- \n
- Le succ\u00e8s<\/strong>, not\u00e9
![\"S\"](\"https:\/\/coetlogon.fr\/wp-content\/ql-cache\/quicklatex.com-9bc0c4e5c963cee9337cebbcf21817cf_l3.png\" "\\"Rendered")
, de probabilit\u00e9 ![\"p\"](\"https:\/\/coetlogon.fr\/wp-content\/ql-cache\/quicklatex.com-ebf20cd1bbe8dcdebf9aaa1523889d40_l3.png\" "\\"Rendered")
.<\/li>\n\n\n\n- L’\u00e9chec<\/strong>, not\u00e9
![\"\bar{S}\"](\"https:\/\/coetlogon.fr\/wp-content\/ql-cache\/quicklatex.com-c6da7bfd2599b730807edc043b5d778d_l3.png\" "\\"Rendered")
, de probabilit\u00e9 ![\"q = 1 - p\"](\"https:\/\/coetlogon.fr\/wp-content\/ql-cache\/quicklatex.com-dde3f821552b9a9f638b3a37714cc448_l3.png\" "\\"Rendered")
.<\/li>\n<\/ul>\n\n\n\nD\u00e9finition :<\/strong><\/p>\n\n\n\n
Un sch\u00e9ma de Bernoulli<\/strong> de param\u00e8tres
![\"n\"](\"https:\/\/coetlogon.fr\/wp-content\/ql-cache\/quicklatex.com-446376b77457ba020f6f46ed77df3b3d_l3.png\" "\\"Rendered")
et est la r\u00e9p\u00e9tition de \u00e9preuves de Bernoulli de param\u00e8tre , de mani\u00e8re identique et ind\u00e9pendante<\/strong>.<\/p>\n\n\n\n
\n\n\n\nII. La Loi Binomiale
![\"\mathcal{B}(n, p)\"](\"https:\/\/coetlogon.fr\/wp-content\/ql-cache\/quicklatex.com-f5f286b65dee9e64f8a07e71c87e23bf_l3.png\" "\\"Rendered")
<\/h3>\n\n\n\nD\u00e9finition :<\/strong><\/p>\n\n\n\n
\n
On consid\u00e8re un sch\u00e9ma de Bernoulli de param\u00e8tres
et . Soit ![\"X\"](\"https:\/\/coetlogon.fr\/wp-content\/ql-cache\/quicklatex.com-48a71fe637998f080a1f57b5d5b08fd4_l3.png\" "\\"Rendered")
la variable al\u00e9atoire qui compte le nombre de succ\u00e8s obtenus au cours des \u00e9preuves.<\/p>\n\n\n\nOn dit que
suit la loi binomiale<\/strong> de param\u00e8tres et . On note :<\/p>\n\n\n\n<\/span> <\/span>
![\"\[X \sim \mathcal{B}(n, p).\]\"](\"https:\/\/coetlogon.fr\/wp-content\/ql-cache\/quicklatex.com-6417a5e7ecbb0d398907bfcdde4889fe_l3.png\" "\\"Rendered")
<\/p><\/p>\n<\/blockquote>\n\n\n\nL’univers image :<\/strong> La variable al\u00e9atoire
peut prendre toutes les valeurs enti\u00e8res de ![\"0\"](\"https:\/\/coetlogon.fr\/wp-content\/ql-cache\/quicklatex.com-fb06cd20964624e42366981135663e5c_l3.png\" "\\"Rendered")
\u00e0 . On note ![\"X(\Omega) = \{0; 1; \dots; n\}\"](\"https:\/\/coetlogon.fr\/wp-content\/ql-cache\/quicklatex.com-0bb1c16696f3b10cce779079c555fc78_l3.png\" "\\"Rendered")
.<\/p>\n\n\n\n
\n\n\n\nIII. Calculs de probabilit\u00e9s<\/h3>\n\n\n\n
Th\u00e9or\u00e8me :<\/strong><\/p>\n\n\n\n
Pour tout entier
![\"k \in \{0; 1; \dots; n\}\"](\"https:\/\/coetlogon.fr\/wp-content\/ql-cache\/quicklatex.com-e2776858d060bca8cb43c337b705d7e9_l3.png\" "\\"Rendered")
