{"id":78574,"date":"2020-11-25T20:12:33","date_gmt":"2020-11-25T19:12:33","guid":{"rendered":"https:\/\/groupe-reussite.fr\/?page_id=78574"},"modified":"2022-07-12T15:31:02","modified_gmt":"2022-07-12T14:31:02","slug":"cours-en-ligne-exercices-corriges-variables-aleatoires-maths-sup","status":"publish","type":"page","link":"https:\/\/groupe-reussite.fr\/ressources\/cours-en-ligne-exercices-corriges-variables-aleatoires-maths-sup\/","title":{"rendered":"Exercices corrig\u00e9s variables al\u00e9atoires"},"content":{"rendered":"
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\n\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
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Cours en ligne Maths en Maths Sup<\/a><\/p>\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t

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Chapitres Maths en MPSI, PCSI, MP2I, PTSI<\/p>\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t

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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tR\u00e9currence<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tSommes et produits<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tNombres complexes<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tTrigonom\u00e9trie<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tNombres r\u00e9els<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tEnsembles et applications<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tFonctions (intro)<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tFonctions usuelles<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tPrimitives<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\t\u00c9quations diff\u00e9rentielles<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tSuites num\u00e9riques<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tLimites et continuit\u00e9<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tD\u00e9riv\u00e9es<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tSyst\u00e8mes<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tPolyn\u00f4mes<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tFractions rationnelles<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tArithm\u00e9tique et polyn\u00f4mes<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tArithm\u00e9tique et fractions rationnelles<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tAnalyse asymptotique<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tD\u00e9veloppements limit\u00e9s<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tD\u00e9nombrements<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tEspaces vectoriels<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tMatrices<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tInt\u00e9grations<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tEspaces pr\u00e9hilbertiens<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tEspaces euclidiens<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tS\u00e9ries num\u00e9riques<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tProbabilit\u00e9s<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tD\u00e9terminants<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tVariables al\u00e9atoires<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tCONTACTEZ-NOUS<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
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Cours en ligne Maths en Maths Sup<\/a><\/p>\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t

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Chapitres Maths en MPSI, PCSI, MP2I, PTSI<\/p>\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t

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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tR\u00e9currence<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tSommes et produits<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tNombres complexes<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tTrigonom\u00e9trie<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tNombres r\u00e9els<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tEnsembles et applications<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tFonctions (intro)<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tFonctions usuelles<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tPrimitives<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\t\u00c9quations diff\u00e9rentielles<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tSuites num\u00e9riques<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tLimites et continuit\u00e9<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tD\u00e9riv\u00e9es<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tSyst\u00e8mes<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tPolyn\u00f4mes<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tFractions rationnelles<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tArithm\u00e9tique et polyn\u00f4mes<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tArithm\u00e9tique et fractions rationnelles<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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Exercices corrig\u00e9s sur les variables al\u00e9atoires en Maths Sup<\/h1>\n<\/div>\n

R\u00e9sum\u00e9 de cours<\/a> Exercices et corrig\u00e9s<\/a><\/p>\n

Cours en ligne de Maths en Maths Sup<\/a><\/p>\n

\n

Exercice sur la fonction g\u00e9n\u00e9ratrice en Maths Sup<\/h2>\n

Si \"X\" est une variable al\u00e9atoire<\/strong> \u00e0 valeurs dans \"[\![0,\, n]\!]\", on d\u00e9finit pour tout \"t \in \mathbb{R}, \, G(t) = \displaystyle \sum _ {k = 0} ^n \mathbb{P}( X = k) \, t ^k\".<\/p>\n

\"G\" est appel\u00e9e fonction<\/a> g\u00e9n\u00e9ratrice de la variable al\u00e9atoire<\/strong> \"X\".<\/p>\n

Question 1 :\u00a0 <\/strong><\/p>\n

\"\textrm{E}(X) = G'(1)\".<\/p>\n

Vrai ou Faux ?<\/p>\n

Question 2 :\u00a0<\/strong><\/p>\n

Calculer \"G\" lorsque \"X\" suit une loi uniforme sur \"[\![1,\, n]\!]\" puis une loi \"\mathcal{B}(n ,\, p)\" .<\/p>\n

Exercice sur l’utilisation de la fonction g\u00e9n\u00e9ratrice en Maths Sup<\/h2>\n

Soit \"n \in \mathbb{N}^*\", on effectue \"n\" tirages sans remise de jetons num\u00e9rot\u00e9s de 1 \u00e0 \"n\".<\/p>\n

On note \"X_n\" la variable al\u00e9atoire<\/strong> \u00e9gale au nombre de fois o\u00f9 l’on obtient un num\u00e9ro strictement sup\u00e9rieur aux num\u00e9ros d\u00e9j\u00e0 obtenus.<\/p>\n

Par exemple si \"n = 4\" et si l’on obtient \"1,3,4,2\" (resp \"3,2,4,1\"), \"X_4\" est \u00e9gal \u00e0 3 (resp. 2).
\n(le premier num\u00e9ro tir\u00e9 donne toujours un maximum).<\/p>\n

