diff --git a/module2/exo1/toy_notebook_en.ipynb b/module2/exo1/toy_notebook_en.ipynb
index 78e09b34a09fa815810cc6886f62a6e45dc76cc1..0b3ffbd99d9e79970834dcec0a593b062e98b585 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",
     "My computer tells me that $\\pi$ is *approximatively*"
    ]
@@ -33,7 +38,7 @@
    "metadata": {},
    "source": [
     "## Buffon's needle\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__"
    ]
   },
   {
@@ -69,7 +74,7 @@
    "source": [
     "## Using a surface fraction argument\n",
     "A method that is easier to understand and does not make use of the sin function is based on the  \n",
-    "fact that if X ~ *U*(0,1) and Y ~ *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:"
+    "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:"
    ]
   },
   {