, la probabilit\u00e9 d’obtenir exactement ![\"k\"](\"https:\/\/coetlogon.fr\/wp-content\/ql-cache\/quicklatex.com-09b299f3aa31f833cb7801685dd1bcea_l3.png\" "\\"Rendered")
succ\u00e8s est :<\/p>\n\n\n\n<\/span> <\/span>
![\"\[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\]\"](\"https:\/\/coetlogon.fr\/wp-content\/ql-cache\/quicklatex.com-306fab5de0b493dc7242a88a1861244a_l3.png\" "\\"Rendered")
<\/p><\/p>\n\n\n\n- \n
![\"\binom{n}{k}\"](\"https:\/\/coetlogon.fr\/wp-content\/ql-cache\/quicklatex.com-6a59a008daf0fabdc9444916629558d0_l3.png\" "\\"Rendered")
est le coefficient binomial<\/strong> (lu \u00a0\u00bb parmi \u00ab\u00a0), repr\u00e9sentant le nombre de chemins menant \u00e0 succ\u00e8s dans l’arbre pond\u00e9r\u00e9.<\/li>\n\n\n\n![\"p^k\"](\"https:\/\/coetlogon.fr\/wp-content\/ql-cache\/quicklatex.com-985cf3a80135ad41b3a2213deb6ce1e8_l3.png\" "\\"Rendered")
correspond \u00e0 la probabilit\u00e9 des succ\u00e8s.<\/li>\n\n\n\n![\"(1-p)^{n-k}\"](\"https:\/\/coetlogon.fr\/wp-content\/ql-cache\/quicklatex.com-b1e56a3e88fff1a3fc5e128f854768de_l3.png\" "\\"Rendered")
correspond \u00e0 la probabilit\u00e9 des ![\"n-k\"](\"https:\/\/coetlogon.fr\/wp-content\/ql-cache\/quicklatex.com-334705cc39b8511fd5dca3d30c965fe8_l3.png\" "\\"Rendered")
\u00e9checs.<\/li>\n<\/ul>\n\n\n\nPropri\u00e9t\u00e9 :<\/strong><\/p>\n\n\n\n
La somme des probabilit\u00e9s est \u00e9gale \u00e0 1 :
<\/span> <\/span>
![\"\[\sum_{k=0}^{n} P(X=k) = 1.\]\"](\"https:\/\/coetlogon.fr\/wp-content\/ql-cache\/quicklatex.com-a94d07d07beb5ea9ec51e320683e3015_l3.png\" "\\"Rendered")
<\/p><\/p>\n\n\n\n
\n\n\n\nIV. Repr\u00e9sentation graphique et indicateurs<\/h3>\n\n\n\n
Indicateurs de position et de dispersion :<\/strong><\/p>\n\n\n\n
Si
![\"X \sim \mathcal{B}(n, p)\"](\"https:\/\/coetlogon.fr\/wp-content\/ql-cache\/quicklatex.com-b0a534014e593c3407562bd19a9fc42c_l3.png\" "\\"Rendered")
, alors :<\/p>\n\n\n\n- \n
- Esp\u00e9rance :<\/strong>
![\"E(X) = np\"](\"https:\/\/coetlogon.fr\/wp-content\/ql-cache\/quicklatex.com-a14af2999ec1952b15a3656ee8645913_l3.png\" "\\"Rendered")
<\/li>\n\n\n\n- Variance :<\/strong>
![\"V(X) = np(1-p)\"](\"https:\/\/coetlogon.fr\/wp-content\/ql-cache\/quicklatex.com-a2eda9c2d5cfe13cfe6a2d5dad78917c_l3.png\" "\\"Rendered")
<\/li>\n\n\n\n- \u00c9cart-type :<\/strong>
![\"\sigma(X) = \sqrt{np(1-p)}\"](\"https:\/\/coetlogon.fr\/wp-content\/ql-cache\/quicklatex.com-41fbff26ff60d9f257fa81e7c0277326_l3.png\" "\\"Rendered")
<\/li>\n<\/ul>\n\n\n\nRemarque sur la forme :<\/strong> Lorsque
est grand, le diagramme en b\u00e2tons de la loi binomiale prend une forme de \u00ab\u00a0cloche\u00a0\u00bb sym\u00e9trique autour de l’esp\u00e9rance.<\/p>\n\n\n\nV. Exercices d’application (D\u00e9tails \/ Accord\u00e9on)<\/h3>\n\n\n\n
