{"id":26674,"date":"2021-03-29T13:00:04","date_gmt":"2021-03-29T13:00:04","guid":{"rendered":"http:\/\/toposuranos.com\/material\/?p=26674"},"modified":"2024-05-21T10:24:21","modified_gmt":"2024-05-21T10:24:21","slug":"theoremes-utiles-pour-le-calcul-des-probabilites","status":"publish","type":"post","link":"http:\/\/toposuranos.com\/material\/fr\/theoremes-utiles-pour-le-calcul-des-probabilites\/","title":{"rendered":"Th\u00e9or\u00e8mes Utiles pour le Calcul des Probabilit\u00e9s"},"content":{"rendered":"<div style=\"background-color:#F3F3F3; padding:20px;\">\n<center><\/p>\n<h1>Th\u00e9or\u00e8mes Utiles pour le Calcul des Probabilit\u00e9s<\/h1>\n<p><\/p>\n<p style=\"text-align:center;\"><strong>R\u00e9sum\u00e9<\/strong><br \/><em>Dans ce cours, nous pr\u00e9sentons des exercices r\u00e9solus qui d\u00e9montrent certains th\u00e9or\u00e8mes utiles pour le calcul des probabilit\u00e9s, y compris des d\u00e9monstrations et des d\u00e9ductions. Les exercices couvrent des sujets tels que la probabilit\u00e9 compl\u00e9mentaire, l&#8217;inclusion des ensembles et la convergence des \u00e9v\u00e9nements. Compl\u00e9ter ces exercices vous fournira une base solide pour approfondir l&#8217;\u00e9tude de la th\u00e9orie des probabilit\u00e9s.<\/em><\/p>\n<p><\/center><br \/>\n<\/p>\n<p style=\"text-align:center;\"><strong>OBJECTIFS D&#8217;APPRENTISSAGE :<\/strong><br \/>\nEn terminant ce cours, l&#8217;\u00e9tudiant sera capable de :\n<\/p>\n<ol>\n<li><strong>D\u00e9montrer<\/strong> les propri\u00e9t\u00e9s de base des probabilit\u00e9s<\/li>\n<\/ol>\n<p><center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/SSkPFP5FpKM\" 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 style=\"text-align: justify; color: #000000;\">Ce qui suit est un guide d&#8217;exercices (r\u00e9solus) o\u00f9 l&#8217;objectif est de d\u00e9montrer certains th\u00e9or\u00e8mes utiles pour la th\u00e9orie des probabilit\u00e9s. Essayez de les r\u00e9soudre, puis comparez vos r\u00e9sultats &gt;:D<\/p>\n<ol style=\"text-align: justify; color: #000000;\">\n<li><strong><a href=\"https:\/\/www.youtube.com\/watch?v=SSkPFP5FpKM&amp;t=111s\" target=\"_blank\" rel=\"noopener\"><span style=\"color: #ff0000;\">D\u00e9montrez que <span class=\"katex-eq\" data-katex-display=\"false\">P(A^c) = 1 -P(A)<\/span> et,<\/span><\/a> \u00e0 partir de cela, faites une d\u00e9duction qui permet d&#8217;argumenter que <span class=\"katex-eq\" data-katex-display=\"false\">P(\\emptyset) = 0<\/span><\/strong><br \/>\n<span class=\"collapseomatic \" id=\"id6a58b77818618\"  tabindex=\"0\" title=\"MONTRER LA SOLUTION\"    >MONTRER LA SOLUTION<\/span><div id=\"target-id6a58b77818618\" class=\"collapseomatic_content \">\n<p style=\"text-align: justify; color: #000000;\">D&#8217;apr\u00e8s la d\u00e9finition de <a href=\"https:\/\/toposuranos.com\/el-espacio-de-probabilidades-medida-de-probabilidad\/\" target=\"_blank\" rel=\"noopener\">mesure de probabilit\u00e9<\/a>, si <span class=\"katex-eq\" data-katex-display=\"false\">A<\/span> et <span class=\"katex-eq\" data-katex-display=\"false\">B<\/span> sont des \u00e9v\u00e9nements mesurables quelconques, alors il sera vrai que :<\/p>\n<table>\n<tbody>\n<tr>\n<td>[a]<\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">0\\leq P(A) \\leq 