diff --git a/module2/exo1/toy_document_en.Rmd b/module2/exo1/toy_document_en.Rmd
index eb907c63e2cec03e7e7cbdc95d661a6ab65a2baa..d54411471d8c85ab106d1f54e4d292d1ac6c3fe5 100644
--- a/module2/exo1/toy_document_en.Rmd
+++ b/module2/exo1/toy_document_en.Rmd
@@ -5,7 +5,6 @@ date: "20/05/2021"
 output: html_document
 ---
 
-
 ```{r setup, include=FALSE}
 knitr::opts_chunk$set(echo = TRUE)
 ```
@@ -15,7 +14,8 @@ 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__
 
@@ -27,7 +27,6 @@ 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 $X\sim U(0,1)$ 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:
 
@@ -38,10 +37,11 @@ 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()
+
 ```	
 
 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)
-
 ```