From 8d9b98e1325a9ce1ddfb302cd93cecd6fb1e01f4 Mon Sep 17 00:00:00 2001
From: a1b871d0ecf115713a1a78157e18e70b
 <a1b871d0ecf115713a1a78157e18e70b@app-learninglab.inria.fr>
Date: Thu, 6 May 2021 16:40:48 +0000
Subject: [PATCH] This one is just nitpicking

---
 module2/exo1/toy_notebook_en.ipynb | 13 +++++--------
 1 file changed, 5 insertions(+), 8 deletions(-)

diff --git a/module2/exo1/toy_notebook_en.ipynb b/module2/exo1/toy_notebook_en.ipynb
index 42bd493..2e53c55 100644
--- a/module2/exo1/toy_notebook_en.ipynb
+++ b/module2/exo1/toy_notebook_en.ipynb
@@ -37,8 +37,8 @@
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "## Buffon’s needle\n",
-    "Applying the method of [Buffon’s needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the **approximation**"
+    "## Buffon's needle\n",
+    "Applying the method of [Buffon's needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the __approximation__"
    ]
   },
   {
@@ -71,13 +71,12 @@
    "metadata": {},
    "source": [
     "## Using a surface fraction argument\n",
-    "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 $[X^2 + Y^2 \\le 1] = \\pi/4$ (see [\"Monte Carlo method\"\n",
-    "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 $X\\sim U(0,1)$ 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:"
    ]
   },
   {
    "cell_type": "code",
-   "execution_count": 3,
+   "execution_count": 5,
    "metadata": {},
    "outputs": [
     {
@@ -94,7 +93,6 @@
     }
    ],
    "source": [
-    "%matplotlib inline\n",
     "import matplotlib.pyplot as plt\n",
     "\n",
     "np.random.seed(seed=42)\n",
@@ -115,8 +113,7 @@
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "It is then straightforward to obtain a (not really good) approximation to $\\pi$ by counting how\n",
-    "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