1<\/span><\/td>\n<\/tr>\n<tr>\n<td>[b]<\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">A\\cap B = \\emptyset \\rightarrow P(A\\cup B) = P(A) + P(B)<\/span><\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">P(\\Omega) = 1<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: justify; color: #000000;\">Maintenant, comme <span class=\"katex-eq\" data-katex-display=\"false\">A\\cap A^c = \\emptyset<\/span>, d&#8217;apr\u00e8s la partie [b], il sera vrai que :<\/p>\n<table>\n<tbody>\n<tr>\n<td>[d]<\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">A\\cap A^c = \\emptyset \\rightarrow P(A\\cup A^c) = P(A) + P(A^c)<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: justify; color: #000000;\">Comme <span class=\"katex-eq\" data-katex-display=\"false\">A\\cap A^c = \\emptyset<\/span> est toujours vrai et <span class=\"katex-eq\" data-katex-display=\"false\">A\\cup A^c = \\Omega<\/span>, alors il sera vrai que :<\/p>\n<p style=\"text-align: center; color: #000000;\"><span class=\"katex-eq\" data-katex-display=\"false\">1=P(\\Omega) = P(A\\cup A^c) = P(A) + P(A^c)<\/span>\n<p style=\"text-align: justify; color: #000000;\">Et, par cons\u00e9quent :<\/p>\n<p style=\"text-align: center; color: #000000;\"><span class=\"katex-eq\" data-katex-display=\"false\">P(A^c) = 1-P(A)<\/span>\n<p style=\"text-align: justify; color: #000000;\">C&#8217;est ce que nous voulions d\u00e9montrer.<\/p>\n<p style=\"text-align: justify; color: #000000;\">Pour d\u00e9montrer que <span class=\"katex-eq\" data-katex-display=\"false\">P(\\emptyset)=0<\/span>, il suffit de prendre <span class=\"katex-eq\" data-katex-display=\"false\">A=\\Omega<\/span> dans la relation que nous venons de prouver et il sera vrai que :<\/p>\n<p style=\"text-align: center; color: #000000;\"><span class=\"katex-eq\" data-katex-display=\"false\">P(\\emptyset) = P(\\Omega^c) =1 - P(\\Omega) = 1-1 = 0<\/span>\n<\/div><\/li>\n<li><strong><a href=\"https:\/\/www.youtube.com\/watch?v=SSkPFP5FpKM&amp;t=534s\" target=\"_blank\" rel=\"noopener\"><span style=\"color: #ff0000;\">a) D\u00e9montrez que <span class=\"katex-eq\" data-katex-display=\"false\">A\\subseteq B \\rightarrow P(B\\setminus A) = P(B) - P(A)<\/span><\/span><\/a>, L&#8217;\u00e9galit\u00e9 est-elle vraie en g\u00e9n\u00e9ral ?<\/strong><strong>b) D\u00e9montrez que <span class=\"katex-eq\" data-katex-display=\"false\">A\\subseteq B \\rightarrow P(B)\\leq P(A)<\/span><\/strong><span class=\"collapseomatic \" id=\"id6a58b778188d4\"  tabindex=\"0\" title=\"MONTRER LA SOLUTION a)\"    >MONTRER LA SOLUTION a)<\/span><div id=\"target-id6a58b778188d4\" class=\"collapseomatic_content \">\n<table>\n<tbody>\n<tr>\n<td>(1)<\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">A\\subseteq B<\/span> ; Pr\u00e9misse<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">\\equiv A\\cap B = A<\/span><\/td>\n<\/tr>\n<tr>\n<td>(2)<\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">B\\setminus A = B\\cap A^c<\/span> ; D\u00e9finition de la Th\u00e9orie des Ensembles<\/td>\n<\/tr>\n<tr>\n<td>(3)<\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">B= (B\\cap A) \\cup (B\\cap A^c)<\/span> ; Propri\u00e9t\u00e9 des ensembles<\/td>\n<\/tr>\n<tr>\n<td>(4)<\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">(B\\cap