Exercice 1 : Calcul direct<\/strong><\/p>\n\n\n\n
Un tireur \u00e0 l’arc touche sa cible avec une probabilit\u00e9
![\"p = 0,7\"](\"https:\/\/coetlogon.fr\/wp-content\/ql-cache\/quicklatex.com-563c950197fbe5808b71e464620ef9d0_l3.png\" "\\"Rendered")
. Il tire 5 fl\u00e8ches de mani\u00e8re ind\u00e9pendante. Soit le nombre de fl\u00e8ches touchant la cible.<\/p>\n\n\n\n- \n
- Justifier que suit une loi binomiale dont on pr\u00e9cisera les param\u00e8tres.<\/li>\n\n\n\n
- Calculer
![\"P(X=3)\"](\"https:\/\/coetlogon.fr\/wp-content\/ql-cache\/quicklatex.com-3a1c4742642921fdbcc035f845486fa3_l3.png\" "\\"Rendered")
.<\/li>\n\n\n\n - Calculer
![\"P(X \geq 1)\"](\"https:\/\/coetlogon.fr\/wp-content\/ql-cache\/quicklatex.com-86c9651949693eca4c35531067b20f38_l3.png\" "\\"Rendered")
.<\/li>\n<\/ol>\n\n\n\nExercice 2 : Utilisation de la calculatrice<\/strong><\/p>\n\n\n\n
Soit
![\"X \sim \mathcal{B}(50; 0,2)\"](\"https:\/\/coetlogon.fr\/wp-content\/ql-cache\/quicklatex.com-96287ff1d3fd58f1c94a8c8c345df7af_l3.png\" "\\"Rendered")
.<\/p>\n\n\n\n- \n
- Calculer
![\"E(X)\"](\"https:\/\/coetlogon.fr\/wp-content\/ql-cache\/quicklatex.com-3dc336daa49a87e58c2f6d27c45eab76_l3.png\" "\\"Rendered")
et ![\"\sigma(X)\"](\"https:\/\/coetlogon.fr\/wp-content\/ql-cache\/quicklatex.com-5c60e0cb616fb7b8f775b6b28a94bfd6_l3.png\" "\\"Rendered")
.<\/li>\n\n\n\n - D\u00e9terminer \u00e0 l’aide de la calculatrice
![\"P(X \leq 8)\"](\"https:\/\/coetlogon.fr\/wp-content\/ql-cache\/quicklatex.com-465f51a2a8408494ea4c969a1609f162_l3.png\" "\\"Rendered")
et ![\"P(X > 12)\"](\"https:\/\/coetlogon.fr\/wp-content\/ql-cache\/quicklatex.com-b80f6f5873ad50f02903cfc0ec1c2d6d_l3.png\" "\\"Rendered")
.<\/li>\n<\/ol>\n\n\n\nConstruction des coefficients binomiaux<\/h3>\n\n\n\n
Le Triangle de Pascal :<\/strong><\/p>\n\n\n\n
Pour calculer les coefficients
, on utilise la relation de Pascal :<\/p>\n\n\n\n<\/span> <\/span>
![\"\[\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}\]\"](\"https:\/\/coetlogon.fr\/wp-content\/ql-cache\/quicklatex.com-1dba2266f7090be6722754c3a9e4cbaf_l3.png\" "\\"Rendered")
<\/p><\/p>\n\n\n\nVoici les premi\u00e8res lignes du triangle :<\/p>\n\n\n\n
<\/span> <\/span>
![\"\[\begin{array}{|c|cccccc|} \hline n \setminus k & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline 0 & 1 & & & & & \\ 1 & 1 & 1 & & & & \\ 2 & 1 & 2 & 1 & & & \\ 3 & 1 & 3 & 3 & 1 & & \\ 4 & 1 & 4 & 6 & 4 & 1 & \\ 5 & 1 & 5 & 10 & 10 & 5 & 1 \\ \hline \end{array}\]\"](\"https:\/\/coetlogon.fr\/wp-content\/ql-cache\/quicklatex.com-6f33246608df63ce3b7867dcd09c9e20_l3.png\" "\\"Rendered")
