diff --git a/module2/exo1/toy_document_orgmode_python_en.org b/module2/exo1/toy_document_orgmode_python_en.org
index a7082eafc2e96332152e5982dbaff42eb57515ac..f807f2b73b3abc915e5a81290f3b8f3dab2f061d 100644
--- a/module2/exo1/toy_document_orgmode_python_en.org
+++ b/module2/exo1/toy_document_orgmode_python_en.org
@@ -1,49 +1,42 @@
-#+TITLE: Mr.
+
+
+#+TITLE: On the computation of pi
#+AUTHOR: Bikash Adhikari
#+DATE: 23/03/2020
#+LANGUAGE: en
-# #+PROPERTY: header-args :eval never-export
-#+HTML_HEAD:
-#+HTML_HEAD:
-#+HTML_HEAD:
-#+HTML_HEAD:
-#+HTML_HEAD:
-#+HTML_HEAD:
+ #+PROPERTY: header-args :session :export both
-* Table of Contents
- * [[1. Asking the math library]]
- * [[2. *Buffon's needle]]
- * [[3. Using a surface fraction argument]]
-
-* 1. Asking the math library
-My computer tells me that \pi is /approximately/
-#+begin_src python :results output :exports both
+* Asking the math library
+My computer tells me that $\pi$ is /approximately/
+#+begin_src python :results value :session *python* :export both
from math import *
pi
#+end_src
-* 2. *Buffon's needle
-Applying the method of [[https://en.wikipedia.org/wiki/Buffon%27s_needle_problem][_Buffon's needle_]], we get the *approximation*
-#+begin_src python :results output :exports both
+
+
+* * Buffon's needle
+Applying the method of [[https://en.wikipedia.org/wiki/Buffon%27s_needle_problem][_Buffon's needle_]], we get the *approximation
+#+begin_src python :results value :session *python* :export both
import numpy as np
np.random.seed(seed=42)
-N=10000
-x=np.random.uniform(size=N, low =0, high=pi/2)
-theta=np.random.uniform(size=N, low=0, high = pi/2)
+N = 10000
+x = np.random.uniform(size=N, low=0, high=1)
+theta = np.random.uniform(size=N, low=0, high=pi/2)
2/(sum((x+np.sin(theta))>1)/N)
#+end_src
-
-* 3. Using a surface fraction argument
+* 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 use this approach:
+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 use this approach:
-#+begin_src python :results output :exports both
+#+begin_src python :results value :session *python* :export both
import matplotlib.pyplot as plt
np.random.seed(seed=42)
@@ -68,7 +61,6 @@ print(matplot_lib_filename)
It is then straightforward to obtain a (not really good) approximation
of \pi by counting how many time, on average, $X^2$ + $^{}Y^2$ is smaller
than 1:
-#+begin_src python :results output :exports both
+#+begin_src python :results value :session *python* :export both
4*np.mean(accept)
#+end_src
-