724 lines
20 KiB
Plaintext
724 lines
20 KiB
Plaintext
<section xml:id="sec-Probability" xmlns:xi="http://www.w3.org/2001/XInclude">
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<title>Definition of Probability</title>
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<p>
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Now that we have the language to refer to outcomes and events of an
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experiment, we want to start quantifying how likely those outcomes/events
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are to occur.
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</p>
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<definition xml:id="def-probability-distribution">
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<statement>
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<p>
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A <term>probability distribution</term> on a sample space <m>\Omega</m>
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assigns probabilities to every event, satisfying the following
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conditions:
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<ol>
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<li>
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<p>
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<m>\Pr(\Omega) = 1</m>.
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</p>
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</li>
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<li>
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<p>
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<m>0 \leq \Pr(A) \leq 1</m> for any event <m>A</m>.
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</p>
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</li>
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<li>
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<p>
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If <m>A \cap B = \emptyset</m>, then
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<m>\Pr(A\cup B) = \Pr(A) + \Pr(B)</m>.
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</p>
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</li>
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</ol>
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</p>
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</statement>
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</definition>
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<p>
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For small probability spaces (i.e., with finitely many outcomes in the
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sample space), we'll usually assign probabilities to each individual
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outcome, and perhaps list them in a table.
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Then, to find the probability of any event, simply add together the
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probabilities of each outcome in that event.
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</p>
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<example>
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<statement>
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<p>
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An experiment consists of rolling a standard 6-sided die.
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The sample space is <m>\Omega = \{1, 2, 3, 4, 5, 6\}</m>.
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The probability distribution (assuming a fair die) is shown below.
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</p>
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<table>
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<title>Distribution for a fair die</title>
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<tabular halign="center">
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<row bottom="minor">
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<cell><m>x</m></cell>
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<cell><m>\Pr(x)</m></cell>
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</row>
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<row>
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<cell>1</cell>
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<cell><m>1/6</m></cell>
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</row>
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<row>
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<cell>2</cell>
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<cell><m>1/6</m></cell>
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</row>
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<row>
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<cell>3</cell>
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<cell><m>1/6</m></cell>
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</row>
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<row>
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<cell>4</cell>
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<cell><m>1/6</m></cell>
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</row>
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<row>
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<cell>5</cell>
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<cell><m>1/6</m></cell>
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</row>
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<row>
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<cell>6</cell>
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<cell><m>1/6</m></cell>
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</row>
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</tabular>
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</table>
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<p>
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One possible event is <m>A = \{2, 4, 6\}</m>, i.e., the event that the
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result of the roll is even.
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The probability of <m>A</m> is:
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<md>
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<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>
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</md>
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</p>
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</statement>
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</example>
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<definition xml:id="def-discrete-uniform-distribution">
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<statement>
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<p>
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Let <m>\Omega = \{x_1, x_2, \dotsc, x_n\}</m> be a sample space and
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<m>A\subset \Omega</m> an event.
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We refer to the distribution in which <m>\Pr(x_i) = \frac{1}{n}</m> for
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all <m>i</m> as the <term>uniform distribution</term>.
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In this case, it follows that <m>\Pr(A) = \frac{|A|}{|\Omega|}</m>.
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</p>
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</statement>
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</definition>
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<p>
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When we talk about a fair coin flip or a fair die roll, the word "fair" is
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indicating a uniform distribution.
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However, don't make the mistake of assuming that all distributions are
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uniform by default.
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</p>
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<example>
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<statement>
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<p>
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A person picks a random number from 1 to 10.
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What is the probability that they picked 3?
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</p>
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</statement>
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<answer>
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<p>
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Without assuming the distribution is fair (i.e., that each value
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<m>1, 2, \dotsc, 10</m> ) has probability <m>1/10</m> of occurring), we
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don't have enough information to answer this question.
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</p>
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<p>
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In fact, the situation is even more vague than that: the sample space
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itself is unclear.
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Are we only allowed to pick integer values?
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What about fractions like <m>7/2</m>?
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What about irrational numbers like <m>\pi</m>?
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</p>
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</answer>
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</example>
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<exercises xml:id="exercises-Probability">
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<exercise>
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<statement>
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<p>
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Consider the sample space <m>\Omega = \{1, 2, 3, 4, 5, 6, 7, 8\}</m>
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with probability distribution below.
