diff --git a/module2/exo1/toy_document_en.Rmd b/module2/exo1/toy_document_en.Rmd
index 8108e81e14cc289bd3724c6163e4e19e56af3fcf..bae961c411daa333225198b6f6c328c180735ec9 100644
--- a/module2/exo1/toy_document_en.Rmd
+++ b/module2/exo1/toy_document_en.Rmd
@@ -1,50 +1,27 @@
 ---
-title: "On the computation of pi"
-Author: Pedro Mota
-Date: 14.01.2021
-output:
-  html_document:
-    df_print: paged
+title: "How"
+Author: Meiling WU
+Date: 30.05.2022
+output: html_document:
 ---
 
-## Asking the math library 
 
-My computer me that π is *approximatively*
-
-```{r}
-pi
+```{r setup, include=FALSE} 
+knitr::opts_chunk$set(echo = TRUE)
 ```
 
-## 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))
-```
+## Some explanations
 
-## 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)
+```{r cars}
+summary(cars)
 ```
 
 
 
+```{r pressure, echo=FALSE}
+plot(pressure)
+```