Question 1 :<\/strong><\/p>\n

Montrer que si \"n \geqslant 2\" et \"j \in [\![2,\, n]\!]\",<\/p>\n

\"\mathbb{P}(X_n = j) = \displaystyle \frac 1 n \,\mathbb{P}(X _ {n - 1} = j - 1 )\" \"\qquad \qquad \qquad \displaystyle + \,\frac {n - 1} n \, \mathbb{P}(X _ {n - 1} = j )\".<\/p>\n

Question 2 :<\/strong><\/p>\n

Si \"n \in \mathbb{N} ^*\" et \"t \in \mathbb {R}\", on note \"\qquad \quad G_n(t) = \displaystyle \sum _ {k = 1} ^n \mathbb{P}( X_n = k) \, t ^k\".
\n\"\forall\, n \geqslant 2\", \"\forall\, t \in \mathbb{R}\", \"\qquad G_n(t) = \displaystyle \frac {t + n - 1} n \; G_{n - 1} (t)\" .<\/p>\n

Vrai ou Faux ?<\/p>\n

Question 3 :<\/strong><\/p>\n

En d\u00e9duire la valeur de \"G_n\,\" .<\/p>\n

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COURS DE MATHS<\/h3>\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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Les meilleurs professeurs particuliers<\/h4>\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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Pour progresser et r\u00e9ussir<\/h4>\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tCours particuliers maths<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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Avis Google France <\/b>\u2605\u2605\u2605\u2605\u2605<\/b><\/span> 4,9 sur 5<\/b><\/p>\t\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t

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Exercice sur l’ajout d’une boule de couleur oppos\u00e9e en Maths Sup<\/h2>\n

Une urne contient initialement une boule blanche et une boule noire.<\/p>\n

On effectue une succession de tirages dans cette urne. Apr\u00e8s chaque tirage, on remet la boule tir\u00e9e dans l’urne et on ajoute dans l’urne une boule de couleur oppos\u00e9e \u00e0 celle qui vient d’\u00eatre tir\u00e9e.<\/p>\n

On effectue en tout \"n\" tirages et on suppose qu’il existe un univers probabilis\u00e9 fini \"(\Omega, \, \mathbb{P})\" associ\u00e9 \u00e0 cette suite<\/a> d’\u00e9preuves.<\/p>\n

Pour tout \"k \in [\![0 , n]\!]\", on note \"X_k\" le nombre de boules blanches pr\u00e9sentes dans l’urne juste avant le \"k + 1\" -\u00e8me tirage. En particulier, on a \"X_0 = 1\".<\/p>\n

Question 1 :\u00a0<\/strong><\/p>\n

Calculer \"\mathbb{P}(X _{k + 1} = i \mid X_k = j)\"<\/p>\n

Question 2 :<\/strong><\/p>\n

Montrer que \"\forall \, k \in [\![0 , n - 1]\!]\", si \"i \in \mathbb{N}^*\", \"P(X _ {k + 1} = i ) = \displaystyle \frac i {k + 2} \,\mathbb{P}(X_k = i)\" \"\qquad \qquad \displaystyle +\, \frac {3 + k - i} {k + 2} \, \mathbb{P}(X_k = i - 1)\".<\/p>\n

Question 3 :\u00a0<\/strong><\/p>\n

\u00c0 l’aide de la formule pr\u00e9c\u00e9dente, donner la loi de \"X_3\,\".<\/p>\n

\"\mathbb{P}(X_3 = 3 ) =\" ?<\/p>\n

Exercice sur le tirage de la premi\u00e8re et la derni\u00e8re en Maths Sup<\/h2>\n

Une urne contient \"n\" boules noires et \"n\" boules blanches. On fait des tirages sans remise et on note \"X\" le rang d’apparition de la premi\u00e8re blanche et \"Y\" le rang d’apparition de la derni\u00e8re blanche.<\/p>\n

Question 1 :<\/strong><\/p>\n

En d\u00e9duire la valeur de \"\displaystyle \sum_{p= n} ^{2\, n + 1} \binom p n\".<\/p>\n

Question 2 :<\/strong><\/p>\n

\"\textrm{E}(Y) = \displaystyle \frac {n \, (a \, n + 1)} {n + 1}\".<\/p>\n

avec \"a =\" ?<\/p>\n

Question 3 :<\/strong><\/p>\n

Calculer \"\textrm{E}(Y(Y + 1) )\" .<\/p>\n

Trouver \"b\" tel que \"\qquad \quad \textrm{V}(Y) = \displaystyle \frac {n^2 (2\, n + b) } {(n + 2)(n + 1)^2 }\".<\/p>\n

\"b =\" ?<\/p>\n

Exercice sur la loi Hyperg\u00e9om\u00e9trique en Maths Sup<\/h2>\n

On consid\u00e8re une urne contenant \"N\" boules dont \"b\" blanches et \"r\" boules rouges indiscernables au toucher.
\nOn effectue \"n\" tirages sans remise et on note \"X\" le nombre de boules blanches tir\u00e9es.<\/p>\n

Question : <\/strong><\/p>\n

Calculer \"\textrm{E}(X)\".<\/p>\n

Exercice sur le maximum et minimum de 2 tirages en Maths Sup<\/h2>\n

On effectue 2 tirages avec remise dans un ensemble<\/a> de jetons num\u00e9rot\u00e9s de 1 \u00e0 \"n\".
\nOn appelle \"U\" le plus petit num\u00e9ro obtenu et \"V\" le plus grand.<\/p>\n