A)\\cap (B\\cap A^c)=\\emptyset<\/span> ; Propri\u00e9t\u00e9 des ensembles<\/td>\n<\/tr>\n<tr>\n<td>(5)<\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">P(B)= P[(B\\cap A) \\cup (B\\cap A^c)]<\/span> ; De (3)<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">P(B)= P (B\\cap A) + P(B\\cap A^c)<\/span> ; De (4) + D\u00e9f., Mesure de Probabilit\u00e9<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">P(B)= P (B\\cap A) + P(B\\setminus A)<\/span> ; De (2)<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\"> P(B\\setminus A) =P(B) - P (B\\cap A) <\/span><\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">{P(B\\setminus A) =P(B) - P (A) }<\/span> ; De (1)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: center; color: #000000;\">Par cons\u00e9quent <span class=\"katex-eq\" data-katex-display=\"false\">{\\{A\\subseteq B\\}\\vdash P(B\\setminus A) = P(B) - P(A)}.<\/span>\n<p style=\"text-align: justify; color: #000000;\">Notons que cette \u00e9galit\u00e9 n&#8217;est pas vraie en g\u00e9n\u00e9ral, car elle d\u00e9pend du fait que <span class=\"katex-eq\" data-katex-display=\"false\">A\\subseteq B<\/span> soit v\u00e9rifi\u00e9. De ces d\u00e9veloppements, on peut voir que, si cela n&#8217;est pas v\u00e9rifi\u00e9, alors il sera vrai que <span class=\"katex-eq\" data-katex-display=\"false\">P(B\\setminus A) = P(B) - P(A\\cap B).<\/span>\n<\/div>\n<span class=\"collapseomatic \" id=\"id6a58b77818aaf\"  tabindex=\"0\" title=\"MONTRER LA SOLUTION b)\"    >MONTRER LA SOLUTION b)<\/span><div id=\"target-id6a58b77818aaf\" class=\"collapseomatic_content \">\n<p style=\"text-align: justify; color: #000000;\">\u00c0 partir de ce qui a \u00e9t\u00e9 raisonn\u00e9 en a), on a :<\/p>\n<table>\n<tbody>\n<tr>\n<td><span class=\"katex-eq\" data-katex-display=\"false\"> \\{A\\subseteq B\\}\\vdash P(B\\setminus A) = P(B) - P(A)<\/span><\/td>\n<\/tr>\n<tr>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">\\equiv \\{A\\subseteq B\\}\\vdash P(A) + P(B\\setminus A) = P(B)<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: justify; color: #000000;\">Enfin, comme <span class=\"katex-eq\" data-katex-display=\"false\">P<\/span> est une mesure de probabilit\u00e9, on a que <span class=\"katex-eq\" data-katex-display=\"false\">\\forall X (P(X)\\geq 0),<\/span> donc par cons\u00e9quent :<\/p>\n<p style=\"text-align: center; color: #000000;\"><span class=\"katex-eq\" data-katex-display=\"false\">{\\{A\\subseteq B\\}\\vdash P(A) \\leq P(B)}.<\/span>\n<\/div>\n<p>&nbsp;<\/li>\n<li><strong><a href=\"https:\/\/www.youtube.com\/watch?v=SSkPFP5FpKM&amp;t=892s\" target=\"_blank\" rel=\"noopener\"><span style=\"color: #ff0000;\">a) D\u00e9montrez que <span class=\"katex-eq\" data-katex-display=\"false\">P(A\\cup B) = P(A) + P(B) - P(A\\cup B).<\/span><\/span><\/a> Accompagnez la d\u00e9monstration avec un diagramme.