{ "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 }