diff --git a/module2/exo1/toy_notebook_fr.ipynb b/module2/exo1/toy_notebook_fr.ipynb
index 0bbbe371b01e359e381e43239412d77bf53fb1fb..962ff00b285e64681bc8accf20be933b08b3f039 100644
--- a/module2/exo1/toy_notebook_fr.ipynb
+++ b/module2/exo1/toy_notebook_fr.ipynb
@@ -1,5 +1,116 @@
 {
- "cells": [],
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# toy_ntebook_fr"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "March 28, 2019"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "\n",
+    "## A propos du calcul de 'pi'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "\n",
+    "### En demandant à la lib maths\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Mon ordinateur m'indique que pi vaut approximativement"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "scrolled": true
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "3.141592653589793\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import *\n",
+    "print(pi)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### En utilisant la méthode des aiguilles de Buffon"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Mais calculé avec la méthode des aiguilles de Buffon, on obtiendrait comme approximation"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "3.128911138923655"
+      ]
+     },
+     "execution_count": 5,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "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"
+   ]
+  }
+ ],
  "metadata": {
   "kernelspec": {
    "display_name": "Python 3",
@@ -16,10 +127,9 @@
    "name": "python",
    "nbconvert_exporter": "python",
    "pygments_lexer": "ipython3",
-   "version": "3.6.3"
+   "version": "3.6.4"
   }
  },
  "nbformat": 4,
  "nbformat_minor": 2
 }
-