{"id":238085,"date":"2022-05-18T17:10:45","date_gmt":"2022-05-18T15:10:45","guid":{"rendered":"https:\/\/sherpas.com\/blog\/?p=238085"},"modified":"2025-09-29T13:33:25","modified_gmt":"2025-09-29T11:33:25","slug":"variables-aleatoires-discretes-exercice-corrige","status":"publish","type":"post","link":"https:\/\/sherpas.com\/blog\/variables-aleatoires-discretes-exercice-corrige\/","title":{"rendered":"Variables al\u00e9atoires discr\u00e8tes : exercice corrig\u00e9"},"content":{"rendered":"\n

En gal\u00e8re sur un exercice traitant des variables al\u00e9atoires discr\u00e8tes<\/strong> ? Pas de panique ! Gr\u00e2ce \u00e0 cet article vous pourrez pr\u00e9parer sereinement vos interrogations orales et \u00e9crites !<\/p>\n

Si vous souhaitez renforcer davantage vos comp\u00e9tences, nos profs de maths<\/a> sont l\u00e0 pour vous aider \u00e0 ma\u00eetriser les \u00e9preuves de probabilit\u00e9 en pr\u00e9pa. \ud83d\udcc8<\/p>\n\n\n\n

Exercice : Variables al\u00e9atoires discr\u00e8tes<\/h2>\n\n\n\n

\u23f0 Dur\u00e9e : 30 min<\/p>\n\n\n\n

\ud83d\udcaa Difficult\u00e9 : niveau 1\/3<\/p>\n\n\n\n

<\/p>\nSoient \"X, Y\" et \"Z\" trois variables al\u00e9atoires mutuellement ind\u00e9pendantes et d\u00e9finies sur le m\u00eame espace probabilis\u00e9 fini \"\left( \Omega , \mathbb{P} \right)\". On suppose que \"X, Y\" et \"Z\" suivent une loi uniforme sur \"[[1 , n]]\".
\n1.
\n \"(a)\" Donner la loi du couple \"\left( X, Y \right)\".
\n \"(b)\" Montrer que pour tout \"k \in [[2, n + 1]]\", \"\mathbb{P} \left( X + Y = k \right) = \dfrac{k - 1}{n^2}\".
\n \"(c)\" Montrer que pour tout \"k \in [[n + 1 , 2n]]\", \"\mathbb{P} \left( X + Y = k \right) = \dfrac{2n - k + 1}{n^2}\".
\n2. Justifier que \"Z\" et \"X + Y\" sont ind\u00e9pendantes, d\u00e9duire des questions pr\u00e9c\u00e9dentes que :\n

  <\/span>   <\/span>\"\[\mathbb{P} \left( X + Y = Z \right) = \frac{n - 1}{2n^2}.\]\"<\/p>
\n3.
\n \"(a)\" Montrer que la variable al\u00e9atoire \"T = n + 1 - Z\" suit la loi uniforme sur [[1, n]].
\n \"(b)\" Justifier que \"T\" est ind\u00e9pendante de \"X\" et \"Y\". D\u00e9terminer\n

  <\/span>   <\/span>\"\[\mathbb{P} \left( X + Y + Z = n + 1 \right).\]\"<\/p>\n\n\n\n

Corrig\u00e9 de l’exercice : Variables al\u00e9atoires discr\u00e8tes<\/h2>\n\n\n\n

<\/p>\n1.
\n \"(a)\" Il est clair que \"X \left( \Omega \right) = Y \left( \Omega \right) = [[1, n]]\", ainsi \n

  <\/span>   <\/span>\"\[X \left( \Omega \right) \times Y \left( \Omega \right) = [[1, n]] \times [[1, n]].\]\"<\/p> \nSoit \"\left( k , \ell \right) \in [[1, n]] \times [[1, n]]\". Par ind\u00e9pendance, on a :\n

  <\/span>   <\/span>\"\begin{eqnarray*}<\/p>
\n \"(b)\" Soit \"k \in [[2 , n + 1]].\" Nous avons le syst\u00e8me complet d’\u00e9v\u00e9nements suivant : \"\left\{ \left( Y = \ell \right), \, \ell \in [[1, n]] \right\}\", de sorte que, en appliquant la formule des probabilit\u00e9s totales :\n

  <\/span>   <\/span>\"\begin{eqnarray*}<\/p> \nComme \"\ell \in [[1 , k -1]]\", \"k - \ell \in [[1 , k -1]] ,\" d’o\u00f9 \"\mathbb{P} \left( X = k - \ell \right) = \dfrac1n\" puis\n

