bayes exercises

Data/SimulatedDataGroupProjectDynamics.ipynb

@@ -195,7 +195,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 19,
+   "execution_count": 7,
    "metadata": {},
    "outputs": [
     {
@@ -216,7 +216,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 15,
+   "execution_count": 8,
    "metadata": {},
    "outputs": [
     {
@@ -225,10 +225,10 @@
        "$\\hat{y} = 4.712 +0.227 x_{1}$"
       ],
       "text/plain": [
-       "<statwrap.utils.RegressionLine at 0x13c6928d0>"
+       "<statwrap.utils.RegressionLine at 0x13a1d5c10>"
       ]
      },
-     "execution_count": 15,
+     "execution_count": 8,
      "metadata": {},
      "output_type": "execute_result"
     }
@@ -246,7 +246,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 10,
+   "execution_count": 9,
    "metadata": {},
    "outputs": [],
    "source": [
@@ -256,7 +256,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 11,
+   "execution_count": 10,
    "metadata": {},
    "outputs": [
     {
@@ -265,10 +265,10 @@
        "$\\hat{y} = 0.047 +1.491 x_{1}$"
       ],
       "text/plain": [
-       "<statwrap.utils.RegressionLine at 0x13c6696d0>"
+       "<statwrap.utils.RegressionLine at 0x138653690>"
       ]
      },
-     "execution_count": 11,
+     "execution_count": 10,
      "metadata": {},
      "output_type": "execute_result"
     }
@@ -279,7 +279,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 12,
+   "execution_count": 11,
    "metadata": {},
    "outputs": [
     {
@@ -288,10 +288,10 @@
        "$\\hat{y} = 0.056 +0.659 x_{1}$"
       ],
       "text/plain": [
-       "<statwrap.utils.RegressionLine at 0x13c576490>"
+       "<statwrap.utils.RegressionLine at 0x13a3db790>"
       ]
      },
-     "execution_count": 12,
+     "execution_count": 11,
      "metadata": {},
      "output_type": "execute_result"
     }
@@ -302,7 +302,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 20,
+   "execution_count": 12,
    "metadata": {},
    "outputs": [
     {

book/_build/html/_sources/bayes.md

@@ -122,6 +122,107 @@ name: bayesTest
 
 $$\mathbb{P}(Q \mid \cdot U) = \frac{\mathbb{P}(QU)}{\mathbb{P}(\cdot U)}$$
 
-$\mathbb{P}(\cdot U)$ is found by summing the probability for each of the paths that terminate in $U$. This is $\frac{1}{26} + \frac{25}{676}$. 
+$\mathbb{P}(\cdot U)$ is found by summing the probability for each of the paths that terminate in $U$. This is $\frac{1}{26} + \frac{25}{676}$,
 
