diff --git a/module2/exo1/toy_document_orgmode_python_en.org b/module2/exo1/toy_document_orgmode_python_en.org
index 52dc9e798a2b20e30306ca44c706d2cdb7045693..1184cc3ac717bd36de92c4a47b48a700a9497b2f 100644
--- a/module2/exo1/toy_document_orgmode_python_en.org
+++ b/module2/exo1/toy_document_orgmode_python_en.org
@@ -1,4 +1,4 @@
-#+TITLE:  On the computation of pi
+#+TITLE: On the computation of pi
 #+LANGUAGE: en
 
 #+HTML_HEAD: <link rel="stylesheet" type="text/css" href="http://www.pirilampo.org/styles/readtheorg/css/htmlize.css"/>
@@ -8,12 +8,10 @@
 #+HTML_HEAD: <script type="text/javascript" src="http://www.pirilampo.org/styles/lib/js/jquery.stickytableheaders.js"></script>
 #+HTML_HEAD: <script type="text/javascript" src="http://www.pirilampo.org/styles/readtheorg/js/readtheorg.js"></script>
 
-# #+PROPERTY: header-args :eval never-export
-
-* Table of Contents
+# #+PROPERTY: header-args :session :exports both
 
 * Asking the math library
-My computer tells me that \pi is /approximatively/
+My computer tells me that $\pi$ is /approximatively/
 
 #+begin_src python :results value :session :exports both
 from math import *
@@ -24,8 +22,7 @@ pi
 : 3.141592653589793
 
 * * Buffon's needle
-Applying the method of
-[[https://en.wikipedia.org/wiki/Buffon%2527s_needle_problem][Buffon's needle]] we get the *approximation*
+Applying the method of [[https://en.wikipedia.org/wiki/Buffon%2527s_needle_problem][Buffon's needle]], we get the *approximation*
 
 #+begin_src python :results value :session :exports both
 import numpy as np
@@ -42,10 +39,7 @@ theta = np.random.uniform(size=N, low=0, high=pi/2)
 * Using a surface fraction argument
 
 A method that is easier to understand and does not make use of the
-$\sin$ function is based on the fact that if $X ~ U(0,1)$ and $Y ~
-U(0,1)$, then $P[X^2 + Y^2 \leq 1] = \pi/4$ (see
-[[https://en.wikipedia.org/wiki/Monte_Carlo_method]["Monte Carlo
-Method" on Wikipedia]]). The following code uses this approach:
+$\sin$ 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 [[https://en.wikipedia.org/wiki/Monte_Carlo_method]["Monte Carlo Method" on Wikipedia]]). The following code uses this approach:
 
 #+begin_src python :results output file :session :var matplot_lib_filename="figure_pi_mc2.png" :exports both
 import matplotlib.pyplot as plt