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<span id="cb7-9"><a href="#cb7-9" aria-hidden="true" tabindex="-1"></a> gt<span class="sc">::</span><span class="fu">cols_width</span>(<span class="fu">everything</span>() <span class="sc">~</span> <span class="fu">px</span>(<span class="dv">100</span>))</span></code><button title="Copy to Clipboard" class="code-copy-button"><i class="bi"></i></button></pre></div>
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@ -733,11 +733,60 @@ Probability and Likelihood
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</tr>
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</tbody>
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</table>
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<p>A couple things to note about our table (1) + (2) = .4 and (2) + (4) = .6. (1) + (2) + (3) + (4) = 1.</p>
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<ol type="1">
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<li><p><span class="math inline">\(P(A \cap B) = P(A|B)P(B)\)</span> we know the likelihood of <span class="math inline">\(L(B|A) = P(A|B)\)</span> and we also know the prior so we insert these to get <span class="math display">\[ P(A \cap B) = P(A|B)P(B) = .267 \times .4 = .1068\]</span></p></li>
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<li><p><span class="math inline">\(P(A^c \cap B) = P(A^c|B)P(B)\)</span> in this case we do know the prior <span class="math inline">\(P(B) = .4\)</span>, but we don’t directly know the value of <span class="math inline">\(P(A^c|B)\)</span>, however, we note that <span class="math inline">\(P(A|B) + P(A^c|B) = 1\)</span>, therefore we compute <span class="math inline">\(P(A^c|B) = 1 - P(A|B) = 1 - .267 = .733\)</span> <span class="math display">\[ P(A^c \cap B) = P(A^c|B)P(B) = .733 \times .4 = .2932\]</span></p></li>
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</ol>
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<p>A couple things to note about our table (1) + (3) = .4 and (2) + (4) = .6. (1) + (2) + (3) + (4) = 1.</p>
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<p>(1.) <span class="math inline">\(P(A \cap B) = P(A|B)P(B)\)</span> we know the likelihood of <span class="math inline">\(L(B|A) = P(A|B)\)</span> and we also know the prior so we insert these to get <span class="math display">\[ P(A \cap B) = P(A|B)P(B) = .267 \times .4 = .1068\]</span></p>
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<p>(3.) <span class="math inline">\(P(A^c \cap B) = P(A^c|B)P(B)\)</span> in this case we do know the prior <span class="math inline">\(P(B) = .4\)</span>, but we don’t directly know the value of <span class="math inline">\(P(A^c|B)\)</span>, however, we note that <span class="math inline">\(P(A|B) + P(A^c|B) = 1\)</span>, therefore we compute <span class="math inline">\(P(A^c|B) = 1 - P(A|B) = 1 - .267 = .733\)</span> <span class="math display">\[ P(A^c \cap B) = P(A^c|B)P(B) = .733 \times .4 = .2932\]</span></p>
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<p>we now can confirm that <span class="math inline">\(.1068 + .2932 = .4\)</span></p>
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<p>Moving on to (2), (4)</p>
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<p>(2.) <span class="math inline">\(P(A \cap B^c) = P(A|B^c)P(B^c)\)</span>. In this case know the likelihood <span class="math inline">\(L(B^c|A) = P(A|B^c)\)</span> and we know the prior <span class="math inline">\(P(B^c)\)</span> therefore, <span class="math display">\[P(A \cap B^c) = P(A|B^c)P(B^c) = .022 \times .6 = .0132\]</span></p>
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<p>(4.) <span class="math inline">\(P(A^c \cap B^c) = P(A^c|B^c)P(B^c) = (1 - .022) \times .6 = .5868\)</span></p>
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<p>and can confirm that <span class="math inline">\(.0132 + .5868 = .6\)</span></p>
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<p>and we can fill the rest of the table:</p>
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<table class="table">
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<thead>
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<tr class="header">
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<th></th>
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<th><span class="math inline">\(B\)</span></th>
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<th><span class="math inline">\(B^c\)</span></th>
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<th>Total</th>
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</tr>
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</thead>
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<tbody>
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<tr class="odd">
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<td><span class="math inline">\(A\)</span></td>
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<td>.1068</td>
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<td>.0132</td>
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<td>.12</td>
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</tr>
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<tr class="even">
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<td><span class="math inline">\(A^c\)</span></td>
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<td>.2932</td>
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<td>.5868</td>
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<td>.88</td>
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</tr>
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<tr class="odd">
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<td>Total</td>