Question 1 :<\/strong><\/p>\n

\"n + 1 - U\" et \"V\" suivent la m\u00eame loi<\/p>\n

Vrai ou Faux ?<\/p>\n

Question 2 :<\/strong><\/p>\n

Calculer l’esp\u00e9rance et la variance de \"V\"<\/p>\n

Exercice sur deux variables de loi uniforme en Maths Sup<\/h2>\n

Soit \"n \in \mathbb{N}\", \"n \geqslant 2\".<\/p>\n

On tire un num\u00e9ro \"X\" au hasard entre 1 et \"n\" puis un num\u00e9ro \"Y\" entre 1 et \"X\".<\/p>\n

Question 1 :\u00a0\u00a0<\/strong><\/p>\n

Calculer \"\textrm{V}(Y)\".<\/p>\n

Question 2 :\u00a0<\/strong><\/p>\n

\"X - Y + 1\" a m\u00eame loi que \"Y\".<\/p>\n

Vrai ou Faux ?<\/p>\n

Correction de l’exercice sur la fonction g\u00e9n\u00e9ratrice en Maths Sup<\/h2>\n

Question 1 :\u00a0<\/strong><\/p>\n

Vrai<\/p>\n

On d\u00e9rive<\/a> la fonction polyn\u00f4me<\/a> \"G\",<\/p>\n

\"G'(t) = \displaystyle \sum _ {k = 1} ^n k\, \mathbb{P}( X = k) \, t ^{k - 1}\".<\/p>\n

\"G'(1) = \displaystyle \sum _ {k = 1} ^n k\, \mathbb{P}( X = k)\"<\/p>\n

On ajoute un terme nul, \"G'(1) \displaystyle = \sum _ {k = 0} ^n k\, \mathbb{P}( X = k) = \textrm{E}(X)\".<\/p>\n

Question 2 :\u00a0<\/strong><\/p>\n

\"\bullet\" Lorsque \"X\" suit une loi uniforme sur \"[[1,\, n]]\",<\/p>\n

\"\forall\, t \in \mathbb{R}, \, G(t) = \displaystyle \sum _{k = 1} ^n \mathbb{P}(X = k)\, t ^k\"<\/p>\n

\"G(t) =\displaystyle \sum _ {k = 1} ^n \frac 1 n \, t ^k\".<\/p>\n

\"G(1) = 1\"<\/p>\n

\"\forall\, t \neq 1\", \"G(t) = \displaystyle \frac t n \sum _ {k = 1} ^n \, t ^{k - 1} = \frac t n \sum _ {i = 0} ^{n - 1} \, t ^{i}\"<\/p>\n

\"G(t) = \displaystyle \frac {t(t ^n - 1) } {n(t - 1)}\".<\/p>\n

\"\bullet\" Si \"X \hookrightarrow \mathcal{B}(n ,\, p)\", on note \"q = 1 - p\".<\/p>\n

\"\forall\, t \in \mathbb{R}, \, G(t) = \displaystyle \sum _ {k = 0} ^n \binom n k\, p ^k \, q ^{n - k} \, t ^k\"<\/p>\n

\"G(t) =(p\, t + q) ^n\".<\/p>\n

Correction de l’exercice sur l’utilisation de la fonction g\u00e9n\u00e9ratrice<\/h2>\n

Question 1 :\u00a0<\/strong><\/p>\n

Il est important de remarquer que si l’on tire \"n\" au rang \"k\", tous les num\u00e9ros tir\u00e9s ensuite ne permettent pas d’obtenir un maximum.<\/p>\n

Les \u00e9v\u00e9nements \"(Y_n = n)\" et \"(Y_n < n)\" forment un syst\u00e8me<\/a> complet d’\u00e9v\u00e9nements de probabilit\u00e9 non nulle. Par la formule des probabilit\u00e9s totales,<\/p>\n

\"\mathbb{P}(X_n = j) =\" \"\qquad \displaystyle \,\mathbb{P}(X _ {n } = j \mid Y_n = n )\, \mathbb{P}( Y_n = n )\" \"\displaystyle\qquad + \, \mathbb{P}(X _ {n } = j \mid Y_n < n )\, \mathbb{P}( Y_n < n )\"<\/p>\n

\"\bullet\" Si \"(Y_n < n)\", le dernier tirage donne un num\u00e9ro strictement inf\u00e9rieur \u00e0 \"n\", r\u00e9aliser en tout \"j\" maximums entre le temps 1 et le temps \"n\" revient donc \u00e0
\nobtenir \"j\" maximums en tirant les \"n - 1\" premiers num\u00e9ros (tous les num\u00e9ros de 1 \u00e0 \"n\" sauf le num\u00e9ro qui sort au rang \"n\"), donc<\/p>\n

\"\mathbb{P}(X _ {n } = j \mid Y_n < n ) = \mathbb{P}(X _ {n - 1} = j )\"<\/p>\n

\"\bullet\" Si \"(Y_n = n)\", le dernier tirage donne \"n\" donc un num\u00e9ro strictement sup\u00e9rieur aux \"n - 1\" num\u00e9ros d\u00e9j\u00e0 tir\u00e9s tous entre 1 et \"n - 1\".<\/p>\n