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Calculate the probabilities of <m>A = \{1, 3, 7, 8\}</m>,
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<m>B = \{2, 3, 6, 7\}</m>, <m>A\cup B</m>, and <m>A \cap B</m>.
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</p>
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<table>
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<title/>
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<tabular halign="center">
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<row bottom="minor">
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<cell><m>x</m></cell>
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<cell><m>\Pr(x)</m></cell>
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</row>
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<row>
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<cell>1</cell>
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<cell>0.1</cell>
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</row>
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<row>
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<cell>2</cell>
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<cell>0.05</cell>
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</row>
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<row>
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<cell>3</cell>
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<cell>0.2</cell>
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</row>
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<row>
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<cell>4</cell>
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<cell>0.15</cell>
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</row>
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<row>
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<cell>5</cell>
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<cell>0.15</cell>
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</row>
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<row>
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<cell>6</cell>
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<cell>0.1</cell>
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</row>
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<row>
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<cell>7</cell>
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<cell>0.05</cell>
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</row>
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<row>
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<cell>8</cell>
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<cell>0.1</cell>
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</row>
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<row>
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<cell>9</cell>
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<cell>0.1</cell>
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</row>
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</tabular>
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</table>
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</statement>
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<hint>
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<p>
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Remember that, to calculate the probability of an event, you should
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add up the probabilities of each outcome in the event.
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</p>
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</hint>
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<answer>
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<p>
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<m>\Pr(A) = 0.45, \Pr(B) = 0.4, \Pr(A \cup B) = 0.6, \Pr(A \cap B) = 0.25.</m>
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</p>
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</answer>
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<solution>
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<p>
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<md>
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<mrow> \Pr(A) \amp = \Pr(1) + \Pr(3) + \Pr(7) + \Pr(8) </mrow>
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<mrow> \amp = 0.1 + 0.2 + 0.05 + 0.1 </mrow>
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<mrow> \amp = 0.45 </mrow>
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<mrow> \Pr(B) \amp = \Pr(2) + \Pr(3) + \Pr(6) + \Pr(7) </mrow>
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<mrow> \amp = 0.05 + 0.2 + 0.1 + 0.05 </mrow>
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<mrow> \amp = 0.4 </mrow>
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</md>
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The other events are <m>A\cup B = \{1, 2, 3, 6, 7, 8\}</m> and
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<m>A\cap B = \{3, 7\}</m>, so:
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<md>
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<mrow> \Pr(A \cup B) \amp = \Pr(1) + \Pr(2) + \Pr(3) + \Pr(6) + \Pr(7) + \Pr(8) </mrow>
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<mrow> \amp = 0.1 + 0.05 + 0.2 + 0.1 + 0.05 + 0.1 </mrow>
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<mrow> \amp = 0.6 </mrow>
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<mrow> \Pr(A \cap B) \amp = \Pr(3) + \Pr(7) </mrow>
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<mrow> \amp = 0.2 + 0.05 </mrow>
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<mrow> \amp = 0.25 </mrow>
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</md>
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</p>
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</solution>
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</exercise>
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<exercise>
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<introduction>
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<p>
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Suppose we flip a coin two times.
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Answer the questions below.
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What about three flips?
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What about four flips?
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</p>
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</introduction>
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<task>
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<statement>
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<p>
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Write all outcomes in the sample space <m>\Omega</m>.
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</p>
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</statement>
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<answer>
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<p>
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<m>\Omega = \{HH, HT, TH, TT\}.</m>
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</p>
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</answer>
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</task>
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<task>
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<statement>
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<p>
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Make a probability distribution table for <m>\Omega</m> assuming the
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coin is fair.
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</p>
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</statement>
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<answer>
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<table>
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<title>Probability Distribution for Two Fair Coin Flips</title>
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<tabular halign="center">
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<row bottom="minor">
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<cell><m>x</m></cell>
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<cell><m>\Pr(x)</m></cell>
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</row>
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<row>
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<cell><m>HH</m></cell>
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<cell>0.25</cell>
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</row>
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<row>
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<cell><m>HT</m></cell>
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<cell>0.25</cell>
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</row>
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<row>
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<cell><m>TH</m></cell>
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<cell>0.25</cell>
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</row>
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<row>
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<cell><m>TT</m></cell>
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<cell>0.25</cell>
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</row>
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</tabular>
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</table>
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</answer>
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</task>
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<task>
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<statement>
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<p>
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Make a probability distribution table assuming the coin comes up
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heads with probability 0.3.
