diff --git a/module2/exo1/toy_document_en b/module2/exo1/toy_document_en
index c2abb93927a60dec5d3bcb12d4ea0a03c6b1b589..0119720d909a9c2aa1d65ecaa26e1795b4ddd8d8 100644
--- a/module2/exo1/toy_document_en
+++ b/module2/exo1/toy_document_en
@@ -1,10 +1,48 @@
-# On the computation of pi
+---
+title: "On the computation of pi"
+output:
+  html_document:
+    df_print: paged
+---
 
-## Asking the maths library
+## Asking the math library 
 
-My document tells me that π is *approximately*
+My computer me that π 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 **aproximation**
+
+```{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
+
+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 $P[X2+Y2≤1]=π/4$ (see ["Monte Carlo method" on Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method)). The Following code use this approach:  
+
+```{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 π by counting how many times, on average, $X2+Y2$ is smaller than 1 :
+
+```{r}
+4*mean(df$Accept)
+```
+