47 lines
3.8 KiB
HTML
47 lines
3.8 KiB
HTML
<!DOCTYPE html>
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<html lang="en-US" dir="ltr">
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<head>
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<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
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<meta name="robots" content="noindex, nofollow">
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</head>
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<body class="ignore-math">
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<article class="example example-like"><h2 class="heading">
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<span class="type">Example</span><span class="space heading-divison-mark heading-divison-mark__space"> </span><span class="codenumber">1.3.2</span><span class="period heading-divison-mark heading-divison-mark__period">.</span>
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</h2>
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<div class="para logical">
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<div class="para">An experiment consists of rolling a fair 6-sided die two times. Let <span class="process-math">\(A\)</span> be the event that the sum of the rolls is at least 10. To find <span class="process-math">\(\Pr(A)\text{,}\)</span> we note that <span class="process-math">\(A = \{(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)\}\text{,}\)</span> which has 6 elements. Since <span class="process-math">\(|\Omega| = 36\text{,}\)</span> we have</div>
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<div class="displaymath process-math">
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\begin{gather*}
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\Pr(A) = \frac{|A|}{|\Omega|} = \frac{6}{36} = \frac{1}{6}\text{.}
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\end{gather*}
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</div>
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<div class="autopermalink" aria-hidden="true" data-description="Paragraph"><a tabindex="-1" href="#example-rolls-conditional-1-1" title="Copy heading and permalink for Paragraph" aria-label="Copy heading and permalink for Paragraph">🔗</a></div>
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</div>
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<div class="para logical">
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<div class="para">Let <span class="process-math">\(B\)</span> be the event that the first roll is 6. If we pause after the first die roll seeing the value of 6, we might be more inclined to expect a sum of at least 10. The evidence that we’ve already seen changes our understanding of the situation. Since <span class="process-math">\(|B| = 6\text{,}\)</span> <span class="process-math">\(\Pr(B) = \frac{6}{36} = \frac{1}{6}\text{.}\)</span> Also, <span class="process-math">\(A\cap B = \{(6, 4), (6, 5), (6, 6)\}\text{,}\)</span> so <span class="process-math">\(\Pr(A\cap B) = \frac{3}{36} = \frac{1}{12}\text{.}\)</span> Finally:</div>
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<div class="displaymath process-math">
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\begin{gather*}
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\Pr(A \mid B) = \frac{\Pr(A\cap B)}{\Pr(B)} = \frac{1/12}{1/6} = \frac{1}{2}
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\end{gather*}
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</div>
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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>
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<div class="autopermalink" aria-hidden="true" data-description="Paragraph"><a tabindex="-1" 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>
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<div class="autopermalink" aria-hidden="true" data-description="Example 1.3.2"><a tabindex="-1" 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>
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</article><span class="incontext"><a class="internal" href="sec-Conditional-Probability.html#example-rolls-conditional">In Context</a></span>
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</body>
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