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</p>
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</statement>
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<answer>
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<table>
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<title>Probability Distribution for Two Fair Coin Flips</title>
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<tabular halign="center">
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<row bottom="minor">
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<cell><m>x</m></cell>
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<cell><m>\Pr(x)</m></cell>
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</row>
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<row>
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<cell><m>HH</m></cell>
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<cell>0.09</cell>
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</row>
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<row>
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<cell><m>HT</m></cell>
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<cell>0.21</cell>
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</row>
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<row>
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<cell><m>TH</m></cell>
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<cell>0.21</cell>
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</row>
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<row>
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<cell><m>TT</m></cell>
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<cell>0.49</cell>
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</row>
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</tabular>
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</table>
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</answer>
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</task>
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</exercise>
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<exercise>
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<introduction>
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<p>
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Suppose we roll a fair 6-sided die two times.
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Answer the questions below.
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</p>
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</introduction>
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<task>
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<statement>
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<p>
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Write all outcomes in the sample space <m>\Omega</m>.
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</p>
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</statement>
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<answer>
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<p>
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<md>
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<mrow> \Omega = \{\amp (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), </mrow>
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<mrow> \amp (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), </mrow>
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<mrow> \amp (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), </mrow>
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<mrow> \amp (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), </mrow>
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<mrow> \amp (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), </mrow>
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<mrow> \amp (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)\} </mrow>
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</md>
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</p>
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</answer>
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</task>
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<task>
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<statement>
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<p>
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Make a probability distribution table for <m>\Omega</m> assuming the
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die is fair.
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</p>
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</statement>
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<answer>
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<p>
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We'll avoid an overly large table and note that, since the die is
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fair, every outcome is equally likely.
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Therefore, <m>\Pr(x) = \frac{1}{36}</m> for every
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<m>x\in \Omega</m>.
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</p>
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</answer>
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</task>
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<task>
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<statement>
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<p>
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Let <m>A</m> be the event that the second roll is higher than the
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first, and let <m>B</m> be the event that the first roll is even.
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Find <m>\Pr(A), \Pr(B)</m>, and <m>\Pr(A \cap B)</m>.
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</p>
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</statement>
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<answer>
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<p>
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<m>\Pr(A) = \frac{5}{12}, \Pr(B) = \frac{1}{2}, \Pr(A \cap B) = \frac{1}{6}</m>.
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</p>
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</answer>
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<solution>
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<p>
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<md>
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<mrow> A = \{ \amp (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), </mrow>
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<mrow> \amp (2, 3), (2, 4), (2, 5), (2, 6), </mrow>
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<mrow> \amp (3, 4), (3, 5), (3, 6), </mrow>
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<mrow> \amp (4, 5), (4, 6), </mrow>
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<mrow> \amp (5, 6)\} </mrow>
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<mrow> B = \{ \amp (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), </mrow>
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<mrow> \amp (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), </mrow>
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<mrow> \amp (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)\} </mrow>
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<mrow> A \cap B = \{ \amp (2, 3), (2, 4), (2, 5), (2, 6), </mrow>
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<mrow> \amp (4, 5), (4, 6)\} </mrow>
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</md>
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Therefore
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<m>\Pr(A) = \frac{15}{36} = \frac{5}{12}, \Pr(B) = \frac{18}{36} = \frac{1}{2}, \Pr(A \cap B) = \frac{6}{36} = \frac{1}{6}.</m>
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</p>
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</solution>
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</task>
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</exercise>
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<exercise>
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<statement>
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<p>
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Suppose a die has the values <m>1, 2, 3, 4, 5, 6</m> on the faces, but
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the die is not fair.
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Instead, the probabilities scale by the same amount as the face
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values.
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For example, a result of 4 is twice as likely as a result of 2, since
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4 is twice as large as 2; a result of 6 is six times more likely than
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a result of 1; and so on.
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Write a probability distribution table for this die.
