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@ -185,7 +185,7 @@ Mean : 43117.75, Median : 44627.5, Mode : 4978
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<h1>
Day 2 - Probability, Compound Event Probability
</h1>
<ol class="breadcrumb">
<li>
<time class="published" datetime="2018-11-09T20:01:00+01:00">
09 novembre 2018
</time>
</li>
<li>Blog</li>
<li>Basics</li>
</ol>
</header>
<div class='article_content'>
<h2>Basic probability with dices</h2>
<h3>Problem</h3>
<p>In this challenge, we practice calculating probability. In a single toss of 2 fair (evenly-weighted) six-sided dice, find the probability that their sum will be at most 9.</p>
<h3>Mathematical explanation</h3>
<p>A nice way to think about sums-of-two-dice problems is to lay out the sums in a 6-by-6 grid in the obvious manner.
<div>
<style scoped>
.dataframe tbody tr th:only-of-type {
vertical-align: middle;
}</p>
<div class="highlight"><pre><span></span><span class="na">.dataframe</span> <span class="no">tbody</span> <span class="no">tr</span> <span class="no">th</span> <span class="err">{</span>
<span class="nl">vertical-align:</span> <span class="nf">top</span><span class="c">;</span>
<span class="err">}</span>
<span class="na">.dataframe</span> <span class="no">thead</span> <span class="no">th</span> <span class="err">{</span>
<span class="nl">text-align:</span> <span class="nf">right</span><span class="c">;</span>
<span class="err">}</span>
</pre></div>
<p></style>
<table border="1" class="dataframe">
<thead>
<tr style="text-align: right;">
<th></th>
<th>1</th>
<th>2</th>
<th>3</th>
<th>4</th>
<th>5</th>
<th>6</th>
</tr>
</thead>
<tbody>
<tr>
<th>1</th>
<td>2</td>
<td>3</td>
<td>4</td>
<td>5</td>
<td>6</td>
<td>7</td>
</tr>
<tr>
<th>2</th>
<td>3</td>
<td>4</td>
<td>5</td>
<td>6</td>
<td>7</td>
<td>8</td>
</tr>
<tr>
<th>3</th>
<td>4</td>
<td>5</td>
<td>6</td>
<td>7</td>
<td>8</td>
<td>9</td>
</tr>
<tr>
<th>4</th>
<td>5</td>
<td>6</td>
<td>7</td>
<td>8</td>
<td>9</td>
<td>10</td>
</tr>
<tr>
<th>5</th>
<td>6</td>
<td>7</td>
<td>8</td>
<td>9</td>
<td>10</td>
<td>11</td>
</tr>
<tr>
<th>6</th>
<td>7</td>
<td>8</td>
<td>9</td>
<td>10</td>
<td>11</td>
<td>12</td>
</tr>
</tbody>
</table>
</div></p>
<p>We see that the identic values are on the same diagonal. The number of elements on the diagonal varies from 1 to 6 and then back to 1. </p>
<p>let's call A &lt; x the event : the sum all the 2 tosses is at most x.
</p>
<div class="math">$$P(A\leq9)=\sum_{i=2}^{9} P(A = i)$$</div>
<div class="math">$$P(A\leq9)=1-P(A\gt9)$$</div>
<div class="math">$$P(A\leq9)=1-\sum_{i=10}^{12} P(A = i)$$</div>
<p>The value of <span class="math">\(P(A = i) = \frac{i-1}{36}\)</span> if <span class="math">\(i \leq 7\)</span> and <span class="math">\(P(A = i) = \frac{13-i}{36}\)</span></p>
<p>hence
</p>
<div class="math">$$P(A\leq9)=1-\sum_{i=10}^{12} \frac{13-i}{36}$$</div>
<div class="math">$$P(A\leq9)= 1-\frac{6}{36}$$</div>
<div class="math">$$P(A\leq9)= \frac{5}{6}$$</div>
<h3>Let's program it</h3>
<div class="highlight"><pre><span></span><span class="nb">sum</span><span class="p">([</span><span class="mi">1</span> <span class="k">for</span> <span class="n">d1</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span><span class="mi">7</span><span class="p">)</span> <span class="k">for</span> <span class="n">d2</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span><span class="mi">7</span><span class="p">)</span> <span class="k">if</span> <span class="n">d1</span><span class="o">+</span><span class="n">d2</span><span class="o">&lt;=</span><span class="mi">9</span><span class="p">])</span> <span class="o">/</span> <span class="mi">36</span>
</pre></div>
<div class="highlight"><pre><span></span>0.8333333333333334
</pre></div>
<h2>More dices</h2>
<h3>Problem</h3>
<p>In a single toss of 2 fair (evenly-weighted) six-sided dice, find the probability that the values rolled by each die will be different and the two dice have a sum of 6. </p>
<h3>Mathematical explanation</h3>
<p>Let's consider 2 events : A and B. A compound event is a combination of 2 or more simple events. If A and B are simple events, then A∪B denotes the occurence of either A or B. A∩B denotes the occurence of A and B together.</p>
<p>We denote A the event "the values of each dice is different". The opposit event is A' "the values of each dice is the same".
