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{
"cells": [
{
"cell_type": "markdown",
"id": "9b604e23-db98-4106-b0c0-74e72625dfd5",
"metadata": {},
"source": [
"# Exercice 1\n",
"$n>=2, n \\in \\mathbb{N}, X_1, X_2,\\dots, X_n$ des VAR mutuellement indépendantes de loi $\\mathcal{B}(p)$, $p \\in ]0;1[$\n",
"\n",
"## 1\n",
"$$\n",
"S_n = \\sum_{i=1}^{N}X_i\n",
"$$\n",
"$S_n$ suit une loi Binomiale\n",
"\n",
"## 2\n",
"$S_n \\approx \\mathcal{P}(1)$\n",
"\n",
"## 3\n"
]
},
{
"cell_type": "code",
"execution_count": 12,
"id": "da4ff45a-49e3-4b51-904d-2cdd31a1b2ff",
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"[0.7 0. 0.9 0.4 0.8 0.5 0.2 1. 0.7 0.3 0.1 0.3 0. 0. 0. 0.1 0. 0.3\n",
" 0.8 0.2 0.1 0.9 0.2 0.5 0.3 0.3 0.8 0.4 0.5 1. 0.4 0.3 0.2 0.9 0.4 0.6\n",
" 0.7 0.7 1. 0.1 0.5 0.4 0. 0.7 0. 0.2 0.3 0. 0.1 0.6]\n",
"18\n"
]
}
],
"source": [
"import random\n",
"import numpy as np\n",
"\n",
"n = 50 #int(input(\"nombre essais\"))\n",
"p = 0.3 #float(input(\"proba succès générique\"))\n",
"def binom(n, p):\n",
" U = np.zeros(n)\n",
" for k in range(n):\n",
" U[k] = random.randint(0,10) / 10\n",
" S = sum([U[i] < p for i in range(n)])\n",
" return S"
]
},
{
"cell_type": "code",
"execution_count": 14,
"id": "f89c7bbe-ad75-4c23-a504-4b9c36df0895",
"metadata": {},
"outputs": [],
"source": [
"def poiss(l):\n",
" n = 1e6\n",
" p = l/n\n",
" X = Binom(n,p)\n",
" return X"
]
}
],
"metadata": {
"kernelspec": {
"display_name": "Python 3 (ipykernel)",
"language": "python",
"name": "python3"
},
"language_info": {
"codemirror_mode": {
"name": "ipython",
"version": 3
},
"file_extension": ".py",
"mimetype": "text/x-python",
"name": "python",
"nbconvert_exporter": "python",
"pygments_lexer": "ipython3",
"version": "3.13.11"
}
},
"nbformat": 4,
"nbformat_minor": 5
}
|