R\u00e9aliser en tout \"j\" maximums entre le temps 1 et le temps \"n\" revient donc \u00e0\u00a0 obtenir \"j - 1\" maximums en tirant les \"n - 1\" premiers num\u00e9ros, donc<\/p>\n

\"\mathbb{P}(X _ {n } = j \mid Y_n = n ) = \mathbb{P}(X _ {n - 1} = j - 1)\"<\/p>\n

On a donc obtenu<\/p>\n

\"\mathbb{P}(X_n = j) = \displaystyle \frac 1 n \,\mathbb{P}(X _ {n - 1} = j - 1 )\" \"\qquad \qquad \qquad \qquad \displaystyle + \, \frac {n - 1} n \, \mathbb{P}(X _ {n - 1} = j )\"<\/p>\n

Question 2 :\u00a0<\/strong><\/p>\n

Vrai<\/p>\n

En sommant pour \"j \in [[1 , n]]\" les relations d\u00e9duites de Q4,<\/p>\n

\"\mathbb{P}(X_n = j) \, t ^j = \displaystyle \frac 1 n \,\mathbb{P}(X _ {n - 1} = j - 1 )\, t ^{j}\" \"\qquad \qquad \quad \displaystyle + \, \frac {n - 1} n \, \mathbb{P}(X _ {n - 1} = j ) \, t ^j\"<\/p>\n

on obtient \"G_n(t) = \displaystyle \frac 1 n \sum _ {j = 1}^{n } \,\mathbb{P}(X _ {n - 1} = j - 1 )\, t ^{j}\" \"\qquad \qquad \displaystyle +\, \frac {n - 1} n \, \sum _ {j = 1} ^n \mathbb{P}(X _ {n - 1} = j )\, t ^{j}\"<\/p>\n

On pose \"k = j - 1\" dans la premi\u00e8re somme<\/a> et on utilise \"\qquad \mathbb{P}(X_n = 0) = \mathbb{P}(X _ {n - 1} = n ) = 0\"
\npour obtenir<\/p>\n

\"G_n(t) = \displaystyle \frac 1 n \sum _ {k = 0}^{n - 1 } \,\mathbb{P}(X _ {n - 1} = k )\, t ^{k}\times t\" \"\qquad \qquad \displaystyle + \, \frac {n - 1} n \,\sum _ {j = 1} ^{n - 1} \mathbb{P}(X _ {n - 1} = j )\, t ^{j}\"<\/p>\n

soit \"G_n (t) = \displaystyle \frac 1 n \,t \, G_{n - 1}(t) + \frac {n - 1} n \, G_{n - 1}(t)\"<\/p>\n

soit \"\boxed{G_n(t) = \displaystyle \frac {t + n - 1} n \; G_{n - 1} (t)}\" .<\/p>\n

Question 3 :\u00a0<\/strong><\/p>\n

On note si \"n \in \mathbb{N}^*\", \"\qquad H_n : G_n(t) = \displaystyle \frac 1 {n!} \prod_{j = 0} ^{n - 1} (t + j)\".<\/p>\n

\"\ast\" \"G_1(t) = t\", donc \"H_1\" est vraie.<\/p>\n

\"\ast\" On suppose que \"H_n\" est vraie.<\/p>\n

Comme \"G_{n + 1} (t) = \displaystyle \frac {t + n} {n + 1} \, G_n(t)\"<\/p>\n

alors \"\displaystyle G_{n + 1} (t) = \frac 1 {n!} \prod_{j = 0} ^{n - 1} (t + j)\times \frac {t + n} {n + 1}\"<\/p>\n

\"\displaystyle G_{n + 1} (t) = \frac 1 {(n + 1) !} \prod_{j = 0} ^n (t + j)\".<\/p>\n

La propri\u00e9t\u00e9 est vraie par r\u00e9currence<\/a> sur \"n\".<\/p>\n

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COURS PARTICULIERS EN LIGNE<\/h3>\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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Nous avons s\u00e9lectionn\u00e9 pour vous les meilleurs professeurs particuliers.<\/h4>\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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POUR ACC\u00c9L\u00c9RER MA PROGRESSION<\/h4>\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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Avis Google France <\/b>\u2605\u2605\u2605\u2605\u2605<\/b><\/span> 4,9 sur 5<\/b><\/p>\t\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t

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Corrig\u00e9 sur l’ajout d’une boule de couleur oppos\u00e9e en Maths Sup<\/h2>\n

Question 1 :\u00a0<\/strong><\/p>\n

\"\mathbb{P}(X _{k + 1} = i \mid X_k = j) = 0\" si \"i < j\" ou \"i > j+ 2\" car, entre deux tirages, le nombre de blanches est stable ou augmente d’une unit\u00e9.<\/p>\n

Si \"(X_k = j)\" est r\u00e9alis\u00e9, on a \"j\" boules blanches et \"k + 2 - j\" noires,<\/p>\n

\"\mathbb{P}(X _{k + 1} = j \mid X_k = j)\" \"\qquad =\mathbb{P}(B_{k + 1} \mid X_k = j) = \displaystyle \frac {j } {k + 2}\"<\/p>\n

et\u00a0 \"\mathbb{P}(X _{k + 1} = j \mid X_k = j)\" \"\qquad = \mathbb{P}(N_{k + 1} \mid X_k = j) =\displaystyle \frac {k + 2 - j } {k + 2}\".<\/p>\n