-$$\mathbb{P}(Q \mid \cdot U) = \frac{\frac{1}{26}}{\frac{1}{26} + \frac{25}{676}} = \frac{26}{51}$$
+$$\mathbb{P}(Q \mid \cdot U) = \frac{\frac{1}{26}}{\frac{1}{26} + \frac{25}{676}} = \frac{26}{51}.$$
+
+
+## Exercises
+
+```{exercise-start}
+:label: boxes
+```
+
+A box contains two tickets, labeled $H$ or $T$. There is a 25% chance the box contains two $H$s. There is a 25% chance the box contains two $T$s. There is a 50% chance the box contains one $H$ and one $T$. 
+
+1. What is the chance of drawing an $H$? 
+
+2. Suppose you draw an $H$. What is the chance that the box contained two $T$s?
+
+3. Suppose you draw an $H$. What is the chance that the box contained two $H$s?
+
+4. After replacing the $H$, what is the chance of selecting another $H$? 
+
+```{exercise-end}
+```
+
+```{exercise-start}
+:label: bayesraredisease
+```
+Consider a rare disease that affects 1 in 10,000 people in a population. A medical test for the disease has a 99% chance of correctly identifying a diseased person (true positive) and a 99% chance of correctly identifying a non-diseased person (true negative).
+
+If a person from this population tests positive for the disease, what is the probability that they actually have the disease?
+
+Given:
+
+$$\mathbb{P}(\text{Disease}) = \frac{1}{10,000}$$
+
+$$\mathbb{P}(\text{No disease}) = 1 - \mathbb{P}(D)$$
+
+$$\mathbb{P}(\text{Positive} | \text{Disease}) = 0.99$$
+
+$$\mathbb{P}(\text{Negative} | \text{No Disease}) = 0.99$$
+```{exercise-end}
+```
+
+
+```{exercise-start}
+:label: hatcoins
+```
+
+You're playing basketball at the park when your team picks up an unknown player. The unknown player is equally likely to be a scrub or a baller. A baller makes 90% of their shots and each shot is independent. A scrub makes 10% of their shots and each shot is independent. 
+
+1. What is the chance an unknown makes their first shot? 
+2. What is the chance that the player makes their second shot if they made their first? 
+3. Are the first and second shot outcomes independent from *your* perspective? 
+
+
+```{exercise-end}
+```
+
+
+```{exercise-start}
+:label: troll
+```
+
+A pilgrim, traveling home, is wandering through a strange land when a troll appears:
+
+> *Woe, to pass, pilgrim choose of these doors two. <br> Which is which, I cannot reveal to you. <br> Home with chance 7 or 73. <br> It depends on your choice and fate's decree. <br> Independent but certainty you lack. <br> Take now two draws before I turn thee back.*
+
+1. What is the probability the pilgrim opens a door that leads home on the first draw?
+
+2. The troll, old in his years, has seen 200 million other pilgrims pass through. Each has chosen a door randomly to start and then the other door second if the first didn't take them home. Finish filling in the table below with the expected counts.
+
+|                   | Door 7 second | Door 73 second | Home after first |
+|-------------------|---------------|----------------|------------------|
+| Door 7 first      | 0 million     |               |                  |
+| Door 73 first     |               | 0 million     | 73 million       |
+
+3. If the first draw does not lead home, what is the probability the pilgrim opened the door that leads home with chance 7%? What is the probability the pilgrim opened the door that leads home with chance 73%?
+
+4. If the first draw does not lead home, should the pilgrim open a different door on the next draw or try the same door again? Or does it not matter?
+
+5. What is the probability that the pilgrim remains in exile?
+
+6. The pilgrim, alarmed by the risk of remaining in exile, bargains with the troll to replace the two doors with one 40% chance door. This averages the chances. Is this wise?
+
+```{exercise-end}
+```
+
+```{exercise-start}
+:label: bayesReview
+```
+
+Suppose that a product is sold on Amazon and it has either high quality ($H$) or low quality ($L$). We observe a single product review, which can either be good or bad. Reviewers can be of two types: fake or truth-teller. A fake reviewer always leaves a positive review, regardless of the product quality. A truth-teller reviewer leaves a positive review when the product is high quality and leaves a negative review if the product is low quality.
+
+Assume $\mathbb{P}(H) = \frac{1}{2}$ and that each type of reviewer is equally likely.
+
+- a.) What is $\mathbb{P}(\text{good review})$?
+- b.) What is $\mathbb{P}(\text{good review} \mid H)$?
+- c.) What is $\mathbb{P}(H \mid \text{good review})$?
+- d.) Draw a probability tree that summarizes the probabilities based on product quality, review type, and reviewer type.
+- e.) Are the events of "high quality product" and "good review" independent or dependent? Explain.
+- f.) Suppose the truth-teller is replaced by a joker who leaves a negative review if the product is high quality and a positive review if the product is low quality. What is $\mathbb{P}(H \mid \text{bad review})$?
+
+```{exercise-end}
+```

book/_build/html/_sources/probability.md

@@ -333,8 +333,7 @@ $$ \frac{n!}{k!(n-k)!} p^k (1-p)^{n-k}.$$
 
 Think of $p^k (1-p)^{n-k}$ as the probability of a sequence of $k$ heads followed by $n-k$ tails. The binomial coefficient in front then adjusts that probability to allow for all of the other ways to get $k$ heads–$n-k$ tails followed by $k$ heads for example. This only works for coin flips or similar processes where the individual trials are independent and the probability of a heads or some substitutable event of interest is the same from one trial to the next. These trial outcomes are said to be *independent and identically distributed*, or *iid*. 
 
