diff --git a/module2/exo1/toy_notebook_en.ipynb b/module2/exo1/toy_notebook_en.ipynb
index e0ba4038e80cccbea0c9838b50d939dff7a5714d..87c2a9f8c10c86b697b42c00156161e30ea72d75 100644
--- a/module2/exo1/toy_notebook_en.ipynb
+++ b/module2/exo1/toy_notebook_en.ipynb
@@ -1,5 +1,24 @@
 {
  "cells": [
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "# On the computation of $\\pi$"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "## Asking the maths library\n",
+    "My computer tells me that $\\pi$ is *approximatively*"
+   ]
+  },
   {
    "cell_type": "code",
    "execution_count": 1,
@@ -18,6 +37,24 @@
     " print(pi)"
    ]
   },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "## Buffon's needle"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "Applying the method of [Buffon's needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the __approximation__"
+   ]
+  },
   {
    "cell_type": "code",
    "execution_count": 2,
@@ -43,6 +80,24 @@
     " 2/(sum((x+np.sin(theta))>1)/N)"
    ]
   },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "## Using a surface fraction argument"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "A method that is easier to understand and does not make use of the $\\sin$ function is based on the fact that if $X\\sim U(0,1)$ and $Y\\sim U(0,1)$, then $P[X^2+Y^2\\leq 1] = \\pi/4$ (see [\"Monte Carlo method\" on Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method)). The following code uses this approach:"
+   ]
+  },
   {
    "cell_type": "code",
    "execution_count": 3,
@@ -77,6 +132,15 @@
     " ax.set_aspect('equal')"
    ]
   },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "It is then straightforward to obtain a (not really good) approximation to $\\pi$ by counting how many times, on average, $X^2 + Y^2$ is smaller than 1:"
+   ]
+  },
   {
    "cell_type": "code",
    "execution_count": 4,