Question 2 :\u00a0<\/strong><\/p>\n

La famille \"(X_k = j) _{1 \leqslant j \leqslant k + 1}\" est un syst\u00e8me complet d’\u00e9v\u00e9nements.<\/p>\n

Par la formule des probabilit\u00e9s<\/a> totales :
\n\"\displaystyle \mathbb{P} (X_{k + 1} = i) =\" \"\quad \displaystyle \sum_{j = 1} ^{k + 1} \mathbb{P}(X _{k + 1} = i \mid X_k = j) \, \mathbb{P}(X _{k} = j)\"<\/p>\n

seuls les termes pour \"j = i - 1\" et \"j = i\" sont \u00e0 conserver<\/p>\n

si \"i \in \mathbb{N}^*\",<\/p>\n

\"\displaystyle \mathbb{P} (X_{k + 1} = i) =\" \"\mathbb{P}(X _{k + 1} = i \mid X_k = i) \, \mathbb{P}(X _{k} = i)\" \"+ \, \mathbb{P}(X _{k + 1} = i \mid X_k = i - 1 ) \, \mathbb{P}(X _{k} = i - 1)\"<\/p>\n

Par la question 3,<\/p>\n

\"\displaystyle \mathbb{P} (X_{k + 1} = i) = \frac {i} {k + 2} \, \mathbb{P}(X _{k} = i)\" \"\displaystyle \qquad \qquad +\, \frac { 2 + k - i + 1} {k + 2} \, \mathbb{P}(X _{k} = i - 1)\"<\/p>\n

\"\displaystyle \mathbb{P} (X_{k + 1} = i) = \frac {i} {k + 2} \, \mathbb{P}(X _{k} = i)\" \"\displaystyle \qquad \qquad \qquad \; +\, \frac { 3 + k - i} {k + 2} \, \mathbb{P}(X _{k} = i - 1)\"<\/p>\n

Question 3 :\u00a0<\/strong><\/p>\n

\"\mathbb{P}(X_3 = 3 ) = \dfrac{11}{24}\"<\/p>\n

\"X_3(\Omega) = [[1, 4]]\".<\/p>\n

La question pr\u00e9c\u00e9dente donne:<\/p>\n

\"\mathbb{P}(X _ 3 = i) =\" \"\quad \displaystyle \frac i 4 \,\mathbb{P}(X _ 2 = i) + \frac {5 - i} 4 \,\mathbb{P}(X _2 = i - 1)\"<\/p>\n

\"\mathbb{P}(X _ 3 = 1) = \displaystyle \frac 1 4 \,\mathbb{P}(X _ 2 = 1) = \frac 1 {24}\"<\/p>\n

\"\mathbb{P}(X _ 3 = 2) = \displaystyle \frac 2 4 \,\mathbb{P}(X _ 2 = 2) + \frac {3 } 4 \,\mathbb{P}(X _2 = 1)\" \"\displaystyle \mathbb{P}(X_3 = 2) = \frac 1 2 . \frac 2 3 + \frac 3 4 \, \frac 1 6 = \frac {11} {24}\"<\/p>\n

\"\mathbb{P}(X _ 3 = 3) = \displaystyle \frac 3 4 \,\mathbb{P}(X _ 2 = 3) + \frac {2 } 4 \,\mathbb{P}(X _2 = 2)\" \"\mathbb{P}(X_3 = 3) \displaystyle = \frac 3 4 \, \frac 1 6 + \frac 1 2 \frac 2 3 = \frac {11} {24}\"<\/p>\n

\"\mathbb{P}(X _ 3 = 4) = \displaystyle \frac 1 4 \,\mathbb{P}(X _ 2 = 3) = \frac 1 {24}\".<\/p>\n

Correction sur le tirage de la premi\u00e8re et la derni\u00e8re en Maths Sup<\/h2>\n

Question 1 :\u00a0<\/strong><\/p>\n

On traduit \"\displaystyle \sum _ {k = n} ^{2\, n} \mathbb{P}(Y = k) = 1\" pour obtenir \"\displaystyle \sum _ {k = n} ^{2\, n} \binom {k - 1} {n - 1} = \binom {2\, n} n\"<\/p>\n

en posant \"p = k - 1\",<\/p>\n

\"\displaystyle \sum _ {p = n - 1 } ^{2\, n - 1 } \binom {p} {n - 1} = \binom {2\, n} n\"<\/p>\n

puis en rempla\u00e7ant \"n\" par \"n + 1\"<\/p>\n

\"\displaystyle \boxed{ \sum _ {p = n } ^{2\, n + 1 } \binom {p} {n} = \binom {2\, n + 2 } {n + 1} }\".<\/p>\n

Question 2 :\u00a0<\/strong><\/p>\n

avec \"a = 2\"<\/p>\n

\"\textrm{E}(Y) = \displaystyle \sum _ {k = n } ^ {2 \, n} k \, \mathbb{P}(Y = k)\"<\/p>\n

donc \"\displaystyle \binom {2\, n} n \, \textrm{E}(Y) = \displaystyle \sum _ {k = n } ^ {2 \, n} k \, \binom {k - 1} {n - 1}\"<\/p>\n