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</p>
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</statement>
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<answer>
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<table>
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<title>Probability Distribution for a Linearly Scaled Die</title>
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<tabular halign="center">
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<row bottom="minor">
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<cell><m>x</m></cell>
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<cell><m>\Pr(x)</m></cell>
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</row>
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<row>
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<cell>1</cell>
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<cell><m>1/21</m></cell>
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</row>
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<row>
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<cell>2</cell>
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<cell><m>2/21</m></cell>
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</row>
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<row>
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<cell>3</cell>
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<cell><m>3/21</m></cell>
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</row>
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<row>
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<cell>4</cell>
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<cell><m>4/21</m></cell>
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</row>
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<row>
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<cell>5</cell>
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<cell><m>5/21</m></cell>
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</row>
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<row>
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<cell>6</cell>
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<cell><m>6/21</m></cell>
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</row>
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</tabular>
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</table>
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</answer>
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<solution>
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<p>
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Let <m>\Pr(1) = x</m>.
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Then <m>\Pr(2) = 2 \Pr(1) = 2x</m>, and <m>\Pr(3) = 3\Pr(1) = 3x</m>,
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and so on.
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So the total probability in the space is:
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<md>
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<mrow> Pr(\Omega) \amp = \Pr(1) + \Pr(2) + \Pr(3) + \Pr(4) + \Pr(5) + \Pr(6) </mrow>
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<mrow> \amp = x + 2x + 3x + 4x + 5x + 6x </mrow>
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<mrow> \amp = 21x </mrow>
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</md>
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Since the total probability must add up to 1, we have <m>1 = 21x</m>,
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so <m>x = \frac{1}{21}</m>, and we can calculate the rest of the
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probabilities from there.
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</p>
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<table>
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<title>Probability Distribution for a Linearly Scaled Die</title>
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<tabular halign="center">
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<row bottom="minor">
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<cell><m>x</m></cell>
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<cell><m>\Pr(x)</m></cell>
|
|
</row>
|
|
|
|
<row>
|
|
<cell>1</cell>
|
|
<cell><m>1/21</m></cell>
|
|
</row>
|
|
|
|
<row>
|
|
<cell>2</cell>
|
|
<cell><m>2/21</m></cell>
|
|
</row>
|
|
|
|
<row>
|
|
<cell>3</cell>
|
|
<cell><m>3/21</m></cell>
|
|
</row>
|
|
|
|
<row>
|
|
<cell>4</cell>
|
|
<cell><m>4/21</m></cell>
|
|
</row>
|
|
|
|
<row>
|
|
<cell>5</cell>
|
|
<cell><m>5/21</m></cell>
|
|
</row>
|
|
|
|
<row>
|
|
<cell>6</cell>
|
|
<cell><m>6/21</m></cell>
|
|
</row>
|
|
</tabular>
|
|
</table>
|
|
</solution>
|
|
</exercise>
|
|
|
|
<exercise>
|
|
<statement>
|
|
<p>
|
|
Suppose a die has the values <m>1, 2, 3, 4, 5, 6</m> on the faces, but
|
|
the die is not fair.
|
|
Instead, each even value has an equal probability, each odd value has
|
|
an equal probability, and the even values are each twice as likely as
|
|
the odd values to appear on a roll.
|
|
Write a probability distribution table for this die.
|
|
</p>
|
|
</statement>
|
|
|
|
<answer>
|
|
<table>
|
|
<title>Probability Distribution for an Even-biased 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>1/9</m></cell>
|
|
</row>
|
|
|
|
<row>
|
|
<cell>2</cell>
|
|
<cell><m>2/9</m></cell>
|
|
</row>
|
|
|
|
<row>
|
|
<cell>3</cell>
|
|
<cell><m>1/9</m></cell>
|
|
</row>
|
|
|
|
<row>
|
|
<cell>4</cell>
|
|
<cell><m>2/9</m></cell>
|
|
</row>
|
|
|
|
<row>
|
|
<cell>5</cell>
|
|
<cell><m>1/9</m></cell>
|
|
</row>
|
|
|
|
<row>
|
|
<cell>6</cell>
|
|
<cell><m>2/9</m></cell>
|
|
</row>
|
|
</tabular>
|
|
</table>
|
|
</answer>
|
|
|
|
<solution>
|
|
<p>
|
|
Let <m>\Pr(1) = x</m>.
