From 2e54e11b112e9f97800ad9852a936278ee2e2ac5 Mon Sep 17 00:00:00 2001
From: 8388c1425d3a684d1ec014af187ba020
 <8388c1425d3a684d1ec014af187ba020@app-learninglab.inria.fr>
Date: Tue, 24 Mar 2020 11:56:14 +0000
Subject: [PATCH] Update toy_document_orgmode_python_en.org

---
 module2/exo1/toy_document_orgmode_python_en.org | 6 +++---
 1 file changed, 3 insertions(+), 3 deletions(-)

diff --git a/module2/exo1/toy_document_orgmode_python_en.org b/module2/exo1/toy_document_orgmode_python_en.org
index bec678a..e9e120d 100644
--- a/module2/exo1/toy_document_orgmode_python_en.org
+++ b/module2/exo1/toy_document_orgmode_python_en.org
@@ -27,7 +27,7 @@ pi
 #+RESULTS:
 
 * 2. *Buffon's needle
-Applying the method of [[https://en.wikipedia.org/wiki/Buffon%27s_needle_problem][Buffon's needle]], we get the *approximation*
+Applying the method of [[https://en.wikipedia.org/wiki/Buffon%27s_needle_problem][_Buffon's needle_]], we get the *approximation*
 #+begin_src python :results output :exports both
 import numpy as np
 np.random.seed(seed=42)
@@ -42,8 +42,8 @@ theta=np.random.uniform(size=N, low=0, high = pi/2)
 * 3.  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 [[https://en.wikipedia.org/wiki/Monte_Carlo_method][Monte Carlo method on
-Wikipedia]]). THe following code use this approach:
+U(0,1)$, then $P[X^2 + Y^2 \leq 1] = \pi/4$ (see [[https://en.wikipedia.org/wiki/Monte_Carlo_method][_"Monte Carlo method on
+Wikipedia"_]]). THe following code use this approach:
 
 #+begin_src python :results output :exports both
 import matplotlib.pyplot as plt
-- 
2.18.1