<\/strong><strong>b) En utilisant le r\u00e9sultat pr\u00e9c\u00e9dent, d\u00e9montrez que <span class=\"katex-eq\" data-katex-display=\"false\">P(A\\cup B) \\leq P(A) + P(B)<\/span><\/strong><span class=\"collapseomatic \" id=\"id6a58b77818b9b\"  tabindex=\"0\" title=\"MONTRER LA SOLUTION PARTIE a)\"    >MONTRER LA SOLUTION PARTIE a)<\/span><div id=\"target-id6a58b77818b9b\" class=\"collapseomatic_content \">\n<table>\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\">A\\cup B = (A\\triangle B) \\cup (A\\cap B)<\/span> ; Propri\u00e9t\u00e9 des ensembles<\/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\">A\\triangle B := (A\\setminus B) \\cup (B\\setminus A)<\/span> ; D\u00e9finition de la diff\u00e9rence sym\u00e9trique<\/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\">(A\\triangle B)\\cap ( A \\cap B) = \\emptyset<\/span> ; propri\u00e9t\u00e9 des ensembles<\/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\">(A\\setminus B)\\cap (B\\setminus A) = \\emptyset<\/span> ; propri\u00e9t\u00e9 des ensembles<\/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\\cup B) = P[(A\\triangle B) \\cup (A\\cap B)]<\/span> ; De (1)<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">P(A\\cup B) = P(A\\triangle B) + P(A\\cap B)]<\/span> ; De (3)<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">P(A\\cup B) = P(A\\setminus B) + P(B\\setminus A) + P(A\\cap B)]<\/span> ; De (2,4)<\/td>\n<\/tr>\n<tr>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">(6)<\/span><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">P(B\\setminus A) = P(B) - P(A\\cap B)<\/span> ; Application du r\u00e9sultat de l&#8217;exercice 2<\/td>\n<\/tr>\n<tr>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">(7)<\/span><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">P(A\\setminus B) = P(A) - P(A\\cap B)<\/span> ; La m\u00eame chose que (6)<\/td>\n<\/tr>\n<tr>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">(8)<\/span><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">P(A\\cup B) = P(A) - P(A\\cap B) + P(B) - P(A\\cap B) + P(A\\cap B)<\/span> ; de(5,6,7)<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">{P(A\\cup B) = P(A) + P(B) - P(A\\cap B)}. <\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<span class=\"collapseomatic \" id=\"id6a58b77818cf8\"  tabindex=\"0\" title=\"MONTRER LA SOLUTION PARTIE b)\"    >MONTRER LA SOLUTION PARTIE b)<\/span><div id=\"target-id6a58b77818cf8\" class=\"collapseomatic_content \">\n<table>\n<tbody>\n<tr>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">\\vdash P(A\\cup B) = P(A) + P(B) - P(A\\cap B)<\/span> ; r\u00e9sultat de la partie a)<\/td>\n<\/tr>\n<tr>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">\\equiv\\; \\vdash P(A\\cup B) + P(A\\cap B) = P(A) + P(B) <\/span><\/td>\n<\/tr>\n<tr>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">\\equiv\\; \\vdash {P(A\\cup B) \\leq P(A) + P(B)} <\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<p>&nbsp;<\/li>\n<li><strong><a href=\"https:\/\/www.youtube.com\/watch?v=SSkPFP5FpKM&amp;t=1625s\" target=\"_blank\" rel=\"noopener\"><span style=\"color: #ff0000;\">Si <span class=\"katex-eq\" data-katex-display=\"false\">\\{E_n\\}<\/span> est une famille infinie d&#8217;\u00e9v\u00e9nements<\/span> <\/a>tels que <span class=\"katex-eq\" data-katex-display=\"false\">E_1\\supseteq E_2 \\supseteq E_2 \\supseteq \\cdots \\supseteq E_n \\supseteq E_{n+1}\\supseteq \\cdots.