-**Example** 
-A trick coin comes up heads with probability $p = \frac{2}{3}$. Out of four flips, what is the probability of two heads? 
+**Example**: A trick coin comes up heads with probability $p = \frac{2}{3}$. Out of four flips, what is the probability of two heads? 
 
 ```{dropdown} Two heads
 
@@ -355,6 +354,7 @@ $$ 4\cdot \frac{8}{27}\cdot\frac{1}{3} + \frac{16}{81} = \frac{48}{81} = \frac{1
 ```
 
 
+
 ## Exercises 
 
 ```{exercise-start}

book/_build/html/bayes.html

@@ -426,6 +426,7 @@ <h2> Contents </h2>
 </li>
 </ul>
 </li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#exercises">Exercises</a></li>
 </ul>
             </nav>
         </div>
@@ -527,12 +528,114 @@ <h4>Bayes Theorem with Trees<a class="headerlink" href="#bayes-theorem-with-tree
 <p><a class="reference internal" href="#qwordtree"><span class="std std-numref">Fig. 42</span></a> can help us find the probability that a word starts with <span class="math notranslate nohighlight">\(Q\)</span>, given that the second letter is <span class="math notranslate nohighlight">\(U\)</span>. Maybe that’s helpful if you’re on Wheel of Fortune.</p>
 <div class="math notranslate nohighlight">
 \[\mathbb{P}(Q \mid \cdot U) = \frac{\mathbb{P}(QU)}{\mathbb{P}(\cdot U)}\]</div>
-<p><span class="math notranslate nohighlight">\(\mathbb{P}(\cdot U)\)</span> is found by summing the probability for each of the paths that terminate in <span class="math notranslate nohighlight">\(U\)</span>. This is <span class="math notranslate nohighlight">\(\frac{1}{26} + \frac{25}{676}\)</span>.</p>
+<p><span class="math notranslate nohighlight">\(\mathbb{P}(\cdot U)\)</span> is found by summing the probability for each of the paths that terminate in <span class="math notranslate nohighlight">\(U\)</span>. This is <span class="math notranslate nohighlight">\(\frac{1}{26} + \frac{25}{676}\)</span>,</p>
 <div class="math notranslate nohighlight">
-\[\mathbb{P}(Q \mid \cdot U) = \frac{\frac{1}{26}}{\frac{1}{26} + \frac{25}{676}} = \frac{26}{51}\]</div>
+\[\mathbb{P}(Q \mid \cdot U) = \frac{\frac{1}{26}}{\frac{1}{26} + \frac{25}{676}} = \frac{26}{51}.\]</div>
 </section>
 </section>
 </section>
+<section id="exercises">
+<h2>Exercises<a class="headerlink" href="#exercises" title="Permalink to this heading">#</a></h2>
+<div class="exercise admonition" id="boxes">
+
+<p class="admonition-title"><span class="caption-number">Exercise 32 </span></p>
+<section id="exercise-content">
+<p>A box contains two tickets, labeled <span class="math notranslate nohighlight">\(H\)</span> or <span class="math notranslate nohighlight">\(T\)</span>. There is a 25% chance the box contains two <span class="math notranslate nohighlight">\(H\)</span>s. There is a 25% chance the box contains two <span class="math notranslate nohighlight">\(T\)</span>s. There is a 50% chance the box contains one <span class="math notranslate nohighlight">\(H\)</span> and one <span class="math notranslate nohighlight">\(T\)</span>.</p>
+<ol class="arabic simple">
+<li><p>What is the chance of drawing an <span class="math notranslate nohighlight">\(H\)</span>?</p></li>
+<li><p>Suppose you draw an <span class="math notranslate nohighlight">\(H\)</span>. What is the chance that the box contained two <span class="math notranslate nohighlight">\(T\)</span>s?</p></li>
+<li><p>Suppose you draw an <span class="math notranslate nohighlight">\(H\)</span>. What is the chance that the box contained two <span class="math notranslate nohighlight">\(H\)</span>s?</p></li>
+<li><p>After replacing the <span class="math notranslate nohighlight">\(H\)</span>, what is the chance of selecting another <span class="math notranslate nohighlight">\(H\)</span>?</p></li>
+</ol>
+</section>
+</div>
+<div class="exercise admonition" id="bayesraredisease">
+
+<p class="admonition-title"><span class="caption-number">Exercise 33 </span></p>