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<td>.4</td>
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<td>.6</td>
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<td>1</td>
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</tr>
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</tbody>
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</table>
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<p>An important concept we implemented in above is the idea of <strong>total probability</strong></p>
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<div class="callout-tip callout callout-style-default no-icon callout-captioned">
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<div class="callout-header d-flex align-content-center">
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<div class="callout-icon-container">
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<i class="callout-icon no-icon"></i>
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</div>
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<div class="callout-caption-container flex-fill">
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total probability
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</div>
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</div>
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<div class="callout-body-container callout-body">
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<p>The <strong>total probability</strong> of observing a real article is made up the sum of its parts. Namely</p>
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<p><span class="math display">\[P(B^c) = P(A \cap B^c) + P(A^c \cap B^c)\]</span> <span class="math display">\[=P(A|B^c)P(B^c) + P(A^c|B^c)P(B^c)\]</span> <span class="math display">\[=.0132 + .5868 = .6\]</span></p>
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</div>
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</div>
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</section>
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</main>
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R/ch2.qmd
37
R/ch2.qmd
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@ -140,13 +140,44 @@ probability table:
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A couple things to note about our table (1) + (3) = .4 and (2) + (4) = .6.
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(1) + (2) + (3) + (4) = 1.
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(1) $P(A \cap B) = P(A|B)P(B)$ we know the likelihood of $L(B|A) = P(A|B)$ and we also
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(1.) $P(A \cap B) = P(A|B)P(B)$ we know the likelihood of $L(B|A) = P(A|B)$ and we also
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know the prior so we insert these to get
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$$ P(A \cap B) = P(A|B)P(B) = .267 \times .4 = .1068$$
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(3) $P(A^c \cap B) = P(A^c|B)P(B)$ in this case we do know the prior $P(B) = .4$, but we
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(3.) $P(A^c \cap B) = P(A^c|B)P(B)$ in this case we do know the prior $P(B) = .4$, but we
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don't directly know the value of $P(A^c|B)$, however, we note that $P(A|B) + P(A^c|B) = 1$,
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therefore we compute $P(A^c|B) = 1 - P(A|B) = 1 - .267 = .733$
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$$ P(A^c \cap B) = P(A^c|B)P(B) = .733 \times .4 = .2932$$
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we now can confirm that $.1068 + .2932 = .4$
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we now can confirm that $.1068 + .2932 = .4$
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Moving on to (2), (4)
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(2.) $P(A \cap B^c) = P(A|B^c)P(B^c)$. In this case know the likelihood $L(B^c|A) = P(A|B^c)$ and
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we know the prior $P(B^c)$ therefore,
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$$P(A \cap B^c) = P(A|B^c)P(B^c) = .022 \times .6 = .0132$$
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(4.) $P(A^c \cap B^c) = P(A^c|B^c)P(B^c) = (1 - .022) \times .6 = .5868$
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and can confirm that $.0132 + .5868 = .6$
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and we can fill the rest of the table:
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| | $B$ | $B^c$ | Total |
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|------|---- |------ |-------|
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|$A$ | .1068 | .0132 | .12 |
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|$A^c$ | .2932 | .5868 | .88 |
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|Total | .4 | .6 | 1 |
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An important concept we implemented in above is the idea of **total probability**
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:::{.callout-tip}
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## total probability
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The **total probability** of observing a real article is made up the sum of its
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parts. Namely
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$$P(B^c) = P(A \cap B^c) + P(A^c \cap B^c)$$
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$$=P(A|B^c)P(B^c) + P(A^c|B^c)P(B^c)$$
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$$=.0132 + .5868 = .6$$
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:::
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