<\/span> D\u00e9montrez que :<\/strong>\n<p style=\"text-align: center; color: #000000;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle P\\left( \\bigcap_{n=1}^\\infty E_n \\right) = \\lim_{n\\to \\infty}P(E_n)<\/span>\n<span class=\"collapseomatic \" id=\"id6a58b77818ddc\"  tabindex=\"0\" title=\"MONTRER LA SOLUTION\"    >MONTRER LA SOLUTION<\/span><div id=\"target-id6a58b77818ddc\" class=\"collapseomatic_content \">\n<table>\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\">E_n \\supseteq E_{n+1}<\/span> ; Hypoth\u00e8se<\/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\">E_n^c \\subseteq E_{n+1}^c<\/span> ; Par compl\u00e9mentation depuis (1)<\/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\">\\displaystyle(E_n^c \\subseteq E_{n+1}^c) \\rightarrow P\\left( \\bigcup_{n=1}^\\infty E_n^c \\right)= \\lim_{n\\to\\infty}P(E_n^c)<\/span> ; Propri\u00e9t\u00e9 de Continuit\u00e9<\/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\">\\displaystyle \\bigcup_{n=1}^\\infty E_n^c = \\left( \\bigcap_{n=1}^\\infty E_n \\right)^c<\/span> ; Lois de DeMorgan<\/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\\left(E_n^c\\right) = 1 - P(E_n)<\/span> ; D\u00e9montr\u00e9 dans l&#8217;exercice 1<\/td>\n<\/tr>\n<tr>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">(6)<\/span><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle P\\left(\\left[ \\bigcap_{n=1}^\\infty E_n\\right]^c \\right) = \\lim_{n\\to\\infty}[1-P(E_n)] = 1 - \\lim_{n\\to\\infty}P(E_n)<\/span> ; de (2,4,5) appliqu\u00e9 sur (3)<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle 1 - P\\left( \\bigcap_{n=1}^\\infty E_n \\right) = 1 - \\lim_{n\\to\\infty}P(E_n) <\/span><\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">{\\displaystyle P\\left( \\bigcap_{n=1}^\\infty E_n \\right) = \\lim_{n\\to\\infty}P(E_n)} <\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: center; color: #000000;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\\therefore\\{E_n \\supseteq E_{n+1}\\}\\vdash P\\left( \\bigcap_{n=1}^\\infty E_n \\right) = \\lim_{n\\to\\infty}P(E_n).<\/span>\n<p>&nbsp;<\/li>\n<\/ol>\n<\/div>\n<p style=\"text-align: justify; color: #000000;\">En r\u00e9solvant ces exercices, vous compl\u00e9terez un premier pilier qui vous servira de support pour poursuivre l&#8217;\u00e9tude de la th\u00e9orie des probabilit\u00e9s.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Th\u00e9or\u00e8mes Utiles pour le Calcul des Probabilit\u00e9s R\u00e9sum\u00e9Dans ce cours, nous pr\u00e9sentons des exercices r\u00e9solus qui d\u00e9montrent certains th\u00e9or\u00e8mes utiles pour le calcul des probabilit\u00e9s, y compris des d\u00e9monstrations et des d\u00e9ductions. Les exercices couvrent des sujets tels que la probabilit\u00e9 compl\u00e9mentaire, l&#8217;inclusion des ensembles et la convergence des \u00e9v\u00e9nements. Compl\u00e9ter ces exercices vous fournira [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":26394,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"iawp_total_views":15,"footnotes":""},"categories":[569,682],"tags":[],"class_list":["post-26674","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 v27.4 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>Th\u00e9or\u00e8mes Utiles pour le Calcul des Probabilit\u00e9s - toposuranos.com\/material<\/title>\n<meta name=\"description\" content=\"Explorez des th\u00e9or\u00e8mes cl\u00e9s de la probabilit\u00e9 pr\u00e9sent\u00e9s avec des exercices r\u00e9solus. 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