<\/p><\/p>\n\n\n\nAlternativement, on peut utiliser le r\u00e9sultat suivant.<\/p>\n\n\n\n
Th\u00e9or\u00e8me<\/h3>\n\n\n\n
Pour tout entier naturel
et tout entier tel que ![\"0\leqslant k\leqslant n\"](\"https:\/\/coetlogon.fr\/wp-content\/ql-cache\/quicklatex.com-860cb1d07c3514afbb9a7c1006b02f60_l3.png\" "\\"Rendered")
,<\/span> <\/span>
![\"\[\begin{pmatrix}n\\k\end{pmatrix}=\dfrac{n!}{k!(n-k)!}.\]\"](\"https:\/\/coetlogon.fr\/wp-content\/ql-cache\/quicklatex.com-f312ed4e21de9b705cfa561a1763a9d7_l3.png\" "\\"Rendered")
<\/p><\/p>\n\n\n\nExercices d’entra\u00eenement : Loi Binomiale<\/h3>\n\n\n\n
Exercice 1 : Application directe et calculs de probabilit\u00e9s<\/h4>\n\n\n\n
On consid\u00e8re une variable al\u00e9atoire
suivant la loi binomiale ![\"\mathcal{B}(10 \ ; \ 0,4)\"](\"https:\/\/coetlogon.fr\/wp-content\/ql-cache\/quicklatex.com-6a35847e1bec2992d56acef545e02a05_l3.png\" "\\"Rendered")
.<\/p>\n\n\n\n- \n
- Calculer l’esp\u00e9rance et l’\u00e9cart-type de cette loi.<\/li>\n\n\n\n
- Calculer, \u00e0
![\"10^{-4}\"](\"https:\/\/coetlogon.fr\/wp-content\/ql-cache\/quicklatex.com-f33fbca15615809f14cce33c594ab6ae_l3.png\" "\\"Rendered")
pr\u00e8s, la probabilit\u00e9 ![\"P(X = 4)\"](\"https:\/\/coetlogon.fr\/wp-content\/ql-cache\/quicklatex.com-1ac263c2188bd4eea7358cc3ee5b26e0_l3.png\" "\\"Rendered")
.<\/li>\n\n\n\n - D\u00e9terminer la probabilit\u00e9 d’obtenir au moins un succ\u00e8s, .<\/li>\n\n\n\n
- \u00c0 l’aide de la calculatrice, d\u00e9terminer
![\"P(X \leq 6)\"](\"https:\/\/coetlogon.fr\/wp-content\/ql-cache\/quicklatex.com-c8f57959688711f9fd78426f09894eeb_l3.png\" "\\"Rendered")
.<\/li>\n<\/ol>\n\n\n\nSolution Ex 1 :<\/strong><\/summary>\n
- \n
![\"E(X) = 10 \times 0,4 = 4\"](\"https:\/\/coetlogon.fr\/wp-content\/ql-cache\/quicklatex.com-e40196ab2631228cec43db01487021ed_l3.png\" "\\"Rendered")
; ![\"\sigma(X) = \sqrt{10 \times 0,4 \times 0,6} = \sqrt{2,4} \approx 1,549\"](\"https:\/\/coetlogon.fr\/wp-content\/ql-cache\/quicklatex.com-a4f85b40c8ac06afd5a738b20e1ef554_l3.png\" "\\"Rendered")
.<\/li>\n\n\n\n![\"P(X=4) = \binom{10}{4} \times 0,4^4 \times 0,6^6 \approx 0,2508\"](\"https:\/\/coetlogon.fr\/wp-content\/ql-cache\/quicklatex.com-05cb39c3a64653e57f130bd0630bf60d_l3.png\" "\\"Rendered")
.<\/li>\n\n\n\n![\"P(X \geqslant 1) = 1 - P(X=0) = 1 - 0,6^{10} \approx 0,9940\"](\"https:\/\/coetlogon.fr\/wp-content\/ql-cache\/quicklatex.com-22126b54ea6ad9ae41b197d8fad288b0_l3.png\" "\\"Rendered")
.<\/li>\n<\/ol>\n<\/details>\n\n\n\nExercice 2 : Probl\u00e8me contextualis\u00e9 (Contr\u00f4le qualit\u00e9)<\/h4>\n\n\n\n
