From d959696b041f3a036f4b8fc22c29bee16b30caee Mon Sep 17 00:00:00 2001
From: 287eb2433e490772bca1c5b6aae5009d
 <287eb2433e490772bca1c5b6aae5009d@app-learninglab.inria.fr>
Date: Fri, 27 Nov 2020 11:44:51 +0000
Subject: [PATCH] no commit message

---
 module2/exo1/toy_notebook_en.ipynb | 5 +++--
 1 file changed, 3 insertions(+), 2 deletions(-)

diff --git a/module2/exo1/toy_notebook_en.ipynb b/module2/exo1/toy_notebook_en.ipynb
index cf76847..1215d2e 100644
--- a/module2/exo1/toy_notebook_en.ipynb
+++ b/module2/exo1/toy_notebook_en.ipynb
@@ -6,6 +6,7 @@
    "source": [
     "# On the computation of $\\pi$ \n",
     "## Asking the maths library \n",
+    "\n",
     "My computer tells me that $\\pi$ is *approximatively*"
    ]
   },
@@ -65,7 +66,7 @@
    "metadata": {},
    "source": [
     "## Using a surface fraction argument\n",
-    "A method that is easier to understand and doest not make use of the sin function is based on the fact that if $X \\sim U(0,1)$ $Y \\sim U(0,1)$, then $P[X^{2}+Y^{2} \\le 1] = \\pi /4$ (see [\"Monte Carlo method\" on Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method)). 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 $X \\sim U(0,1)$ $Y \\sim U(0,1)$, then $P[X^{2}+Y^{2} \\le 1] = \\pi /4$ (see [\"Monte Carlo method\" on Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method)). The following code uses this approach:"
    ]
   },
   {
@@ -106,7 +107,7 @@
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "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:"
+    "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:"
    ]
   },
   {
-- 
2.18.1