diff --git a/module2/exo1/toy_document_en.Rmd b/module2/exo1/toy_document_en.Rmd
index 1b6ef528789006119a46378729abc2b250b9cced..74d3b915899a44c55a11128e10427373dbadb490 100644
--- a/module2/exo1/toy_document_en.Rmd
+++ b/module2/exo1/toy_document_en.Rmd
@@ -5,41 +5,45 @@ date: "23 Octobre 2023"
 output: html_document
 ---
 
-{r setup, include=FALSE}
+```{r setup, include=FALSE}
 knitr::opts_chunk$set(echo = TRUE)
+```
 
 ## Asking the maths library
 
-My computer tells me that $\pi$ is approximatively
+My computer tells me that $\pi$ is *approximatively*
 
-{r}
+```{r}
 pi
+```
 
-Buffon's needle
+## Buffon's needle
 
-Applying the method of Buffon's needle, we get the approximation
+Applying the method of [Buffon's needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the **approximation**
 
-{r}
+```{r}
 set.seed(42)
 N = 100000
 x = runif(N)
 theta = pi/2*runif(N)
 2/(mean(x+sin(theta)>1))
+```
 
-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 "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 ["Monte Carlo method" on Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method)). The following code uses this approach:
 
-{r}
+```{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()
+```
 
 It is therefore 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}
+```{r}
 4*mean(df$Accept)
-
+```