diff --git a/module2/exo1/toy_notebook_fr.ipynb b/module2/exo1/toy_notebook_fr.ipynb
index 0bbbe371b01e359e381e43239412d77bf53fb1fb..8e1715e0944c9a927e2669f6afe3ec4a657fbb67 100644
--- a/module2/exo1/toy_notebook_fr.ipynb
+++ b/module2/exo1/toy_notebook_fr.ipynb
@@ -1,5 +1,154 @@
 {
- "cells": [],
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# toy_notebook_fr"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "hideOutput": true
+   },
+   "source": [
+    "March 28, 2019"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## A propos du calcul de *π*"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### En demandant à la lib maths\n",
+    "Mon ordinateur m'indique que *π* vaut *approximativement*"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {},
+   "outputs": [
+    {
+     "ename": "SyntaxError",
+     "evalue": "invalid syntax (<ipython-input-4-4ac309e0cc7a>, line 1)",
+     "output_type": "error",
+     "traceback": [
+      "\u001b[0;36m  File \u001b[0;32m\"<ipython-input-4-4ac309e0cc7a>\"\u001b[0;36m, line \u001b[0;32m1\u001b[0m\n\u001b[0;31m    In[1] : from $\\math$ import *\u001b[0m\n\u001b[0m               ^\u001b[0m\n\u001b[0;31mSyntaxError\u001b[0m\u001b[0;31m:\u001b[0m invalid syntax\n"
+     ]
+    }
+   ],
+   "source": [
+    "In[1] : from $\\math$ import *\n",
+    "    'print(pi)'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##### En utilisant la méthode des aiguilles de Buffon\n",
+    "Mais calculé avec la **méthode** des [aiguilles de Buffon](https://fr.wikipedia.org/wiki/Aiguille_de_Buffon), on obtiendrait comme **approximation** :"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {},
+   "outputs": [
+    {
+     "ename": "SyntaxError",
+     "evalue": "invalid syntax (<ipython-input-2-e0eb810442a1>, line 1)",
+     "output_type": "error",
+     "traceback": [
+      "\u001b[0;36m  File \u001b[0;32m\"<ipython-input-2-e0eb810442a1>\"\u001b[0;36m, line \u001b[0;32m1\u001b[0m\n\u001b[0;31m    In[2]:import numpy as np\u001b[0m\n\u001b[0m               ^\u001b[0m\n\u001b[0;31mSyntaxError\u001b[0m\u001b[0;31m:\u001b[0m invalid syntax\n"
+     ]
+    }
+   ],
+   "source": [
+    "In[2]:import numpy as np\n",
+    "np.random.seed(see=42)\n",
+    "N=10000\n",
+    "x - np.random.uniform(size=N, low=0, high=1)\n",
+    "theta=np.random.uniform(size=N, low=0, high=pi/2)\n",
+    "2/(sum(x+np.sin(theta))>1/N)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Avec un argument \"fréquentiel\" de surface\n",
+    "Sinon une méthode plus simple à comprendre et ne faisant par intervenir d'appel à la fonction sinus se base sur le fait si X ~U(0,1) alors P[X²+Y²<=1] = pi/4 (voir [méthode de Monte Carlo sur Wikipedia](https://fr.wikipedia.org/wiki/M%C3%A9thode_de_Monte-Carlo#D%C3%A9termination_de_la_valeur_de_%CF%80)). Le code suivant illustre ce fait : "
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "in[3] %matplotlib inline\n",
+    "import matplotlib.pyplot as plt"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "np.random.seed(seed=42)\n",
+    "N=1000\n",
+    "x=np.random.uniform(size=N, low=0, high=1)\n",
+    "y=np.random.uniform(size=N, low=0, high=1)\n",
+    "accept=(x*x+y*y)<=1\n",
+    "reject=np.logical_not(accept)"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "fig, ax=plt.subplots(1)\n",
+    "ax.scatter(x[accept], y[accept], c='b', alpha=0,2, edgecolor=None)\n",
+    "ax.scatter(x[reject], y[reject], c='r', alpha=0,2, edgecolor=None)\n",
+    "ax.set_aspect('equal')"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Il est alors aisé d'obtenir une approximation (pas terrible) de *π* en comptant combien de fois, en moyenne, X²+Y² est inférieur à 1 : "
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "In [4] 4*np.mean(accept)"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": []
+  }
+ ],
  "metadata": {
   "kernelspec": {
    "display_name": "Python 3",
@@ -16,10 +165,9 @@
    "name": "python",
    "nbconvert_exporter": "python",
    "pygments_lexer": "ipython3",
-   "version": "3.6.3"
+   "version": "3.6.4"
   }
  },
  "nbformat": 4,
  "nbformat_minor": 2
 }
-