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<h1>
Day 4 - Binomial and geometric distributions
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11 novembre 2018
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<h2>Binomial distribution</h2>
<h3>Problem 1</h3>
<p>The ratio of boys to girls for babies born in Russia is <span class="math">\(r=\frac{N_b}{N_g}=1.09\)</span>. If there is 1 child born per birth, what proportion of Russian families with exactly 6 children will have at least 3 boys?</p>
<h3>Mathematical explanation</h3>
<p>Let's first compute the probability of having a boy :</p>
<div class="math">$$p_b=\frac{N_b}{N_b+N_g}$$</div>
<p>where:
<em> <span class="math">\(N_b\)</span> is the number of boys
</em> <span class="math">\(N_g\)</span> is the number of girls
* <span class="math">\(r=\frac{N_b}{N_g}\)</span></p>
<div class="math">$$p_b=\frac{1}{1+\frac{1}{r}}$$</div>
<div class="math">$$p_b=\frac{r}{r+1}$$</div>
<div class="highlight"><pre><span></span><span class="n">r</span> <span class="o">=</span> <span class="mf">1.09</span>
<span class="n">p_b</span><span class="o">=</span><span class="n">r</span><span class="o">/</span><span class="p">(</span><span class="n">r</span><span class="o">+</span><span class="mi">1</span><span class="p">)</span>
<span class="k">print</span><span class="p">(</span><span class="n">f</span><span class="s2">&quot;The probability of having a boy is p={p_b:3f}&quot;</span><span class="p">)</span>
</pre></div>
<div class="highlight"><pre><span></span>The probability of having a boy is p=0.521531
</pre></div>
<p>The probability of getting 3 boys in 6 children is given by :
</p>
<div class="math">$$b(x=3, n=6, p=p_b)$$</div>
<p>In order to compute the proportion of Russian families with exactly 6 children will have at 3 least boys we need to compute the cumulative probability distribution </p>
<div class="math">$$b(x\geq 3, n=6, p=p_b) = \sum_{i=3}^{6} b(x\geq i, n=6, p=p_b)$$</div>
<h3>Let's code it !</h3>
<div class="highlight"><pre><span></span><span class="kn">import</span> <span class="nn">math</span>
<span class="k">def</span> <span class="nf">bi_dist</span><span class="p">(</span><span class="n">x</span><span class="p">,</span> <span class="n">n</span><span class="p">,</span> <span class="n">p</span><span class="p">):</span>
<span class="n">b</span> <span class="o">=</span> <span class="p">(</span><span class="n">math</span><span class="o">.</span><span class="n">factorial</span><span class="p">(</span><span class="n">n</span><span class="p">)</span><span class="o">/</span><span class="p">(</span><span class="n">math</span><span class="o">.</span><span class="n">factorial</span><span class="p">(</span><span class="n">x</span><span class="p">)</span><span class="o">*</span><span class="n">math</span><span class="o">.</span><span class="n">factorial</span><span class="p">(</span><span class="n">n</span><span class="o">-</span><span class="n">x</span><span class="p">)))</span><span class="o">*</span><span class="p">(</span><span class="n">p</span><span class="o">**</span><span class="n">x</span><span class="p">)</span><span class="o">*</span><span class="p">((</span><span class="mi">1</span><span class="o">-</span><span class="n">p</span><span class="p">)</span><span class="o">**</span><span class="p">(</span><span class="n">n</span><span class="o">-</span><span class="n">x</span><span class="p">))</span>
<span class="k">return</span><span class="p">(</span><span class="n">b</span><span class="p">)</span>
<span class="n">b</span><span class="p">,</span> <span class="n">p</span><span class="p">,</span> <span class="n">n</span> <span class="o">=</span> <span class="mi">0</span><span class="p">,</span> <span class="n">p_b</span><span class="p">,</span> <span class="mi">6</span>
<span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">3</span><span class="p">,</span><span class="mi">7</span><span class="p">):</span>
<span class="n">b</span> <span class="o">+=</span> <span class="n">bi_dist</span><span class="p">(</span><span class="n">i</span><span class="p">,</span> <span class="n">n</span><span class="p">,</span> <span class="n">p</span><span class="p">)</span>
<span class="k">print</span><span class="p">(</span><span class="n">f</span><span class="s2">&quot;probability of getting at least 3 boys in a family with exactly 6 children : {b:.3f}&quot;</span><span class="p">)</span>
</pre></div>
<div class="highlight"><pre><span></span>probability of getting at least 3 boys in a family with exactly 6 children : 0.696
</pre></div>
<h3>Problem 2</h3>
<p>A manufacturer of metal pistons finds that, 12% on average, of the pistons they manufacture are rejected because they are incorrectly sized. What is the probability that a batch of 10 pistons will contain:
<em> No more than 2 rejects?