</p>
<div class="math">$$P(A) = 1-P(A')$$</div>
<div class="math">$$P(A)=1-\frac{6}{36}$$</div>
<div class="math">$$P(A)=\frac{5}{6}$$</div>
<p>We denote B the event "the two dice have a sum of 6", this probability has been computed on the first part of the article :
</p>
<div class="math">$$P(B)=\frac{5}{36}$$</div>
<p>The probability of having 2 dice different of sum 6 is :</p>
<div class="math">$$P(A|B) = 4/5$$</div>
<p>The probability that both A and B occure is equal to P(A∩B).</p>
<p>Since <span class="math">\(P(A|B)=\frac{P(A∩B)}{P(B)}\)</span></p>
<div class="math">$$P(A∩B)=P(B)*P(A|B)$$</div>
<div class="math">$$P(A∩B)=5/36*4/5$$</div>
<div class="math">$$P(A∩B)=1/9$$</div>
<h3>Let's program it</h3>
<div class="highlight"><pre><span></span><span class="nb">sum</span><span class="p">([</span><span class="mi">1</span> <span class="k">for</span> <span class="n">d1</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span><span class="mi">7</span><span class="p">)</span> <span class="k">for</span> <span class="n">d2</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span><span class="mi">7</span><span class="p">)</span> <span class="k">if</span> <span class="p">(</span><span class="n">d1</span><span class="o">+</span><span class="n">d2</span><span class="o">==</span><span class="mi">6</span><span class="p">)</span> <span class="ow">and</span> <span class="p">(</span><span class="n">d1</span><span class="o">!=</span><span class="n">d2</span><span class="p">)])</span> <span class="o">/</span> <span class="mi">36</span>
</pre></div>
<div class="highlight"><pre><span></span>0.1111111111111111
</pre></div>
<h2>Compound Event Probability</h2>
<h3>Problem</h3>
<p>There are 3 urns labeled X, Y, and Z.</p>
<ul>
<li>Urn X contains 4 red balls and 3 black balls.</li>
<li>Urn Y contains 5 red balls and 4 black balls.</li>
<li>Urn Z contains 4 red balls and 4 black balls. </li>
</ul>
<p>One ball is drawn from each of the urns. What is the probability that, of the 3 balls drawn, are 2 red and is 1 black?</p>
<h3>Mathematical explanation</h3>
<p>Let's write the different probabilities:</p>
<div>
<style scoped="">
.dataframe tbody tr th:only-of-type {
vertical-align: middle;
}
.dataframe tbody tr th {
vertical-align: top;
}
.dataframe thead th {
text-align: right;
}
</style>
<table class="dataframe" border="1">
<thead>
<tr style="text-align: right;">
<th></th>
<th>Red ball</th>
<th>Black ball</th>
</tr>
</thead>
<tbody>
<tr>
<th>Urne X</th>
<td>$$\frac{4}{7}$$</td>
<td>$$\frac{3}{7}$$</td>
</tr>
<tr>
<th>Urne Y</th>
<td>$$\frac{5}{9}$$</td>
<td>$$\frac{4}{9}$$</td>
</tr>
<tr>
<th>Urne Z</th>
<td>$$\frac{1}{2}$$</td>
<td>$$\frac{1}{2}$$</td>
</tr>
</tbody>
</table>
</div>
<h4>Addition rule</h4>
<p>A and B are said to be mutually exclusive or disjoint if they have no events in common (i.e., and A∩B=∅ and P(A∩B)=0. The probability of any of 2 or more events occurring is the union (∪) of events. Because disjoint probabilities have no common events, the probability of the union of disjoint events is the sum of the events' individual probabilities. A and B are said to be collectively exhaustive if their union covers all events in the sample space (i.e., A∪B=S and P(A∪B)=1). This brings us to our next fundamental rule of probability: if 2 events, A and B, are disjoint, then the probability of either event is the sum of the probabilities of the 2 events (i.e., P(A or B) = P(A)+P(B))</p>
<h4>Mutliplication rule</h4>
<p>If the outcome of the first event (A) has no impact on the second event (B), then they are considered to be independent (e.g., tossing a fair coin). This brings us to the next fundamental rule of probability: the multiplication rule. It states that if two events, A and B, are independent, then the probability of both events is the product of the probabilities for each event (i.e., P(A and B)= P(A)xP(B)). The chance of all events occurring in a sequence of events is called the intersection (∩) of those events. </p>
<p>The balls drawn from the urns are independant hence : </p>
<p>p = P(2 red (R) and 1 back (B))
</p>
<div class="math">$$p = P(RRB) + P(RBR) + P(BRR)$$</div>
<p>Each of those 3 probability if equal to the product of the probability of drawing each ball
<span class="math">\(P(RRB) = P(R|X) * P(R|Y) * P(B|Z) = 4/7*5/9*1/2\)</span></p>
<ul>
<li>
<p><span class="math">\(P(RRB) = 20/126\)</span></p>
</li>
<li>
<p><span class="math">\(P(RBR) = 16/126\)</span></p>