|
|
Then <m>\Pr(3) = x</m> and <m>\Pr(5) = x</m>, since all odd rolls must
|
|
have the same probability.
|
|
The even rolls must have twice the probability, so
|
|
<m>\Pr(2) = \Pr(4) = \Pr(6) = 2x</m>.
|
|
Now:
|
|
<md>
|
|
<mrow> Pr(\Omega) \amp = \Pr(1) + \Pr(2) + \Pr(3) + \Pr(4) + \Pr(5) + \Pr(6) </mrow>
|
|
<mrow> \amp = x + 2x + x + 2x + x + 2x </mrow>
|
|
<mrow> \amp = 9x </mrow>
|
|
</md>
|
|
Since the total probability must add up to 1, we have <m>1 = 9x</m>,
|
|
so <m>x = \frac{1}{9}</m>, and we can calculate the rest of the
|
|
probabilities from there.
|
|
</p>
|
|
|
|
<table>
|
|
<title>Probability Distribution for an Even-biased 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>1/9</m></cell>
|
|
</row>
|
|
|
|
<row>
|
|
<cell>2</cell>
|
|
<cell><m>2/9</m></cell>
|
|
</row>
|
|
|
|
<row>
|
|
<cell>3</cell>
|
|
<cell><m>1/9</m></cell>
|
|
</row>
|
|
|
|
<row>
|
|
<cell>4</cell>
|
|
<cell><m>2/9</m></cell>
|
|
</row>
|
|
|
|
<row>
|
|
<cell>5</cell>
|
|
<cell><m>1/9</m></cell>
|
|
</row>
|
|
|
|
<row>
|
|
<cell>6</cell>
|
|
<cell><m>2/9</m></cell>
|
|
</row>
|
|
</tabular>
|
|
</table>
|
|
</solution>
|
|
</exercise>
|
|
|
|
<exercise>
|
|
<statement>
|
|
<p>
|
|
A toxin molecule inside a cell has a 0.3 probability of leaving the
|
|
cell during a 1-minute period.
|
|
For each value of <m>n = 1, 2, 3, \dotsc</m>, find the probability of
|
|
the toxin molecule leaving the cell during the <m>n</m> th minute.
|
|
What is the probability of the molecule leaving the cell during the
|
|
first 3 minutes?
|
|
</p>
|
|
</statement>
|
|
|
|
<answer>
|
|
<p>
|
|
Write <m>T</m> for the minute that the toxin molecule leaves the cell.
|
|
Then <m>\Pr(T = n) = (0.7)^{n - 1} (0.3)</m>, and <m>\Pr(T \leq 3) = 0.657</m>.
|
|
</p>
|
|
</answer>
|
|
|
|
<solution>
|
|
<p>
|
|
Write <m>T</m> for the minute that the toxin molecule leaves the cell.
|
|
We're told that the toxin molecule has probability <m>0.3</m> of leaving during each minute.
|
|
Therefore, the probability of remaining during a particular minute is <m>1 - 0.3 = 0.7</m>.
|
|
</p>
|
|
|
|
<p>
|
|
For the toxin molecule to leave the cell during minute <m>n</m>, it must remain for minutes <m>1, 2, \dotsc, n - 1</m>, and then leave during minute <m>n</m>.
|
|
The probability to remain each minute is <m>0.7</m>, so we must multiply <m>n - 1</m> copies of <m>0.7</m>.
|
|
Then the probability of leaving is <m>0.3</m>, so we multiply by a factor of <m>0.3</m>, yielding the formula:
|
|
<md>
|
|
<mrow> \Pr(T = n) = (0.7)^{n - 1}(0.3). </mrow>
|
|
</md>
|
|
</p>
|
|
|
|
<p>
|
|
Then, the probaiblity of the molecule leaving within the first three minutes will be:
|
|
<md>
|
|
<mrow> \Pr(T \leq 3) \amp = \Pr(T = 1) + \Pr(T = 2) + \Pr(T = 3) </mrow>
|
|
<mrow> \amp = 0.3 + (0.7)(0.3) + (0.7)^2(0.3) </mrow>
|
|
<mrow> \amp = 0.657. </mrow>
|
|
</md>
|
|
</p>
|
|
</solution>
|
|
</exercise>
|
|
</exercises>
|
|
</section> |