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2026-01-11 14:43:18 -05:00
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knowl/xref/example-rolls-conditional.html
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@@ -36,6 +36,7 @@
\Pr(A \mid B) = \frac{\Pr(A\cap B)}{\Pr(B)} = \frac{1/12}{1/6} = \frac{1}{2}
\end{gather*}
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<div class="para">Before the experiment, we would have said there was only a <span class="process-math">\(\frac{1}{6}\)</span> chance that the sum of the rolls is at least 10. However, with the additional knowledge of seeing the first roll of 6, we find the probability of a sum of at least 10 to be <span class="process-math">\(\frac{1}{2}\text{,}\)</span> substantially more likely than before.</div>
<div class="autopermalink" data-description="Paragraph"><a href="#example-rolls-conditional-1-2" title="Copy heading and permalink for Paragraph" aria-label="Copy heading and permalink for Paragraph">🔗</a></div>
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<div class="autopermalink" data-description="Example 1.3.2"><a href="#example-rolls-conditional" title="Copy heading and permalink for Example 1.3.2" aria-label="Copy heading and permalink for Example 1.3.2">🔗</a></div></article><span class="incontext"><a class="internal" href="sec-Conditional-Probability.html#example-rolls-conditional">in-context</a></span>