnelle et l&#8217;ind\u00e9pendance des \u00e9v\u00e9nements.<\/p>\n<p style=\"text-align: justify; color: #000000;\"><span style=\"color: #800000;\"><strong>D\u00c9FINITION :<\/strong><\/span>\u00c9tant donn\u00e9 deux \u00e9v\u00e9nements <span class=\"katex-eq\" data-katex-display=\"false\">A<\/span> et <span class=\"katex-eq\" data-katex-display=\"false\">B<\/span>, ils sont dits <strong>ind\u00e9pendants<\/strong> s&#8217;ils satisfont la relation suivante :<\/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>Ind\u00e9pendance entre \u00e9v\u00e9nements et compl\u00e9ments d&#8217;\u00e9v\u00e9nements<\/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;\">L&#8217;ind\u00e9pendance entre deux \u00e9v\u00e9nements<\/span><\/strong><\/a> <span class=\"katex-eq\" data-katex-display=\"false\">A<\/span> et <span class=\"katex-eq\" data-katex-display=\"false\">B<\/span> est prouv\u00e9e \u00e9quivalente avec l&#8217;ind\u00e9pendance de <span class=\"katex-eq\" data-katex-display=\"false\">A<\/span> avec <span class=\"katex-eq\" data-katex-display=\"false\">B^c,<\/span> celle de <span class=\"katex-eq\" data-katex-display=\"false\">A^c<\/span> avec <span class=\"katex-eq\" data-katex-display=\"false\">B,<\/span> et celle de <span class=\"katex-eq\" data-katex-display=\"false\">A^c<\/span> avec <span class=\"katex-eq\" data-katex-display=\"false\">B^c.<\/span>\n<p style=\"text-align: justify; color: #000000;\"><span style=\"color: #000080;\"><strong>D\u00c9MONSTRATION<\/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>; Pr\u00e9somption<\/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>; parce que <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\">Voir le d\u00e9veloppement de l&#8217;exercice 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) et (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) et (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>; En factorisant par <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\">Voir le d\u00e9veloppement de l&#8217;exercice 1<\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: justify; color: #000000;\">Cette derni\u00e8re expression se lit comme suit : \u00abDu fait que <span class=\"katex-eq\" data-katex-display=\"false\">A<\/span> et <span class=\"katex-eq\" data-katex-display=\"false\">B<\/span> sont ind\u00e9pendants, on en d\u00e9duit que <span class=\"katex-eq\" data-katex-display=\"false\">A<\/span> et <span class=\"katex-eq\" data-katex-display=\"false\">B^c<\/span> le sont \u00e9galement.<\/p>\n<p style=\"text-align: justify; color: #000000;\">La d\u00e9monstration dans le sens inverse se fait de mani\u00e8re similaire.<\/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>; Pr\u00e9somption<\/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\">Voir le d\u00e9veloppement de l&#8217;exercice 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>; En r\u00e9alisant le produit de la parenth\u00e8se sur le c\u00f4t\u00e9 droit.<\/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>; Parce que <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\">Voir le d\u00e9veloppement de l&#8217;exercice 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) et (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>; En \u00e9liminant les termes similaires<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: justify; color: #000000;\">Et cette expression se lit comme suit : \u00abDu fait que <span class=\"katex-eq\" data-katex-display=\"false\">A<\/span> et <span class=\"katex-eq\" data-katex-display=\"false\">B^c<\/span> sont ind\u00e9pendants, on en d\u00e9duit que <span class=\"katex-eq\" data-katex-display=\"false\">A<\/span> et <span class=\"katex-eq\" data-katex-display=\"false\">B<\/span> le sont \u00e9galement.<\/p>\n<p style=\"text-align: justify; color: #000000;\">Enfin, \u00e0 partir de ces deux raisonnements, on a prouv\u00e9 l&#8217;\u00e9quivalence entre l&#8217;ind\u00e9pendance de <span class=\"katex-eq\" data-katex-display=\"false\">A<\/span> avec <span class=\"katex-eq\" data-katex-display=\"false\">B<\/span> et celle de <span class=\"katex-eq\" data-katex-display=\"false\">A<\/span> avec <span class=\"katex-eq\" data-katex-display=\"false\">B^c.<\/span>\n<p style=\"text-align: justify; color: #000000;\">Les autres \u00e9quivalences prouv\u00e9es peuvent \u00eatre obtenues de mani\u00e8re similaire. Cela reste comme un d\u00e9fi pour le lecteur &gt;:D<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Probabilit\u00e9 Conditionnelle et Ind\u00e9pendance entre \u00c9v\u00e9nements R\u00e9sum\u00e9Dans cette session, nous explorerons le concept de probabilit\u00e9 conditionnelle et l&#8217;interaction entre les \u00e9v\u00e9nements. Nous allons acqu\u00e9rir les comp\u00e9tences n\u00e9cessaires pour calculer des probabilit\u00e9s conditionnelles et d\u00e9terminer la d\u00e9pendance ou l&#8217;ind\u00e9pendance entre les \u00e9v\u00e9nements. Nous appliquerons des exemples pratiques, comme l&#8217;\u00e9tude de la pr\u00e9valence des caries chez les [&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":1,"footnotes":""},"categories":[569,682],"tags":[],"class_list":["post-26717","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>Probabilit\u00e9 Conditionnelle et Ind\u00e9pendance entre \u00c9v\u00e9nements - toposuranos.com\/material<\/title>\n<meta name=\"description\" content=\"Voulez-vous savoir ce qu&#039;est la probabilit\u00e9 conditionnelle ? 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