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<h1>
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Day 2 - Probability, Compound Event Probability
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</h1>
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<ol class="breadcrumb">
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<li>
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<time class="published" datetime="2018-11-09T20:01:00+01:00">
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09 novembre 2018
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</time>
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</li>
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<li>Blog</li>
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<li>Basics</li>
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</ol>
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</header>
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<div class='article_content'>
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<h2>Basic probability with dices</h2>
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<h3>Problem</h3>
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<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>
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<h3>Mathematical explanation</h3>
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<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.
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<div>
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<style scoped>
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.dataframe tbody tr th:only-of-type {
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vertical-align: middle;
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}</p>
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<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>
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<span class="nl">vertical-align:</span> <span class="nf">top</span><span class="c1">;</span>
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<span class="err">}</span>
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<span class="na">.dataframe</span> <span class="no">thead</span> <span class="no">th</span> <span class="err">{</span>
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<span class="nl">text-align:</span> <span class="nf">right</span><span class="c1">;</span>
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<span class="err">}</span>
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</pre></div>
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<p></style>
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<table border="1" class="dataframe">
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<thead>
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<tr style="text-align: right;">
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<th></th>
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<th>1</th>
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<th>2</th>
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<th>3</th>
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<th>4</th>
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<th>5</th>
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<th>6</th>
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</tr>
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</thead>
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<tbody>
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<tr>
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<th>1</th>
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<td>2</td>
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<td>3</td>
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<td>4</td>
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<td>5</td>
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<td>6</td>
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<td>7</td>
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</tr>
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<tr>
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<th>2</th>
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<td>3</td>
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<td>4</td>
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<td>5</td>
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<td>6</td>
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<td>7</td>
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<td>8</td>
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</tr>
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<tr>
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<th>3</th>
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<td>4</td>
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<td>5</td>
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<td>6</td>
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<td>7</td>
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<td>8</td>
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<td>9</td>
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</tr>
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<tr>
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<th>4</th>
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<td>5</td>
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<td>6</td>
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<td>7</td>
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<td>8</td>
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<td>9</td>
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<td>10</td>
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</tr>
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<tr>
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<th>5</th>
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<td>6</td>
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<td>7</td>
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<td>8</td>
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<td>9</td>
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<td>10</td>
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<td>11</td>
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</tr>
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<tr>
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<th>6</th>
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<td>7</td>
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<td>8</td>
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<td>9</td>
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<td>10</td>
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<td>11</td>
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<td>12</td>
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</tr>
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</tbody>
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</table>
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</div></p>
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<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>
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<p>let's call A < x the event : the sum all the 2 tosses is at most x.
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</p>
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<div class="math">$$P(A\leq9)=\sum_{i=2}^{9} P(A = i)$$</div>
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<div class="math">$$P(A\leq9)=1-P(A\gt9)$$</div>
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<div class="math">$$P(A\leq9)=1-\sum_{i=10}^{12} P(A = i)$$</div>
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<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>
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<p>hence
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</p>
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<div class="math">$$P(A\leq9)=1-\sum_{i=10}^{12} \frac{13-i}{36}$$</div>
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<div class="math">$$P(A\leq9)= 1-\frac{6}{36}$$</div>
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<div class="math">$$P(A\leq9)= \frac{5}{6}$$</div>
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<h3>Let's program it</h3>
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<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"><=</span><span class="mi">9</span><span class="p">])</span> <span class="o">/</span> <span class="mi">36</span>
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</pre></div>
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<div class="highlight"><pre><span></span><span class="err">0.8333333333333334</span>
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</pre></div>
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<h2>More dices</h2>
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<h3>Problem</h3>
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<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>
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<h3>Mathematical explanation</h3>
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<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>
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<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".