Une entreprise produit des composants \u00e9lectroniques. On admet que chaque composant a une probabilit\u00e9
![\"p = 0,02\"](\"https:\/\/coetlogon.fr\/wp-content\/ql-cache\/quicklatex.com-33423b7f1b65f59cc54f4277a8515292_l3.png\" "\\"Rendered")
d’\u00eatre d\u00e9fectueux, ind\u00e9pendamment des autres. On pr\u00e9l\u00e8ve au hasard un lot de 100 composants. Soit ![\"Y\"](\"https:\/\/coetlogon.fr\/wp-content\/ql-cache\/quicklatex.com-5943181a342fb2c10f4dd7df81872400_l3.png\" "\\"Rendered")
la variable al\u00e9atoire comptant le nombre de composants d\u00e9fectueux dans le lot.<\/p>\n\n\n\n- \n
- Justifier que suit une loi binomiale dont on pr\u00e9cisera les param\u00e8tres.<\/li>\n\n\n\n
- Calculer la probabilit\u00e9 qu’il n’y ait aucun composant d\u00e9fectueux dans le lot sous forme exacte, puis sous forme d\u00e9cimale approch\u00e9e \u00e0 pr\u00e8s.<\/li>\n\n\n\n
- Calculer la probabilit\u00e9 qu’il y ait au plus 2 composants d\u00e9fectueux sous forme d\u00e9cimale approch\u00e9e \u00e0 pr\u00e8s.<\/li>\n\n\n\n
- Quel est le nombre moyen de composants d\u00e9fectueux attendus dans un tel lot ?<\/li>\n<\/ol>\n\n\n\n
Solution Ex 2<\/summary>\n
- \n
- On r\u00e9alise 100 tirages ind\u00e9pendants et identiques. Chaque tirage est une \u00e9preuve de Bernouilli, avec deux issues : le composant est d\u00e9fectueux (succ\u00e8s, probabilit\u00e9
![\"p=0,02\"](\"https:\/\/coetlogon.fr\/wp-content\/ql-cache\/quicklatex.com-6302c5dccec5117bc597dba6e88c2bd1_l3.png\" "\\"Rendered")
) ou non (\u00e9chec, probabilit\u00e9 ![\"q=0,98\"](\"https:\/\/coetlogon.fr\/wp-content\/ql-cache\/quicklatex.com-e57e68617d100de565713444e764e743_l3.png\" "\\"Rendered")
). est alors la variable al\u00e9atoire comptant le nombre de succ\u00e8s et suit donc la loi binomiale de param\u00e8tres ![\"n=100\"](\"https:\/\/coetlogon.fr\/wp-content\/ql-cache\/quicklatex.com-b907c5989d24e46cd30e9ae5288dc736_l3.png\" "\\"Rendered")
et :<\/span> <\/span>
![\"\[Y\sim\mathcal{B}(100;0,02).\]\"](\"https:\/\/coetlogon.fr\/wp-content\/ql-cache\/quicklatex.com-82993aadc6624256af91b49cbbf3b16a_l3.png\" "\\"Rendered")
<\/p><\/li>\n\n\n\n![\"P(Y=0)=0,98^{100}\approx0,1326\"](\"https:\/\/coetlogon.fr\/wp-content\/ql-cache\/quicklatex.com-579f88666318959c8168a8489aa8fbe3_l3.png\" "\\"Rendered")
.<\/li>\n\n\n\n![\"P(Y\leqslant2)\approx0,6767\"](\"https:\/\/coetlogon.fr\/wp-content\/ql-cache\/quicklatex.com-c79682a5fc4d359b72113aaf936b48d2_l3.png\" "\\"Rendered")
.<\/li>\n\n\n\n![\"E(Y)=100\times0.02=2\"](\"https:\/\/coetlogon.fr\/wp-content\/ql-cache\/quicklatex.com-a5eca5babbad4834e91daa9f722e4fc6_l3.png\" "\\"Rendered")
.
