diff --git a/module2/exo1/toy_notebook_fr.ipynb b/module2/exo1/toy_notebook_fr.ipynb
index 0bbbe371b01e359e381e43239412d77bf53fb1fb..f52d38ecdfee42addcabfacf8dce3682c667adef 100644
--- a/module2/exo1/toy_notebook_fr.ipynb
+++ b/module2/exo1/toy_notebook_fr.ipynb
@@ -1,6 +1,136 @@
 {
- "cells": [],
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# À propos du calcul de $\\pi$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## En demandant à la lib maths"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Mon ordinateur m’indique que $\\pi$ vaut *approximativement*"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "from math import *\n",
+    "print(pi)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "hideCode": false,
+    "hidePrompt": false
+   },
+   "source": [
+    "## En utilisant la méthode des aiguilles de Buffon"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "hideCode": false,
+    "hidePrompt": false
+   },
+   "source": [
+    "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": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "import numpy as np\n",
+    "np.random.seed(seed=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"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Sinon, une méthode plus simple à comprendre et ne faisant pas intervenir d’appel à la fonction sinus se base sur le fait que si $X \\sim \\mathcal{U}(0,1)$ et $Y \\sim \\mathcal{U}(0,1)$ alors $P[X^2 + Y^2 \\le 1] = \\pi/4$ (voir [méthode de Monte Carlo sur Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method)). Le code suivant illustre ce fait :"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "%matplotlib inline\n",
+    "import matplotlib.pyplot as plt\n",
+    "\n",
+    "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)\n",
+    "\n",
+    "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 $\\pi$ en comptant combien de fois, en moyenne, $X^2 + Y^2$ est inférieur à 1 :"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "3.112"
+      ]
+     },
+     "execution_count": 5,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    " 4*np.mean(accept)"
+   ]
+  }
+ ],
  "metadata": {
+  "hide_code_all_hidden": false,
   "kernelspec": {
    "display_name": "Python 3",
    "language": "python",
@@ -16,10 +146,9 @@
    "name": "python",
    "nbconvert_exporter": "python",
    "pygments_lexer": "ipython3",
-   "version": "3.6.3"
+   "version": "3.6.4"
   }
  },
  "nbformat": 4,
  "nbformat_minor": 2
 }
-