</em> At least 2 rejects?</p>
<h3>Mathematical explanation</h3>
<p>On average 12% of the pistons are rejected, this means that a piston has a probability of <span class="math">\(p_{rejected}=0.12\)</span> to be rejected.</p>
<p>The probability of getting less than 2 faulty pistons in a batch is :
</p>
<div class="math">$$p(rejet&lt;2) = b(x\leq 2, n= 10, p=p_{rejected})$$</div>
<div class="math">$$p(rejet&lt;2) = \sum_{i=0}^{2} b(x\leq i, n=10, p=p_{rejected})$$</div>
<div class="highlight"><pre><span></span><span class="n">b</span><span class="p">,</span> <span class="n">p</span><span class="p">,</span> <span class="n">n</span> <span class="o">=</span> <span class="mi">0</span><span class="p">,</span> <span class="mi">12</span><span class="o">/</span><span class="mi">100</span><span class="p">,</span> <span class="mi">10</span>
<span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span><span class="mi">3</span><span class="p">):</span>
<span class="n">b</span> <span class="o">+=</span> <span class="n">bi_dist</span><span class="p">(</span><span class="n">i</span><span class="p">,</span> <span class="n">n</span><span class="p">,</span> <span class="n">p</span><span class="p">)</span>
<span class="k">print</span><span class="p">(</span><span class="n">f</span><span class="s2">&quot;The probability of getting less than 2 faulty pistons in a batch is : {b:.3f}&quot;</span><span class="p">)</span>
</pre></div>
<div class="highlight"><pre><span></span>The probability of getting less than 2 faulty pistons in a batch is : 0.891
</pre></div>
<p>The probability that a batch of 10 pistons will contain at least 2 rejects :
</p>
<div class="math">$$p(rejet&lt;2) = b(x\geq 2, n= 10, p=p_{rejected})$$</div>
<div class="math">$$p(rejet&lt;2) = \sum_{i=2}^{10} b(x\geq i, n=10, p=p_{rejected})$$</div>
<div class="highlight"><pre><span></span><span class="n">b</span><span class="p">,</span> <span class="n">p</span><span class="p">,</span> <span class="n">n</span> <span class="o">=</span> <span class="mi">0</span><span class="p">,</span> <span class="mi">12</span><span class="o">/</span><span class="mi">100</span><span class="p">,</span> <span class="mi">10</span>
<span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">2</span><span class="p">,</span><span class="mi">11</span><span class="p">):</span>
<span class="n">b</span> <span class="o">+=</span> <span class="n">bi_dist</span><span class="p">(</span><span class="n">i</span><span class="p">,</span> <span class="n">n</span><span class="p">,</span> <span class="n">p</span><span class="p">)</span>
<span class="k">print</span><span class="p">(</span><span class="n">f</span><span class="s2">&quot;The probability of getting at least 2 faulty pistons in a batch is : {b:.3f}&quot;</span><span class="p">)</span>
</pre></div>
<div class="highlight"><pre><span></span>The probability of getting at least 2 faulty pistons in a batch is : 0.342
</pre></div>
<h2>Geometric distribution</h2>
<h3>Problem 1</h3>
<p>The probability that a machine produces a defective product is <span class="math">\(\frac{1}{3}\)</span>. What is the probability that the first defect is found during the fith inspection?</p>
<h3>Mathematical explanation</h3>
<p>In this case, we will use a geometric distribution to evaluate the probability :
<em> <span class="math">\(n=5\)</span>
</em> <span class="math">\(p=\frac{1}{3}\)</span></p>
<p>Hence, the probability that the first defect is found during the fith inspection is <span class="math">\(g(n=5,p=1/3)\)</span></p>
<div class="highlight"><pre><span></span><span class="k">print</span><span class="p">(</span><span class="n">f</span><span class="s2">&quot;The probability that the first defect is found during the fith inspection is {round(((1-p)**(n-1)) * p, 3)}&quot;</span><span class="p">)</span>
</pre></div>
<div class="highlight"><pre><span></span>The probability that the first defect is found during the fith inspection is 0.038
</pre></div>