  <\/span>   <\/span>\"\[\mathbb{P} \left( X + Y = k \right) = \sum_{\ell=1}^{k - 1} \frac{1}{n} \times \frac1n = \frac{k - 1}{n^2}.\]\"<\/p>
\n \"(c)\" Le point de d\u00e9part est le m\u00eame que pour la question pr\u00e9c\u00e9dente : on applique la formule des probabilit\u00e9s totales avec le syst\u00e8me complet d’\u00e9v\u00e9nements \"\left\{ \left( Y = \ell \right), \, \ell \in [[1, n]] \right\}\". Soit \"k \in [[n + 1 , 2n]]\", on a :\n

  <\/span>   <\/span>\"\begin{eqnarray*}<\/p>\nOr, \"\mathbb{P} \left( X = k- \ell \right) = 0 \; \text{si} \; \ell \le k - n - 1\" et pour tout \"\ell \in [[k - n , n]]\", on a \"k - \ell \in [[k - n , n]].\"\n\nDonc, \"\mathbb{P} \left( X = k - \ell \right) = \dfrac1n\" car \"k - n \ge 1\". On a donc :\n

  <\/span>   <\/span>\"\begin{eqnarray*}<\/p>
\n2. Les variables al\u00e9atoires \"X\", \"Y\" et \"Z\" sont mutuellement ind\u00e9pendantes, donc, par le lemme des coalitions, \"X+Y\" et \"Z\" sont ind\u00e9pendantes.
\nOn utilise la formule des probabilit\u00e9s totales avec le syst\u00e8me complet d’\u00e9v\u00e9nements \"\big( ( Z = k )\big)_{k \in [[1, n]]}\". On a donc :\n

  <\/span>   <\/span>\"\[\mathbb{P} \left( X + Y = Z \right) = \sum_{k=1}^n \mathbb{P} \left( \left( X + Y = Z \right) \cap \left( Z = k \right) \right) = \sum_{k=1}^n \mathbb{P} \left( \left( X + Y = k \right) \cap \left( Z = k \right) \right).\]\"<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

<\/p>\n3. \"(a)\" Comme \"Z \left( \Omega \right) = [[1, n]]\", on a bien \"T \left( \Omega \right) = [[1, n]]\". De plus, si \"k \in [[1, n]]\", on a :\n

  <\/span>   <\/span>\"\[\mathbb{P} \left( T = k \right) = \mathbb{P} \left( Z = n + 1 - k \right) = \frac1n.\]\"<\/p>\nAinsi \"T\" suit bien une loi uniforme sur \"[[1, n]]\".
\n \"(b)\" On \u00e9crit \"\mathbb{P} \left( X + Y + Z = n + 1 \right) = \mathbb{P} \left( X + Y = n + 1 - Z \right) = \mathbb{P} \left( X + Y = T \right)\".
\nDe plus, par le lemme des coalitions, \"X+Y\" et \"T\" sont ind\u00e9pendantes, le calcul de la question 2 reste valable et\n

  <\/span>   <\/span>\"\[\mathbb{P} \left( X + Y + Z = n + 1 \right) = \frac{n - 1}{2n^2}.\]\"<\/p>\n\n\n\n

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Cet article est extrait de l’ouvrage Maths MPSI-MP2I. Tout-en-un : cours, m\u00e9thodes, entra\u00eenement et corrig\u00e9s <\/em>(\u00e9ditions Vuibert, juin 2021) \u00e9crit par E. Thomas, S. Bellec, G. Boutard. ISBN n\u00b09782311408720<\/em><\/p>\n\n <\/div>\n <\/section>\n<\/div>\n<\/div>\n\n\n

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En gal\u00e8re sur un exercice traitant des variables al\u00e9atoires discr\u00e8tes ? Pas de panique ! Gr\u00e2ce \u00e0 cet (…)<\/p>\n","protected":false},"author":158,"featured_media":244756,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":true,"footnotes":""},"category":[803,810],"tag":[78,345,783],"class_list":["post-238085","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-apprendre-matiere","category-maths","tag-prepa","tag-prepa-scientifique","tag-probabilites"],"acf":[],"_links":{"self":[{"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/posts\/238085","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/users\/158"}],"replies":[{"embeddable":true,"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/comments?post=238085"}],"version-history":[{"count":0,"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/posts\/238085\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/media\/244756"}],"wp:attachment":[{"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/media?parent=238085"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/category?post=238085"},{"taxonomy":"tag","embeddable":true,"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/tag?post=238085"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}