+<section id="exercise-content">
+<p>Consider a rare disease that affects 1 in 10,000 people in a population. A medical test for the disease has a 99% chance of correctly identifying a diseased person (true positive) and a 99% chance of correctly identifying a non-diseased person (true negative).</p>
+<p>If a person from this population tests positive for the disease, what is the probability that they actually have the disease?</p>
+<p>Given:</p>
+<div class="math notranslate nohighlight">
+\[\mathbb{P}(\text{Disease}) = \frac{1}{10,000}\]</div>
+<div class="math notranslate nohighlight">
+\[\mathbb{P}(\text{No disease}) = 1 - \mathbb{P}(D)\]</div>
+<div class="math notranslate nohighlight">
+\[\mathbb{P}(\text{Positive} | \text{Disease}) = 0.99\]</div>
+<div class="math notranslate nohighlight">
+\[\mathbb{P}(\text{Negative} | \text{No Disease}) = 0.99\]</div>
+</section>
+</div>
+<div class="exercise admonition" id="hatcoins">
+
+<p class="admonition-title"><span class="caption-number">Exercise 34 </span></p>
+<section id="exercise-content">
+<p>You’re playing basketball at the park when your team picks up an unknown player. The unknown player is equally likely to be a scrub or a baller. A baller makes 90% of their shots and each shot is independent. A scrub makes 10% of their shots and each shot is independent.</p>
+<ol class="arabic simple">
+<li><p>What is the chance an unknown makes their first shot?</p></li>
+<li><p>What is the chance that the player makes their second shot if they made their first?</p></li>
+<li><p>Are the first and second shot outcomes independent from <em>your</em> perspective?</p></li>
+</ol>
+</section>
+</div>
+<div class="exercise admonition" id="troll">
+
+<p class="admonition-title"><span class="caption-number">Exercise 35 </span></p>
+<section id="exercise-content">
+<p>A pilgrim, traveling home, is wandering through a strange land when a troll appears:</p>
+<blockquote>
+<div><p><em>Woe, to pass, pilgrim choose of these doors two. <br> Which is which, I cannot reveal to you. <br> Home with chance 7 or 73. <br> It depends on your choice and fate’s decree. <br> Independent but certainty you lack. <br> Take now two draws before I turn thee back.</em></p>
+</div></blockquote>
+<ol class="arabic simple">
+<li><p>What is the probability the pilgrim opens a door that leads home on the first draw?</p></li>
+<li><p>The troll, old in his years, has seen 200 million other pilgrims pass through. Each has chosen a door randomly to start and then the other door second if the first didn’t take them home. Finish filling in the table below with the expected counts.</p></li>
+</ol>
+<table class="table">
+<thead>
+<tr class="row-odd"><th class="head"><p></p></th>
+<th class="head"><p>Door 7 second</p></th>
+<th class="head"><p>Door 73 second</p></th>
+<th class="head"><p>Home after first</p></th>
+</tr>
+</thead>
+<tbody>
+<tr class="row-even"><td><p>Door 7 first</p></td>
+<td><p>0 million</p></td>
+<td><p></p></td>
+<td><p></p></td>
+</tr>
+<tr class="row-odd"><td><p>Door 73 first</p></td>
+<td><p></p></td>
+<td><p>0 million</p></td>
+<td><p>73 million</p></td>
+</tr>
+</tbody>
+</table>
+<ol class="arabic simple" start="3">
+<li><p>If the first draw does not lead home, what is the probability the pilgrim opened the door that leads home with chance 7%? What is the probability the pilgrim opened the door that leads home with chance 73%?</p></li>
+<li><p>If the first draw does not lead home, should the pilgrim open a different door on the next draw or try the same door again? Or does it not matter?</p></li>
+<li><p>What is the probability that the pilgrim remains in exile?</p></li>
+<li><p>The pilgrim, alarmed by the risk of remaining in exile, bargains with the troll to replace the two doors with one 40% chance door. This averages the chances. Is this wise?</p></li>
