diff --git a/module2/exo1/toy_notebook_en.ipynb b/module2/exo1/toy_notebook_en.ipynb
index d9489073fcacb491f653ddaaadc38f1ee95ccb18..c003f68dd037233811194998244b9c419b4eaed6 100644
--- a/module2/exo1/toy_notebook_en.ipynb
+++ b/module2/exo1/toy_notebook_en.ipynb
@@ -4,8 +4,13 @@
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "# On the computation of $\\pi$\n",
-    "\n",
+    "# On the computation of $\\pi$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
     "##  Asking the maths library\n",
     "\n",
     "My computer tells me that $\\pi$ is *approximatively*"
@@ -34,8 +39,7 @@
    "metadata": {},
    "source": [
     "## Buffon’s needle\n",
-    "\n",
-    "Applying the method of [Buffon’s needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the **approximation**"
+    "Applying the method of [Buffon’s needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the __approximation__"
    ]
   },
   {
@@ -68,9 +72,7 @@
    "metadata": {},
    "source": [
     "## Using a surface fraction argument\n",
-    "\n",
-    "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 $P[X^2 + Y^2 \\le 1] = \\pi/4$ (see [\"Monte Carlo method\"\n",
-    "on Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method)). The following code uses this approach:"
+    "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 $P[X^2 + Y^2 \\le 1] = \\pi/4$ (see [\"Monte Carlo method\" on Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method)). The following code uses this approach:"
    ]
   },
   {
@@ -113,8 +115,7 @@
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "It is then straightforward to obtain a (not really good) approximation to $\\pi$ by counting how\n",
-    "many times, on average, $X^2 + Y^2$ is smaller than 1:"
+    "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:"
    ]
   },
   {