diff --git a/module2/exo1/toy_document_orgmode_R_en.org b/module2/exo1/toy_document_orgmode_R_en.org
index 6f86037b556082d3c5ee8477298954491ae50906..a782a5fc4384529d7a6c6a0542f477f4179b39bb 100644
--- a/module2/exo1/toy_document_orgmode_R_en.org
+++ b/module2/exo1/toy_document_orgmode_R_en.org
@@ -18,34 +18,34 @@ My computer tells me that $\pi$ is /approximatively/
 pi
 #+end_src
 
-#+RESULTS:	
-: [1] 3.141593	
-* Buffon's needle	
-Applying the method of [[https://en.wikipedia.org/wiki/Buffon%2527s_needle_problem][Buffon's needle]], we get the *approximation*	
-#+begin_src R :results output :session *R* :exports both	
-set.seed(42)	
-N = 100000	
-x = runif(N)	
-theta = pi/2*runif(N)	
-2/(mean(x+sin(theta)>1))	
-#+end_src	
-#+RESULTS:	
-: [1] 3.14327	
-* 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:	
-#+begin_src R :results output graphics :file figure_pi_mc1.png :exports both :width 600 :height 400 :session *R* 	
-set.seed(42)	
-N = 1000	
-df = data.frame(X = runif(N), Y = runif(N))	
-df$Accept = (df$X**2 + df$Y**2 <=1)	
-library(ggplot2)	
-ggplot(df, aes(x=X,y=Y,color=Accept)) + geom_point(alpha=.2) + coord_fixed() + theme_bw()	
-#+end_src	
-#+RESULTS:	
-[[file:figure_pi_mc1.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:	
-#+begin_src R :results output :session *R* :exports both	
-4*mean(df$Accept)	
-#+end_src	
-#+RESULTS:	
+#+RESULTS:
+: [1] 3.141593
+* Buffon's needle
+Applying the method of [[https://en.wikipedia.org/wiki/Buffon%2527s_needle_problem][Buffon's needle]], we get the *approximation*
+#+begin_src R :results output :session *R* :exports both
+set.seed(42)
+N = 100000
+x = runif(N)
+theta = pi/2*runif(N)
+2/(mean(x+sin(theta)>1))
+#+end_src
+#+RESULTS:
+: [1] 3.14327
+* 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:
+#+begin_src R :results output graphics :file figure_pi_mc1.png :exports both :width 600 :height 400 :session *R* 
+set.seed(42)
+N = 1000
+df = data.frame(X = runif(N), Y = runif(N))
+df$Accept = (df$X**2 + df$Y**2 <=1)
+library(ggplot2)
+ggplot(df, aes(x=X,y=Y,color=Accept)) + geom_point(alpha=.2) + coord_fixed() + theme_bw()
+#+end_src
+#+RESULTS:
+[[file:figure_pi_mc1.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:
+#+begin_src R :results output :session *R* :exports both
+4*mean(df$Accept)
+#+end_src
+#+RESULTS:
 : [1] 3.156
\ No newline at end of file