<h3>Problem 2</h3>
<p>The probability that a machine produces a defective product is <span class="math">\(\frac{1}{3}\)</span>. What is the probability that the first defect is found during the first 5 inspections?</p>
<h3>Mathematical explanation</h3>
<p>In this problem, we need to compute the cumulative distribution function
</p>
<div class="math">$$p(x \leq5) = \sum_{i=1}^{5} g(n=i,p=1/3)$$</div>
<div class="highlight"><pre><span></span><span class="n">p_x5</span> <span class="o">=</span> <span class="mi">0</span>
<span class="n">p</span><span class="o">=</span><span class="mi">1</span><span class="o">/</span><span class="mi">3</span>
<span class="n">n</span><span class="o">=</span><span class="mi">5</span>
<span class="k">for</span> <span class="n">i</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="n">n</span><span class="o">+</span><span class="mi">1</span><span class="p">):</span>
<span class="n">p_x5</span><span class="o">+=</span><span class="p">(</span><span class="mi">1</span><span class="o">-</span><span class="n">p</span><span class="p">)</span><span class="o">**</span><span class="p">(</span><span class="n">i</span><span class="o">-</span><span class="mi">1</span><span class="p">)</span> <span class="o">*</span> <span class="n">p</span>
<span class="k">print</span><span class="p">(</span><span class="n">f</span><span class="s2">&quot;The probability that the first defect is found during the first 5 inspection is {round(p_x5, 3)}&quot;</span><span class="p">)</span>
</pre></div>
<div class="highlight"><pre><span></span>The probability that the first defect is found during the first 5 inspection is 0.868
</pre></div>
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<h1>
Binomial distribution
</h1>
<ol class="breadcrumb">
<li>
<time class="published" datetime="2018-11-11T11:19:00+01:00">
11 novembre 2018
</time>
</li>
<li>Mathematics</li>
<li>Statistics</li>
</ol>
</header>
<div class='article_content'>
<h2>Binomial Experiment</h2>
<p>A binomial experiment (or Bernoulli trial) is a statistical experiment that has the following properties:</p>
<ul>
<li>The experiment consists of n repeated trials.</li>
<li>The trials are independent.</li>
<li>The outcome of each trial is either success (s) or failure (f).</li>
</ul>
<h2>Binomial Distribution</h2>
<p>We define a binomial process to be a binomial experiment meeting the following conditions:</p>
<ul>
<li>The number of successes is <span class="math">\(x\)</span>.</li>
<li>The total number of trials is <span class="math">\(n\)</span>.</li>
<li>The probability of success of 1 trial is <span class="math">\(p\)</span>.</li>
<li>The probability of failure of 1 trial <span class="math">\(q\)</span>, where <span class="math">\(q=1-p\)</span>.</li>
<li><span class="math">\(b(x,n,p)\)</span> is the <em>binomial probability</em>, meaning the probability of having exactly <span class="math">\(x\)</span> successes out of <span class="math">\(n\)</span> trials.</li>
</ul>
<p>The binomial random variable is the number of successes, <span class="math">\(x\)</span>, out of <span class="math">\(n\)</span> trials.</p>
<p>The binomial distribution is the probability distribution for the binomial random variable, given by the following probability mass function:</p>
<div class="math">$$b(x,n,p) = \frac{n!}{x!(n-x)!}p^xq^{n-x}$$</div>
<p>Python code for the Binomial distribution</p>
<div class="highlight"><pre><span></span><span class="kn">import</span> <span class="nn">math</span>
<span class="k">def</span> <span class="nf">bi_dist</span><span class="p">(</span><span class="n">x</span><span class="p">,</span> <span class="n">n</span><span class="p">,</span> <span class="n">p</span><span class="p">):</span>
