diff --git a/module2/exo1/toy_document_orgmode_python_en.org b/module2/exo1/toy_document_orgmode_python_en.org
index 1184cc3ac717bd36de92c4a47b48a700a9497b2f..c8c264d1b418f5fdc816c5df5b7dfb4b4cfe967c 100644
--- a/module2/exo1/toy_document_orgmode_python_en.org
+++ b/module2/exo1/toy_document_orgmode_python_en.org
@@ -8,7 +8,7 @@
 #+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 :session :exports both
+#+PROPERTY: header-args  :session  :exports both
 
 * Asking the math library
 My computer tells me that $\pi$ is /approximatively/
@@ -38,8 +38,7 @@ 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 $X \sim U(0,1)$ 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:
+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 [[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
@@ -65,8 +64,7 @@ print(matplot_lib_filename)
 [[file:figure_pi_mc2.png]]
 
 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:
+to $\pi$ by counting how many times, on average, $X^2 + Y^2$ is smaller than 1:
 
 #+begin_src python :results value :session :exports both
 4*np.mean(accept)