On utilise \"\displaystyle k\, \binom {k - 1} {n - 1} = n\, \binom {k } {n}\"<\/p>\n

\"\displaystyle \binom {2\, n} n \, \textrm{E}(Y) = n \, \sum _ {k = n } ^ {2 \, n} \binom {k } {n}\"<\/p>\n

on peut donc utiliser la question pr\u00e9c\u00e9dente :<\/p>\n

\"\displaystyle \binom {2\, n} n \, \textrm{E}(Y) = n \,\left ( \binom {2 \, n + 2} {n + 1} - \binom {2 n + 1} {n} \right )\" (il manquait le terme pour \"p = 2\, n + 1\" )<\/p>\n

et par la formule du triangle de Pascal,<\/p>\n

\"\displaystyle \binom {2\, n} n \, \textrm{E}(Y) = n \, \binom {2 \, n + 1} {n + 1}\".<\/p>\n

Puis par quotient, \"\textrm{E}(Y) = \displaystyle \frac {n \, (2 \, n + 1)!} {(n + 1)! \, n!} \times \frac {n! \, n!} {(2\, n)!}\"<\/p>\n

\"\boxed{\textrm{E}(Y) = \displaystyle \frac {n \, (2 \, n + 1)} {n + 1}}\".<\/p>\n

Question 3 :\u00a0<\/strong><\/p>\n

\"b = 1\"<\/p>\n

\"\bullet\" Par le th\u00e9or\u00e8me de transfert,<\/p>\n

\"\textrm{E}(Y(Y + 1) ) = \displaystyle \sum _ {k = n } ^ {2 \, n} k \, (k + 1) \, \mathbb{P}(Y = k)\"<\/p>\n

donc \"\displaystyle \binom {2\, n} n \, \textrm{E}(Y(Y + 1) ) =\" \"\qquad \qquad \qquad \displaystyle \sum _ {k = n } ^ {2 \, n} (k + 1) \, k \, \binom {k - 1} {n - 1}\"<\/p>\n

On utilise \"\displaystyle (k + 1)\, k\, \binom {k - 1} {n - 1} = n( n + 1) \, \binom {k + 1 } {n + 1 }\"
\n\"\displaystyle \binom {2\, n} n \, \textrm{E}(Y(Y + 1) ) =\" \"\qquad \qquad \qquad \displaystyle n \, (n + 1) \, \sum _ {k = n } ^ {2 \, n} \binom {k + 1 } {n + 1 }\"<\/p>\n

On pose \"p = k + 1\",<\/p>\n

\"\displaystyle \binom {2\, n} n \, \textrm{E}(Y(Y + 1) ) =\" \"\qquad \qquad \qquad \displaystyle n \, (n + 1) \, \sum _ {p = n + 1 } ^ {2 \, n + 1} \binom {p } {n + 1 }\"<\/p>\n

On peut donc utiliser la question 3 en rempla\u00e7ant \"n\" par \"n + 1\" et en remarquant qu’il manque les termes pour \"p = 2\, n+2\" et \"p = 2\, n + 3\" :<\/p>\n

\"\displaystyle \binom {2\, n} n \, \textrm{E}(Y) =n (n + 1) \times\" \"\quad\displaystyle \left ( \binom {2 \, n + 4} {n + 2} - \binom {2 n + 3} {n+ 1 } -\binom {2 n + 2} {n + 1} \right )\"<\/p>\n

Par le triangle de Pascal :<\/p>\n

\"\displaystyle \binom {2 \, n + 4} {n + 2} - \binom {2 \,n + 3} {n+ 1 } -\binom {2 n + 2} {n + 1}\" \"\displaystyle = \binom {2\, n + 3} {n +2 } - \binom {2 \,n + 2} {n + 1}\"<\/p>\n

\"\displaystyle = \binom {2\, n + 3} {n +1 } - \binom {2 \,n + 2} {n +1 }\"<\/p>\n

et avec le triangle de Pascal<\/p>\n

\"\displaystyle = \binom {2\, n + 2} {n}\".<\/p>\n

\"\displaystyle \binom {2\, n} n \, \textrm{E}(Y) = n ( n + 1) \,\binom {2\, n + 2} {n }\"<\/p>\n

puis par quotient<\/p>\n

\"\displaystyle \textrm{E}(Y(Y + 1) ) =\" \"\qquad \qquad \displaystyle n(n + 1) \frac {(2\, n + 2)!} {(n + 2)! \, n! } \times \frac {n! \, n!} {(2\, n)!}\"<\/p>\n

\"\displaystyle \textrm{E}(Y(Y + 1) ) = n(n + 1)\displaystyle \frac {(2\, n + 2)(2\, n + 1) } {(n + 2)( n + 1) }\"<\/p>\n

\"\displaystyle\boxed{ \textrm{E}(Y(Y + 1)) = \frac {2\, n (n + 1) \, (2 \, n + 1)} {n + 2} }\".<\/p>\n

\"\bullet\" Calcul de la variance.<\/p>\n

\"\textrm{V}(Y) = \textrm{E}(Y(Y + 1)) - \textrm{E} (Y) - (\textrm{E}(Y))^2\"<\/p>\n

