{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"\n",
"\n",
"\n",
"*(C) Copyright Franck CHEVRIER 2019-2021 http://www.python-lycee.com/*\n",
"\n",
" Pour exécuter une saisie Python, sélectionner la cellule et valider avec SHIFT+Entrée.\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Inférence bayésienne \n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"### Sommaire \n",
"\n",
"1. De cause à effet...
\n",
"2. ...et de l'effet à la cause.
\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## 1. De cause à effet..."
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"
\n",
"
\n",
"
\n", " Théorème de Bayes:\n", "\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "__2.2. À l'aide du théorème de Bayes, déterminer la probabilité que la boule ait été tirée de l'urne n°1 sachant qu'elle est rouge.__\n", "
\n", " Étant donné $A$ et $B$ deux événements de probabilités non nulles, on a :
\n", " $$\\displaystyle p_B(A)=\\frac{p_A(B)\\;p(A)}{p(B)}=\\frac{p_A(B)\\;p(A)}{p_A(B)\\;p(A)+p_{\\bar{A}}(B)\\;p(\\bar{A})}$$\n", "
\",\"
\"\n",
" \n",
" #code html pour les boules colorées (Rouges et vertes)\n",
" boule_coloree_img = \"
\",\"
\"\n",
" \n",
" #calculs et génération des affichages (pièces et boules):\n",
" pieces=\"\"\n",
" contenu_urne={True:\"\",False:\"\"}\n",
" nb_faces=0\n",
" nb_rouges=[0,0]\n",
" for couple in res:\n",
" piece,couleur=couple\n",
" pieces+=Piece_img[piece]\n",
" contenu_urne[piece]+=boule_coloree_img[couleur]\n",
" nb_rouges[piece]+=couleur\n",
" nb_faces+=piece\n",
" nb_piles=len(res)-nb_faces\n",
" \n",
" #en-tête du rendu html (avec balise de style) \n",
" html=\"\"\n",
" html+=\"\"\" \n",
" \n",
" \n",
" \n",
" \n",
" \"\"\"\n",
" #corps du rendu html\n",
" html+=\"\"\"\n",
" \n",
" \"\"\"+pieces+\" | \n",
"\n",
" ||
| \n", " | ||
\"\"\"+contenu_urne[True]+\" | \n",
" \"\"\"+contenu_urne[False]+\" | \n",
" |