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Figured out the list error

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andyeisenberg
2025-12-30 16:59:11 +00:00
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commit 82889f6a06
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source/ch-Probability.ptx
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@@ -62,9 +62,7 @@
<statement> <statement>
<p> <p>
Consider sets <m>A</m> and <m>B</m>, each contained inside <m>\Omega</m>. Consider sets <m>A</m> and <m>B</m>, each contained inside <m>\Omega</m>.
We can combine sets in a variety of ways: We can combine sets in a variety of ways: <dl>
</p>
<dl>
<li> <li>
<title>Union</title> <title>Union</title>
@@ -105,6 +103,7 @@
</p> </p>
</li> </li>
</dl> </dl>
</p>
</statement> </statement>
</definition> </definition>
@@ -220,8 +219,6 @@
<statement> <statement>
<p> <p>
A <term>probability distribution</term> on a sample space <m>\Omega</m> assigns probabilities to every event, satisfying the following conditions: A <term>probability distribution</term> on a sample space <m>\Omega</m> assigns probabilities to every event, satisfying the following conditions:
</p>
<ol> <ol>
<li> <li>
<p> <p>
@@ -241,9 +238,74 @@
</p> </p>
</li> </li>
</ol> </ol>
</p>
</statement> </statement>
</definition> </definition>
<p>
For small probability spaces (i.e., with finitely many outcomes in the sample space), we'll usually assign probabilities to each individual outcome, and perhaps list them in a table.
Then, to find the probability of any event, simply add together the probabilities of each outcome in that event.
</p>
<example>
<statement>
<p>
An experiment consists of rolling a standard 6-sided die.
The sample space is <m>\Omega = \{1, 2, 3, 4, 5, 6\}</m>.
The probability distribution (assuming a fair die) is shown below.
</p>
<table>
<title>Distribution for a fair die</title>
<tabular halign="center">
<row bottom="minor">
<cell><m>x</m></cell>
<cell><m>\Pr(x)</m></cell>
</row>
<row>
<cell>1</cell>
<cell><m>1/6</m></cell>
</row>
<row>
<cell>2</cell>
<cell><m>1/6</m></cell>
</row>
<row>
<cell>3</cell>
<cell><m>1/6</m></cell>
</row>
<row>
<cell>4</cell>
<cell><m>1/6</m></cell>
</row>
<row>
<cell>5</cell>
<cell><m>1/6</m></cell>
</row>
<row>
<cell>6</cell>
<cell><m>1/6</m></cell>
</row>
</tabular>
</table>
<p>
One possible event is <m>A = \{2, 4, 6\}</m>, i.e., the event that the result of the roll is even.
The probability of <m>A</m> is:
<md>
<mrow> \Pr(A) = \Pr(2) + \Pr(4) + \Pr(6) = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2} </mrow>
</md>
</p>
</statement>
</example>
<exercises xml:id="exercises-Probability"> <exercises xml:id="exercises-Probability">
<exercise> <exercise>
<statement> <statement>
@@ -255,31 +317,55 @@
<table> <table>
<title></title> <title></title>
<tabular> <tabular halign="center">
<row> <row bottom="minor">
<cell halign="center"><m>x</m></cell> <cell><m>x</m></cell>
<cell halign="center">1</cell> <cell><m>\Pr(x)</m></cell>
<cell halign="center">2</cell>
<cell halign="center">3</cell>
<cell halign="center">4</cell>
<cell halign="center">5</cell>
<cell halign="center">6</cell>
<cell halign="center">7</cell>
<cell halign="center">8</cell>
<cell halign="center">9</cell>
</row> </row>
<row> <row>
<cell halign="center"><m>\Pr(x)</m></cell> <cell>1</cell>
<cell halign="center">0.1</cell> <cell>0.1</cell>
<cell halign="center">0.05</cell> </row>
<cell halign="center">0.2</cell>
<cell halign="center">0.15</cell> <row>
<cell halign="center">0.15</cell> <cell>2</cell>
<cell halign="center">0.1</cell> <cell>0.05</cell>
<cell halign="center">0.05</cell> </row>
<cell halign="center">0.1</cell>
<cell halign="center">0.1</cell> <row>
<cell>3</cell>
<cell>0.2</cell>
</row>
<row>
<cell>4</cell>
<cell>0.15</cell>
</row>
<row>
<cell>5</cell>
<cell>0.15</cell>
</row>
<row>
<cell>6</cell>
<cell>0.1</cell>
</row>
<row>
<cell>7</cell>
<cell>0.05</cell>
</row>
<row>
<cell>8</cell>
<cell>0.1</cell>
</row>
<row>
<cell>9</cell>
<cell>0.1</cell>
</row> </row>
</tabular> </tabular>
</table> </table>
@@ -511,35 +597,9 @@
<exercise> <exercise>
<statement> <statement>
<p> <p>
Let <m>A = \{1, 2, 3\}</m> and <m>B = \{3, 4, 5\}</m>. Let <m>A = \{1, 2, 3\}</m> and <m>B = \{3, 4, 5\}</m> be events in the sample space <m>\Omega = \{1, 2, 3, 4, 5, 6\}</m>.
Fill in the following probability distribution table so that <m>A, B</m> are independent. Create a probability distribution for <m>\Omega</m> so that <m>A, B</m> are independent.
</p> </p>
<table>
<title></title>
<tabular>
<row>
<cell halign="center"><m>x</m></cell>
<cell halign="center">1</cell>
<cell halign="center">2</cell>
<cell halign="center">3</cell>
<cell halign="center">4</cell>
<cell halign="center">5</cell>
<cell halign="center">6</cell>
</row>
<row>
<cell halign="center"><m>\Pr(x)</m></cell>
<cell halign="center"></cell>
<cell halign="center"></cell>
<cell halign="center"></cell>
<cell halign="center"></cell>
<cell halign="center"></cell>
<cell halign="center"></cell>
</row>
</tabular>
</table>
</statement> </statement>
</exercise> </exercise>
</exercises> </exercises>