\"\textrm{V}(Y) = \displaystyle \frac {2\, n \, (n + 1) \, (2 \, n + 1)} {n + 2}\" \"\qquad \qquad \displaystyle - \, \frac {n(2\, n + 1)} {n + 1} - \frac {n^2 (2\, n + 1) ^2} {(n + 1)^2}\"<\/p>\n

\"\textrm{V}(Y) = \displaystyle \frac {n (2\, n + 1) } {(n + 2)(n + 1)^2 }\times A\"<\/p>\n

\"A= 2(n + 1)^3 - (n + 2)(n + 1)\" \"\qquad \qquad \qquad \qquad - \, n(2 \,n + 1) (n + 2)\"<\/p>\n

apr\u00e8s calculs : \"A = n\" donc<\/p>\n

\"\qquad \boxed{\textrm{V}(Y) = \displaystyle \frac {n^2 (2\, n + 1) } {(n + 2)(n + 1)^2 }}\".<\/p>\n

Correction de l’exercice sur la loi Hyperg\u00e9om\u00e9trique en Maths Sup<\/h2>\n

Si \"i \in [[1 , \, b]]\", on note \"X_i\" la variable al\u00e9atoire<\/strong> \u00e9gale \u00e0 \"1\" si l’on tire la boule blanche \"i\" et \"0\" sinon.<\/p>\n

Pour r\u00e9aliser \"(X_i = 1)\", on prend la boule \"i\" et on ajoute \"n - 1\" boules choisies parmi les \"N - 1\" autres boules.<\/p>\n

\"\mathbb{P}(X _ i = 1) = \displaystyle \frac {\displaystyle \binom {N - 1} {n - 1} } {\displaystyle \binom {N} n}\"\"\quad = \displaystyle \frac {(N - 1)! } {(n - 1)! (N - n)!} \times \frac {n! \, (N - n)! } {N!}\" \"\displaystyle \mathbb{P}(X _ i = 1) = \frac n N\".<\/p>\n

Comme \"X = \displaystyle \sum _ {i = 1} ^b X_i\,\", \"\textrm{E}(X) = \displaystyle \sum _ {i = 1} ^b \textrm{E}(X_i) \Rightarrow\" \"\boxed{\textrm{E}(X) = \frac {b\, n } N }\".<\/p>\n

Correction de l’\u00e9xercice sur le maximum et minimum de 2 tirages<\/h2>\n

Question 1 :\u00a0<\/strong><\/p>\n

Vrai<\/p>\n

On note \"Z = n + 1 - U\".
\n\"Z(\Omega) = V(\Omega) = [[1 , \, n]]\".<\/p>\n

Si \"k \in [[1 , n]]\",<\/p>\n

\"\mathbb{P}(Z = k) = \mathbb {P}( U = n + 1 - k)\"<\/p>\n

\"\mathbb{P}(Z = k) = \displaystyle \frac {2 \, n - 2 \, n - 2 + 2\, k +1 } {n ^2}\"<\/p>\n

\"\mathbb{P}(Z = k) = \displaystyle \frac { 2\, k - 1 } {n ^2} = \mathbb{P}(V = k)\"<\/p>\n

Donc \"n + 1 - U\" et \"V\" suivent la m\u00eame loi.<\/p>\n

Question 2 :\u00a0<\/strong><\/p>\n

\"\bullet\" \"\textrm{E}(V) = \displaystyle \sum _ {v = 1} ^n v \, \mathbb{P}(V = v)\"<\/p>\n

\"\textrm {E}(V) = \displaystyle \frac {1} {n ^2} \left ( 2\, \sum _ {v = 1} ^n v ^2 - \sum _ {v = 1 } ^n v \right )\"<\/p>\n

\"\;\;\; = \displaystyle \frac {1} {n ^2} \left (\frac {n(n + 1)(2\, n + 1)} 3 - \frac{n(n + 1)} 2 \right )\"<\/p>\n

\"\boxed{\textrm {E}(V) = \displaystyle \frac {(n + 1)(4\, n - 1) } {6\, n } }\".<\/p>\n

\"\bullet\" \"\textrm{E}(V^2 ) = \displaystyle \sum _ {v = 1} ^n v^2 \, \mathbb{P}(V = v)\"<\/p>\n

\"\textrm {E}(V^2 ) = \displaystyle \frac {1} {n ^2} \left ( 2\, \sum _ {v = 1} ^n v ^3 - \sum _ {v = 1 } ^n v^2 \right )\"<\/p>\n

\"= \displaystyle \frac {1} {n ^2} \left ( \frac {n^2 (n + 1)^2 } 2 - \frac{n(n + 1)(2n + 1 )} 6 \right )\"<\/p>\n

\"\textrm {E}(V^2) = \displaystyle \frac {(n + 1)(3\, n ^2 + n - 1) } {6\, n }\"<\/p>\n

\"\bullet\" \"\textrm{V}(V) = \textrm {E}(V^2) - \left ( \textrm {E}(V) \right ) ^2\"<\/p>\n

\"\textrm{V}(V) = \displaystyle \frac {(n + 1)(3\, n ^2 + n - 1) } {6\, n }\" \"\qquad \qquad \qquad \displaystyle -\, \frac {(n + 1)^2 (4 \, n - 1) ^2 } {36\, n^2 }\"<\/p>\n

apr\u00e8s factorisation et r\u00e9duction au m\u00eame d\u00e9nominateur ,<\/p>\n