</li>
<li>
<p><span class="math">\(P(BRR) = 15/126\)</span></p>
</li>
</ul>
<p>this leads to </p>
<ul>
<li><span class="math">\(p = 51/126\)</span></li>
</ul>
<p>and finally
</p>
<div class="math">$$p = \frac{17}{42}$$</div>
<h3>Let's program it</h3>
<div class="highlight"><pre><span></span><span class="n">X</span> <span class="o">=</span> <span class="mi">3</span><span class="o">*</span><span class="p">[</span><span class="s2">&quot;B&quot;</span><span class="p">]</span><span class="o">+</span><span class="mi">4</span><span class="o">*</span><span class="p">[</span><span class="s2">&quot;R&quot;</span><span class="p">]</span>
<span class="n">Y</span> <span class="o">=</span> <span class="mi">4</span><span class="o">*</span><span class="p">[</span><span class="s2">&quot;B&quot;</span><span class="p">]</span><span class="o">+</span><span class="mi">5</span><span class="o">*</span><span class="p">[</span><span class="s2">&quot;R&quot;</span><span class="p">]</span>
<span class="n">Z</span> <span class="o">=</span> <span class="mi">4</span><span class="o">*</span><span class="p">[</span><span class="s2">&quot;B&quot;</span><span class="p">]</span><span class="o">+</span><span class="mi">4</span><span class="o">*</span><span class="p">[</span><span class="s2">&quot;R&quot;</span><span class="p">]</span>
<span class="n">target</span> <span class="o">=</span> <span class="p">[</span><span class="s2">&quot;BRR&quot;</span><span class="p">,</span> <span class="s2">&quot;RRB&quot;</span><span class="p">,</span> <span class="s2">&quot;RBR&quot;</span><span class="p">]</span>
<span class="nb">sum</span><span class="p">([</span><span class="mi">1</span> <span class="k">for</span> <span class="n">x</span> <span class="ow">in</span> <span class="n">X</span> <span class="k">for</span> <span class="n">y</span> <span class="ow">in</span> <span class="n">Y</span> <span class="k">for</span> <span class="n">z</span> <span class="ow">in</span> <span class="n">Z</span> <span class="k">if</span> <span class="n">x</span><span class="o">+</span><span class="n">y</span><span class="o">+</span><span class="n">z</span> <span class="ow">in</span> <span class="n">target</span><span class="p">])</span><span class="o">/</span><span class="nb">sum</span><span class="p">([</span><span class="mi">1</span> <span class="k">for</span> <span class="n">x</span> <span class="ow">in</span> <span class="n">X</span> <span class="k">for</span> <span class="n">y</span> <span class="ow">in</span> <span class="n">Y</span> <span class="k">for</span> <span class="n">z</span> <span class="ow">in</span> <span class="n">Z</span><span class="p">])</span>
</pre></div>
<div class="highlight"><pre><span></span>0.40476190476190477
</pre></div>
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<h1>
Running the notebook on a remote server
</h1>
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<time class="published" datetime="2018-08-15T11:07:00+02:00">
15 août 2018
</time>
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<li>Python</li>
<li>Basics</li>
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<div class='article_content'>
<h2>Jupyter hub</h2>
<p>With JupyterHub you can create a multi-user Hub which spawns, manages, and proxies multiple instances of the single-user Jupyter notebook server.</p>
<p>Project Jupyter created JupyterHub to support many users. The Hub can offer notebook servers to a class of students, a corporate data science workgroup, a scientific research project, or a high performance computing group.</p>
<p>However, if you are the only one using the server and you just want a simple way to run the notebook on your server and access it through the web interface on a light client without having to install and configure the jupyter hub, you can do the following.</p>
<h2>Problem with jupyter notebook</h2>
<p>On your server, run the command </p>
<div class="highlight"><pre><span></span>jupyter-notebook
</pre></div>
<p>you should get something like :</p>
<div class="highlight"><pre><span></span>[I 11:18:44.514 NotebookApp] Serving notebooks from local directory: /volume2/homes/admin
[I 11:18:44.515 NotebookApp] 0 active kernels
[I 11:18:44.516 NotebookApp] The Jupyter Notebook is running at: http://localhost:8888/?token=357587e3269b0f20f2b7e1918492890ae7573ac7ef1d2023
[I 11:18:44.516 NotebookApp] Use Control-C to stop this server and shut down all kernels (twice to skip confirmation).
