diff --git a/module2/exo1/toy_document_en.Rmd b/module2/exo1/toy_document_en.Rmd
index 06139c543d83076fc35f4c5f3d8d0dc2e0ed0b81..f032a56e9f59bfed27104396b36289dee57a484e 100644
--- a/module2/exo1/toy_document_en.Rmd
+++ b/module2/exo1/toy_document_en.Rmd
@@ -4,15 +4,19 @@ author: "Arnaud Legrand"
 date: "25 juin 2018"
 output: html_document
 ---
-##Asking the maths library
-My computer tells me that π is *approximatively*
+
+```{r setup, include=FALSE}	
+knitr::opts_chunk$set(echo = TRUE)	
+```
+## Asking the maths library
+My computer tells me that $\pi$ is *approximatively*
 
 ```{r}
 pi
 ```
 
-##Buffon’s needle
-Applying the method of [Buffon’s needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the **approximation**
+## Buffon's needle
+Applying the method of [Buffon's needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the __approximation__
 ```{r}
 set.seed(42)
 N = 100000
@@ -20,8 +24,8 @@ x = runif(N)
 theta = pi/2*runif(N)
 2/(mean(x+sin(theta)>1))
 ```
-##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≤1]=π/4 **(see [“Monte Carlo method” on Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method)). The following code uses this approach:
+## 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 ["Monte Carlo method" on Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method)). The following code uses this approach:
 ```{r}
 set.seed(42)
 N = 1000
@@ -30,7 +34,8 @@ 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()
 ```
-It is therefore straightforward to obtain a (not really good) approximation to π by counting how many times, on average, ***X*^2+*Y*^2** is smaller than 1 :
+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:	
+
 ```{r}
 4*mean(df$Accept)
 ```