\"\textrm{V}(V) = \displaystyle \frac {n + 1}{36\, n ^2} \left ( 2 \, n ^3 - 2\, n ^2 + n - 1 \right )\"<\/p>\n

\"\boxed{\textrm{V}(V) = \displaystyle \frac {(n^2 - 1)(2\, n^2 + 1) }{36\, n ^2} }\".<\/p>\n

Correction de l’exercice sur deux variables de loi uniforme<\/h2>\n

Question 1 :\u00a0\u00a0<\/strong><\/p>\n

\"\bullet\" On calcule \"\textrm{E}(Y^2)\" en utilisant le m\u00eame type de raisonnement que dans la question pr\u00e9c\u00e9dente.<\/p>\n

\"\textrm{E}(Y^2) = \displaystyle \sum _ {j = 1} ^n j^2 \,\mathbb{P} (Y = j)\"<\/p>\n

\"\textrm{E}(Y^2 ) = \displaystyle \sum _ {j = 1} ^n \left (\sum _ {k = j} ^n \frac {j ^2} { k\, n } \right)\"
\nque l’on \u00e9crit sous la forme<\/p>\n

\"\textrm{E}(Y) = \displaystyle \sum _ {1 \leqslant j \leqslant k \leqslant n } \frac {j ^2} { k\, n}\"
\nce qui permet d’intervertir les signes \"\sum\"<\/p>\n

\"\textrm{E}(Y^2) = \displaystyle \sum _ {k= 1} ^n \left (\sum _ {j = 1} ^k \frac {j ^2 } { k\, n } \right )\"<\/p>\n

\"\textrm{E}(Y^2) = \displaystyle \sum _ {k= 1} ^n \frac {(k + 1)(2\, k + 1) } { 6\, n}\" \"\quad \displaystyle = \sum _{k = 1} ^n \frac {k ^2 } {3\, n} + \sum _{k = 1} ^n \frac {3\, k} {6 \,n} + \sum _ {k = 1} ^n \frac 1 {6\, n}\"<\/p>\n

\"\quad = \displaystyle \frac {(n + 1)(2\, n + 1) } {18} + \frac {3 ( n + 1)} {12} + \frac 1 6\"<\/p>\n

\"\textrm{E}(Y^2) = \displaystyle \frac {4 \, n ^2 + 15\, n + 17} {36}\".<\/p>\n

\"\bullet\" Variance<\/p>\n

\"\textrm{V}(Y) = \textrm{E}(Y^2) - \left ( \textrm{E}(Y) \right ) ^2\"<\/p>\n

\"\textrm{V}(Y) = \displaystyle \frac {4 \, n ^2 + 15\, n + 17} {36} - \frac { (n + 3)^2} {16}\"<\/p>\n

\"\boxed{\textrm{V}(Y) = \displaystyle \frac {7 \, n ^2 + 6\, n -13} {144} \textrm{V}(Y) \displaystyle = \frac {(n - 1)(7 \,n + 13)} {144} }\".<\/p>\n

Question 2 :\u00a0\u00a0<\/strong><\/p>\n

Vrai<\/p>\n

La variable<\/strong> \"X - Y\" est \u00e0 valeurs dans \"\mathbb{N}\" car \"Y \leqslant X\", donc \"X - Y + 1 \in \mathbb{N}^*\".<\/p>\n

Puis \"1 \leqslant Y\" et \"X \leqslant n \Rightarrow X - Y + 1 \leq n\"<\/p>\n

D’autre part, \"X - Y + 1\" prend la valeur \"k\" lorsque par exemple \"X = k\" et \"Y = 1\".<\/p>\n

Donc \"(X - Y + 1)(\Omega) = [[1 , \, n]]\".<\/p>\n

Si \"k \in [[1 , n]]\",<\/p>\n

\"\mathbb{P}(X - Y + 1 = k ) =\" \"\qquad \quad \displaystyle \sum _{j = 1} ^n \mathbb{P} (X - Y + 1 = k , Y = j)\"
\nen utilisant le syst\u00e8me complet d’\u00e9v\u00e9nements \"(Y=j) _{1\leqslant j \leqslant n}\,\",<\/p>\n

\"\mathbb{P}(X - Y + 1 = k ) =\" \"\qquad \quad \displaystyle \sum _{j = 1} ^n \mathbb{P} (X = k + j - 1 = k , Y = j)\"
\nil ne faut garder que les indices \"j\" v\u00e9rifiant \"1 \leqslant j \leqslant k + j - 1 \leq n\" ssi \"1 \leqslant j \leqslant n - k + 1\"<\/p>\n

\"\mathbb{P}(X - Y + 1 = k ) = \displaystyle \sum _{j = 1} ^{n - k + 1} \frac {1 } {(j + k - 1) \, n}\"<\/p>\n

Puis en posant \"i = j+ k - 1\",<\/p>\n

\"\mathbb{P}(X - Y + 1 = k ) = \displaystyle \sum _{i = k} ^{n} \frac {1 } {i \, n} = \mathbb{P}(Y = k)\"<\/p>\n

donc \"X - Y + 1\" a m\u00eame loi que \"Y\".<\/p>\n

Nos cours en ligne en maths pour les Maths Sup proposent \u00e9galement des exercices sur d’autres chapitres, par exemple :<\/p>\n