Il faut donc s’attendre \u00e0 une moyenne de deux composants d\u00e9fectueux par lot.<\/li>\n<\/ol>\n<\/details>\n\n\n\nExercice 3 : Seuil et r\u00e9p\u00e9tition<\/h4>\n\n\n\n
On lance
fois un d\u00e9 \u00e9quilibr\u00e9 \u00e0 6 faces. On appelle \u00ab\u00a0succ\u00e8s\u00a0\u00bb l’obtention d’un 6.<\/p>\n\n\n\n- \n
- Exprimer en fonction de la probabilit\u00e9
![\"p_n\"](\"https:\/\/coetlogon.fr\/wp-content\/ql-cache\/quicklatex.com-397d1cb026ae61d831050f1a7f237f27_l3.png\" "\\"Rendered")
d’obtenir au moins une fois le chiffre 6.<\/li>\n\n\n\n - D\u00e9terminer le nombre minimal de lancements n\u00e9cessaires pour que
![\"p_n \geq 0,95\"](\"https:\/\/coetlogon.fr\/wp-content\/ql-cache\/quicklatex.com-0e0b9a3bb17ae9d06f7497dcdb249f19_l3.png\" "\\"Rendered")
.<\/li>\n<\/ol>\n\n\n\nSolution Ex 3<\/summary>\n
- \n
![\"p_n = 1 - (\frac{5}{6})^n\"](\"https:\/\/coetlogon.fr\/wp-content\/ql-cache\/quicklatex.com-46fd96720f3bc85fa4f6b970a9cfe920_l3.png\" "\\"Rendered")
.<\/li>\n\n\n\n![\"p_n \geqslant 0,95 \iff1 - \left(\frac{5}{6}\right)^n \geqslant 0,95 \iff \left(\frac{5}{6}\right)^n \leqslant 0,05\"](\"https:\/\/coetlogon.fr\/wp-content\/ql-cache\/quicklatex.com-0e0a797083dc1b2b0c63e78f16a8b48d_l3.png\" "\\"Rendered")
![\"\iff n \ln\left(\frac{5}{6}\right) \leqslant \ln(0,05)\iff n \geqslant \frac{\ln(0,05)}{\ln\left(5/6\right)} \approx 16,43\"](\"https:\/\/coetlogon.fr\/wp-content\/ql-cache\/quicklatex.com-3aae3b43a2abbc35d81d842edf1966d4_l3.png\" "\\"Rendered")
.
Il faut donc au moins 17 lancers<\/strong>.<\/li>\n<\/ol>\n<\/details>\n\n\n\nPoint m\u00e9thode : de l’\u00e9nonc\u00e9 au mod\u00e8le<\/h2>\n\n\n\n
Pour justifier qu’une variable al\u00e9atoire
suit une loi binomiale.<\/p>\n\n\n\n- \n
- \u00c9tape 1 :<\/strong> Identifier l’\u00e9preuve de Bernoulli et le succ\u00e8s (donner sa probabilit\u00e9 ).<\/li>\n\n\n\n
- \u00c9tape 2 :<\/strong> Pr\u00e9ciser que l’on r\u00e9p\u00e8te
fois cette \u00e9preuve de mani\u00e8re identique et ind\u00e9pendante<\/strong>.<\/li>\n\n\n\n- \u00c9tape 3 :<\/strong> D\u00e9finir la variable al\u00e9atoire
comme le nombre de succ\u00e8s.<\/li>\n\n\n\n- \u00c9tape 4 :<\/strong> Conclure que
.<\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"I. \u00c9preuve et sch\u00e9ma de Bernoulli D\u00e9finition : Une \u00e9preuve de Bernoulli est une exp\u00e9rience al\u00e9atoire n’ayant que deux issues possibles : D\u00e9finition : Un sch\u00e9ma de Bernoulli de param\u00e8tres et est la r\u00e9p\u00e9tition de \u00e9preuves de Bernoulli de param\u00e8tre , de mani\u00e8re identique et ind\u00e9pendante. II. La Loi Binomiale D\u00e9finition : On consid\u00e8re un […]<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":19,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":["post-54","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/coetlogon.fr\/index.php?rest_route=\/wp\/v2\/pages\/54","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/coetlogon.fr\/index.php?rest_route=\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/coetlogon.fr\/index.php?rest_route=\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/coetlogon.fr\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/coetlogon.fr\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=54"}],"version-history":[{"count":17,"href":"https:\/\/coetlogon.fr\/index.php?rest_route=\/wp\/v2\/pages\/54\/revisions"}],"predecessor-version":[{"id":71,"href":"https:\/\/coetlogon.fr\/index.php?rest_route=\/wp\/v2\/pages\/54\/revisions\/71"}],"up":[{"embeddable":true,"href":"https:\/\/coetlogon.fr\/index.php?rest_route=\/wp\/v2\/pages\/19"}],"wp:attachment":[{"href":"https:\/\/coetlogon.fr\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=54"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
- \u00c9tape 2 :<\/strong> Pr\u00e9ciser que l’on r\u00e9p\u00e8te
- \u00c9tape 1 :<\/strong> Identifier l’\u00e9preuve de Bernoulli et le succ\u00e8s (donner sa probabilit\u00e9
- On r\u00e9alise 100 tirages ind\u00e9pendants et identiques. Chaque tirage est une \u00e9preuve de Bernouilli, avec deux issues : le composant est d\u00e9fectueux (succ\u00e8s, probabilit\u00e9
- Justifier que
- Calculer l’esp\u00e9rance
- Calculer
- Variance :<\/strong>
- L’\u00e9chec<\/strong>, not\u00e9