+</ol>
+</section>
+</div>
+<div class="exercise admonition" id="bayesReview">
+
+<p class="admonition-title"><span class="caption-number">Exercise 36 </span></p>
+<section id="exercise-content">
+<p>Suppose that a product is sold on Amazon and it has either high quality (<span class="math notranslate nohighlight">\(H\)</span>) or low quality (<span class="math notranslate nohighlight">\(L\)</span>). We observe a single product review, which can either be good or bad. Reviewers can be of two types: fake or truth-teller. A fake reviewer always leaves a positive review, regardless of the product quality. A truth-teller reviewer leaves a positive review when the product is high quality and leaves a negative review if the product is low quality.</p>
+<p>Assume <span class="math notranslate nohighlight">\(\mathbb{P}(H) = \frac{1}{2}\)</span> and that each type of reviewer is equally likely.</p>
+<ul class="simple">
+<li><p>a.) What is <span class="math notranslate nohighlight">\(\mathbb{P}(\text{good review})\)</span>?</p></li>
+<li><p>b.) What is <span class="math notranslate nohighlight">\(\mathbb{P}(\text{good review} \mid H)\)</span>?</p></li>
+<li><p>c.) What is <span class="math notranslate nohighlight">\(\mathbb{P}(H \mid \text{good review})\)</span>?</p></li>
+<li><p>d.) Draw a probability tree that summarizes the probabilities based on product quality, review type, and reviewer type.</p></li>
+<li><p>e.) Are the events of “high quality product” and “good review” independent or dependent? Explain.</p></li>
+<li><p>f.) Suppose the truth-teller is replaced by a joker who leaves a negative review if the product is high quality and a positive review if the product is low quality. What is <span class="math notranslate nohighlight">\(\mathbb{P}(H \mid \text{bad review})\)</span>?</p></li>
+</ul>
+</section>
+</div>
+</section>
 </section>
 
     <script type="text/x-thebe-config">
@@ -607,6 +710,7 @@ <h4>Bayes Theorem with Trees<a class="headerlink" href="#bayes-theorem-with-tree
 </li>
 </ul>
 </li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#exercises">Exercises</a></li>
 </ul>
   </nav></div>
 

book/_build/html/probability.html

@@ -725,8 +725,7 @@ <h3>Binomial Formula<a class="headerlink" href="#id5" title="Permalink to this h
 <div class="math notranslate nohighlight">
 \[ \frac{n!}{k!(n-k)!} p^k (1-p)^{n-k}.\]</div>
 <p>Think of <span class="math notranslate nohighlight">\(p^k (1-p)^{n-k}\)</span> as the probability of a sequence of <span class="math notranslate nohighlight">\(k\)</span> heads followed by <span class="math notranslate nohighlight">\(n-k\)</span> tails. The binomial coefficient in front then adjusts that probability to allow for all of the other ways to get <span class="math notranslate nohighlight">\(k\)</span> heads–<span class="math notranslate nohighlight">\(n-k\)</span> tails followed by <span class="math notranslate nohighlight">\(k\)</span> heads for example. This only works for coin flips or similar processes where the individual trials are independent and the probability of a heads or some substitutable event of interest is the same from one trial to the next. These trial outcomes are said to be <em>independent and identically distributed</em>, or <em>iid</em>.</p>
-<p><strong>Example</strong>
-A trick coin comes up heads with probability <span class="math notranslate nohighlight">\(p = \frac{2}{3}\)</span>. Out of four flips, what is the probability of two heads?</p>
+<p><strong>Example</strong>: A trick coin comes up heads with probability <span class="math notranslate nohighlight">\(p = \frac{2}{3}\)</span>. Out of four flips, what is the probability of two heads?</p>
 <details class="sd-sphinx-override sd-dropdown sd-card sd-mb-3">
 <summary class="sd-summary-title sd-card-header">
 Two heads<div class="sd-summary-down docutils">