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</p>
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<div class="math">$$P(A) = 1-P(A')$$</div>
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<div class="math">$$P(A)=1-\frac{6}{36}$$</div>
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<div class="math">$$P(A)=\frac{5}{6}$$</div>
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<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 :
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</p>
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<div class="math">$$P(B)=\frac{5}{36}$$</div>
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<p>The probability of having 2 dice different of sum 6 is :</p>
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<div class="math">$$P(A|B) = 4/5$$</div>
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<p>The probability that both A and B occure is equal to P(A∩B).</p>
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<p>Since <span class="math">\(P(A|B)=\frac{P(A∩B)}{P(B)}\)</span></p>
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<div class="math">$$P(A∩B)=P(B)*P(A|B)$$</div>
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<div class="math">$$P(A∩B)=5/36*4/5$$</div>
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<div class="math">$$P(A∩B)=1/9$$</div>
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<h3>Let's program it</h3>
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<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>
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</pre></div>
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<div class="highlight"><pre><span></span><span class="err">0.1111111111111111</span>
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</pre></div>
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<h2>Compound Event Probability</h2>
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<h3>Problem</h3>
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<p>There are 3 urns labeled X, Y, and Z.</p>
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<ul>
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<li>Urn X contains 4 red balls and 3 black balls.</li>
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<li>Urn Y contains 5 red balls and 4 black balls.</li>
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<li>Urn Z contains 4 red balls and 4 black balls. </li>
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</ul>
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<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>
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<h3>Mathematical explanation</h3>
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<p>Let's write the different probabilities:</p>
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<div>
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<style scoped="">
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.dataframe tbody tr th:only-of-type {
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||
vertical-align: middle;
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}
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||
|
||
.dataframe tbody tr th {
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||
vertical-align: top;
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}
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.dataframe thead th {
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text-align: right;
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}
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</style>
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<table class="dataframe" border="1">
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<thead>
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<tr style="text-align: right;">
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<th></th>
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<th>Red ball</th>
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<th>Black ball</th>
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</tr>
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</thead>
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<tbody>
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<tr>
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<th>Urne X</th>
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<td>$$\frac{4}{7}$$</td>
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<td>$$\frac{3}{7}$$</td>
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</tr>
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<tr>
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<th>Urne Y</th>
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<td>$$\frac{5}{9}$$</td>
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<td>$$\frac{4}{9}$$</td>
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</tr>
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<tr>
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<th>Urne Z</th>
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<td>$$\frac{1}{2}$$</td>
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<td>$$\frac{1}{2}$$</td>
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</tr>
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</tbody>
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</table>
|
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</div>
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<h4>Addition rule</h4>
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<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>
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<h4>Mutliplication rule</h4>
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<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>
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<p>The balls drawn from the urns are independant hence : </p>
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<p>p = P(2 red (R) and 1 back (B))
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</p>
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<div class="math">$$p = P(RRB) + P(RBR) + P(BRR)$$</div>
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<p>Each of those 3 probability if equal to the product of the probability of drawing each ball
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<span class="math">\(P(RRB) = P(R|X) * P(R|Y) * P(B|Z) = 4/7*5/9*1/2\)</span></p>
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<ul>
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||
<li>
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<p><span class="math">\(P(RRB) = 20/126\)</span></p>
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</li>
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<li>
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<p><span class="math">\(P(RBR) = 16/126\)</span></p>
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</li>
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<li>
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<p><span class="math">\(P(BRR) = 15/126\)</span></p>
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</li>
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</ul>
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<p>this leads to </p>
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<ul>
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<li><span class="math">\(p = 51/126\)</span></li>
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</ul>
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<p>and finally
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</p>
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<div class="math">$$p = \frac{17}{42}$$</div>
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<h3>Let's program it</h3>
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<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">"B"</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">"R"</span><span class="p">]</span>
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<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">"B"</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">"R"</span><span class="p">]</span>
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<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">"B"</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">"R"</span><span class="p">]</span>
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<span class="n">target</span> <span class="o">=</span> <span class="p">[</span><span class="s2">"BRR"</span><span class="p">,</span> <span class="s2">"RRB"</span><span class="p">,</span> <span class="s2">"RBR"</span><span class="p">]</span>
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|
||
<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><span class="err">0.40476190476190477</span>
|
||
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|
||
|
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