<span class="n">b</span> <span class="o">=</span> <span class="p">(</span><span class="n">math</span><span class="o">.</span><span class="n">factorial</span><span class="p">(</span><span class="n">n</span><span class="p">)</span><span class="o">/</span><span class="p">(</span><span class="n">math</span><span class="o">.</span><span class="n">factorial</span><span class="p">(</span><span class="n">x</span><span class="p">)</span><span class="o">*</span><span class="n">math</span><span class="o">.</span><span class="n">factorial</span><span class="p">(</span><span class="n">n</span><span class="o">-</span><span class="n">x</span><span class="p">)))</span><span class="o">*</span><span class="p">(</span><span class="n">p</span><span class="o">**</span><span class="n">x</span><span class="p">)</span><span class="o">*</span><span class="p">((</span><span class="mi">1</span><span class="o">-</span><span class="n">p</span><span class="p">)</span><span class="o">**</span><span class="p">(</span><span class="n">n</span><span class="o">-</span><span class="n">x</span><span class="p">))</span>
<span class="k">return</span><span class="p">(</span><span class="n">b</span><span class="p">)</span>
</pre></div>
<p>Using numpy </p>
<div class="highlight"><pre><span></span><span class="kn">import</span> <span class="nn">numpy</span> <span class="kn">as</span> <span class="nn">np</span>
<span class="n">n</span> <span class="o">=</span> <span class="mi">10</span> <span class="c1">#number of coin toss</span>
<span class="n">p</span> <span class="o">=</span> <span class="mf">0.5</span> <span class="c1">#probability</span>
<span class="n">samples</span> <span class="o">=</span> <span class="mi">1000</span> <span class="c1">#number of samples</span>
<span class="n">s</span> <span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">random</span><span class="o">.</span><span class="n">binomial</span><span class="p">(</span><span class="n">n</span><span class="p">,</span> <span class="n">p</span><span class="p">,</span> <span class="n">samples</span><span class="p">)</span>
</pre></div>
<p>Using the stats module from scipy</p>
<div class="highlight"><pre><span></span><span class="kn">from</span> <span class="nn">scipy.stats</span> <span class="kn">import</span> <span class="n">binom</span>
<span class="n">n</span> <span class="o">=</span> <span class="mi">10</span> <span class="c1">#number of coin toss</span>
<span class="n">p</span> <span class="o">=</span> <span class="mf">0.5</span> <span class="c1">#probability</span>
<span class="n">samples</span> <span class="o">=</span> <span class="mi">1000</span> <span class="c1">#number of samples</span>
<span class="n">s</span> <span class="o">=</span> <span class="n">binom</span><span class="o">.</span><span class="n">rvs</span><span class="p">(</span><span class="n">n</span><span class="p">,</span> <span class="n">p</span><span class="p">,</span> <span class="n">size</span><span class="o">=</span><span class="n">samples</span><span class="p">)</span>
</pre></div>
<h2>Cumulative probabilities</h2>
<p>A cumulative probability refers to the probability that the value of a random variable falls within a specified range. Frequently, cumulative probabilities refer to the probability that a random variable is less than or equal to a specified value. </p>
<p>A fair coin is tossed 10 times. </p>
<h3>Probability of getting 5 heads</h3>
<p>The probability of getting heads is:
</p>
<div class="math">$$b(x=5, n=10, p=0.5)=0.246$$</div>
<h3>Probability of getting at least 5 heads</h3>
<p>The probability of getting at least heads is: </p>
<div class="math">$$b(x\geq 5, n=10, p=0.5)= \sum_{r=5}^{10} b(x=r, n=10, p=0.5)$$</div>
<div class="math">$$b(x\geq 5, n=10, p=0.5)= 0.623$$</div>
<h3>Probability of getting at most 5 heads</h3>
<p>The probability of getting at most heads is: </p>
<div class="math">$$b(x\leq 5, n=10, p=0.5)= \sum_{r=0}^{5} b(x=r, n=10, p=0.5)$$</div>
<div class="math">$$b(x\leq 5, n=10, p=0.5)= 0.623$$</div>
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<h1>
Geometric distribution
</h1>
<ol class="breadcrumb">
<li>
<time class="published" datetime="2018-11-11T21:19:00+01:00">
11 novembre 2018
</time>
</li>