[W 11:18:44.519 NotebookApp] No web browser found: could not locate runnable browser.
[C 11:18:44.520 NotebookApp]
Copy/paste this URL into your browser when you connect for the first time,
to login with a token:
http://localhost:8888/?token=357587e3269b0f20f2b7e1918492890ae7573ac7ef1d2023
</pre></div>
<p>and if you try to connect to your server ip (in my example : http://192.168.1.2:8888/?token=357587e3269b0f20f2b7e1918492890ae7573ac7ef1d2023) you will get an "ERR_CONNECTION_REFUSED" error. This is because, by default, Jupyter Notebook only accepts connections from localhost.</p>
<h2>Allowing connexions from other sources</h2>
<h3>From any IP</h3>
<p>The simplest way to avoid the connection error is to allow the notebook to accept connections from any ip </p>
<div class="highlight"><pre><span></span>jupyter-notebook --ip<span class="o">=</span>*
</pre></div>
<p>you will get something like </p>
<div class="highlight"><pre><span></span>[W 11:26:45.285 NotebookApp] WARNING: The notebook server is listening on all IP addresses and not using encryption. This is not recommended.
[I 11:26:45.342 NotebookApp] Serving notebooks from local directory: /volume2/homes/admin
[I 11:26:45.342 NotebookApp] 0 active kernels
[I 11:26:45.343 NotebookApp] The Jupyter Notebook is running at: http://[all ip addresses on your system]:8888/?token=52af33d628881824968b4031967e8541a27cc28b1720c199
[I 11:26:45.343 NotebookApp] Use Control-C to stop this server and shut down all kernels (twice to skip confirmation).
[W 11:26:45.346 NotebookApp] No web browser found: could not locate runnable browser.
[C 11:26:45.347 NotebookApp]
Copy/paste this URL into your browser when you connect for the first time,
to login with a token:
http://localhost:8888/?token=52af33d628881824968b4031967e8541a27cc28b1720c199
</pre></div>
<p>and if you connect form a remote client (192.168.1.1 in my example), the following line will be added to the output :</p>
<div class="highlight"><pre><span></span>[I 11:26:54.798 NotebookApp] 302 GET /?token=52af33d628881824968b4031967e8541a27cc28b1720c199 (192.168.1.1) 111.17ms
</pre></div>
<p>note that you should only do that if you are the only one using the server because the connection is not encypted.</p>
<h3>From a specific IP</h3>
<p>You can also, explicitly specify the ip of the client</p>
<div class="highlight"><pre><span></span>jupyter-notebook --ip<span class="o">=</span><span class="m">192</span>.168.1.1
</pre></div>
<div class="highlight"><pre><span></span>[I 11:44:58.104 NotebookApp] JupyterLab extension loaded from C:\Users\Guillaume\Miniconda3\lib\site-packages\jupyterlab
[I 11:44:58.104 NotebookApp] JupyterLab application directory is C:\Users\Guillaume\Miniconda3\share\jupyter\lab
[I 11:44:58.244 NotebookApp] Serving notebooks from local directory: C:\Users\Guillaume
[I 11:44:58.245 NotebookApp] 0 active kernels
[I 11:44:58.245 NotebookApp] The Jupyter Notebook is running at: http://192.168.1.1:8888/?token=503576dd8fa87d1f2c416df307e9b900e520b4942e317b32
[I 11:44:58.245 NotebookApp] Use Control-C to stop this server and shut down all kernels (twice to skip confirmation).
[C 11:44:58.258 NotebookApp]
Copy/paste this URL into your browser when you connect for the first time,
to login with a token:
http://192.168.1.1:8888/?token=503576dd8fa87d1f2c416df307e9b900e520b4942e317b32
[I 11:44:59.083 NotebookApp] Accepting one-time-token-authenticated connection from 192.168.1.1
</pre></div>
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