<li>Mathematics</li>
<li>Statistics</li>
</ol>
</header>
<div class='article_content'>
<h2>Negative Binomial Experiment</h2>
<p>A negative binomial experiment is a statistical experiment that has the following properties:</p>
<ul>
<li>The experiment consists of n repeated trials.</li>
<li>The trials are independent.</li>
<li>The outcome of each trial is either success (s) or failure (f).</li>
<li><span class="math">\(P(s)\)</span> is the same for every trial.</li>
<li>The experiment continues until x successes are observed</li>
</ul>
<p>If <span class="math">\(X\)</span> is the number of experiments until the <span class="math">\(x^{th}\)</span> success occures, then <span class="math">\(X\)</span> is a discrete random variable called a negative binomial</p>
<h2>Negative Binomial Distribution</h2>
<p>Consider the following probability mass function:
</p>
<div class="math">$$b^*(x,n,p) = {\binom{n-1}{x-1}}p^xq^{n-x}$$</div>
<p>
The function above is negative binomial and has the following properties:</p>
<ul>
<li>The number of successes to be observed is <span class="math">\(x\)</span>.</li>
<li>The total number of trials is <span class="math">\(n\)</span>.</li>
<li>The probability of success of 1 trial is <span class="math">\(p\)</span>.</li>
<li>The probability of failure of 1 trial <span class="math">\(q\)</span>, where <span class="math">\(q=1-p\)</span>.</li>
<li><span class="math">\(b^*(x,n,p)\)</span> is the <em>negative binomial probability</em>, meaning the probability of having exactly <span class="math">\(x-1\)</span> successes out of <span class="math">\(n-1\)</span> trials and having <span class="math">\(x\)</span> successes after <span class="math">\(n\)</span> trials.</li>
</ul>
<h2>Geometric Distribution</h2>
<p>The geometric distribution is a special case of the negative binomial distribution that deals with the number of Bernoulli trials required to get a success (i.e., counting the number of failures before the first success). Recall that <span class="math">\(X\)</span> is the number of successes in <span class="math">\(n\)</span> independent Bernoulli trials, so for each <span class="math">\(i\)</span> (where $1\leq i\leq n):</p>
<p>$ X_i = </p>
<div class="math">\begin{cases}
1 if the i^{th} trial is a success \\
0 otherwise x
\end{cases}</div>
<p>
$</p>
<p>The geometric distribution is a negative binomial distribution where the number of successes is 1. We express this with the following formula:</p>
<div class="math">$$g(n,p)=q^{n-1}p$$</div>
<h3>Example</h3>
<p>Bob is a high school basketball player. He is a 70% free throw shooter, meaning his probability of making a free throw is 0.7. What is the probability that Bob makes his first free throw on his fifth shot?</p>
<p>For this experiment n=5, p=0.7 and q=0.3
So :
</p>
<div class="math">$$g(n=5, p=0.7)=0.3^4 0.7=0.00567$$</div>
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python/counting.html
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python/dask_distributed_parallelism.html
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python/design_own_libs.html
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python/sorting.html
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@ -192,7 +192,7 @@ Sorted array : [0.13876208 0.19939047 0.40021786 0.43734158 0.46015169]
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python/unique.html
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python/updating_all_python_package_with_anaconda.html
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search.html
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@ -5,7 +5,7 @@ xmlns="http://www.sitemaps.org/schemas/sitemap/0.9">
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sql/Sorting_results.html
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sql/display_table_filter.html
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tipuesearch_content.json
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