1130 lines
30 KiB
XML
1130 lines
30 KiB
XML
<?xml version="1.0" encoding="UTF-8"?>
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<section xml:id="Exam-1-Review">
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<title>Exam 1 Review</title>
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<introduction>
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<p>
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Use the following problems to prepare for the exam.
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There will be in-class review on Thursday, February 12.
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Your recitation this week will also be exam review.
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</p>
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</introduction>
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<subsection>
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<title>Allowed Materials</title>
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<p>
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You will be allowed to use a scientific calculator (<em>not</em> a graphing calculator, <em>not</em> a calculator app on your phone).
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You may not share a calculator with another student; you must use your own calculator.
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</p>
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<p>
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You may bring a standard 3 in x 5 in index card with prepared notes.
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You may use both sides of the notecard.
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You must put your full name in the top right corner of the card, and turn it in along with your exam.
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</p>
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</subsection>
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<exercises>
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<exercise>
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<introduction>
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<p>
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Consider the sets <m>A = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}</m>, <m>B = \{2, 4, 9, 10, 12, 14, 19\}</m>, and <m>C = \{9, 10, 11, 14, 16, 17, 20\}</m>, which are all subsets of <m>\Omega = \{1, 2, 3, \dotsc, 20\}</m>.
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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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Find <m>A - (B \cap C)</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>\{1, 2, 3, 4, 5, 6, 7, 8\}</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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Find <m>|A|</m>, <m>|B|</m>, <m>|C|</m>, <m>|A\cup B|</m>, <m>|A \cap B|</m>, <m>|B\cap C|</m>, <m>|A\cap C|</m>, and <m>|A\cup B\cup C|</m>.
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Is it true that the size of the union of sets is equal to the sum of the sizes of the individual sets?
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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>|A| = 10</m>, <m>|B| = 7</m>, <m>|C| = 7</m>, <m>|A \cup B| = 13</m>, <m>|A \cap B| = 4</m>, <m>|B\cap C| = 3</m>, <m>|A\cap C| = 2</m>, <m>|A\cup B\cup C| = 17</m>. In particular, note that <m>|A\cup B| = 13 \neq 10 + 7 = |A| + |B|</m>, so it is not true in general that the size of the union of sets is the sum of the sizes of the individual sets.
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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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Find <m>A^c</m> and <m>(A\cup B)^c</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>A^c = \{11, 12, 13, 14, 15, 16, 17, 18, 19, 20\}</m>, <m>(A\cup B)^c = \{11, 13, 15, 16, 17, 18, 20\}</m>.
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</p>
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</answer>
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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 we have a 6-sided die that's weighted to roll a 6 half of the time.
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We roll the die two times.
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List the set of all possible results.
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[Note: the result (2, 4)---rolling a 2 and then a 4---is different from the result <m>(4, 2)</m>---rolling a 4 and then a 2.]
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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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Note that <m>\Omega</m> simply lists outcomes with no reference to the probabilities.
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So the answer here is the same as in <xref ref="example-sample-space"/>.
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</p>
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</answer>
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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 we flip a coin two times.
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List the set of all possible results.
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What about flipping three times? Four times? If we flip the coin 10 times, how many possible results will there be?
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</p>
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</statement>
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<answer>
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<p>
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For two flips: <m>\Omega = \{ HH, HT, TH, TT \}</m>.
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</p>
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<p>
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For three flips: <m>\Omega = \{ HHH, HHT, HTH, THH, HTT, THT, TTH, TTT \}</m>.
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</p>
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<p>
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For four flips:
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<md>
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<mrow> \Omega = \{ \amp HHHH, HHHT, HHTH, HTHH, </mrow>
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<mrow> \amp THHH, HHTT, HTHT, HTTH, </mrow>
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<mrow> \amp THHT, THTH, TTHH, HTTT, </mrow>
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<mrow> \amp THTT, TTHT, TTTH, TTTT \}. </mrow>
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</md>
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</p>
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<p>
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Each additional flip doubles the number of outcomes.
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So, with ten flips, we'll have <m>|\Omega| = 2^{10} = 1024</m>.
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</p>
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</answer>
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</exercise>
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<exercise>
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<statement>
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<p>
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If we roll a 6-sided die ten times, how many possible results will there be?
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</p>
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</statement>
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<answer>
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<p>
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Each additional roll will multiply the number of outcomes by 6.
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So, with 10 rolls, we'll have <m>|\Omega| = 6^{10}.</m>
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</p>
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</answer>
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</exercise>
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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> with probability distribution below.
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Calculate the probabilities of <m>A = \{1, 3, 7, 8\}</m>, <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></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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<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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</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 the die is not fair.
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Instead, the probabilities scale by the same amount as the face values.
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For example, a result of 4 is twice as likely as a result of 2, since 4 is twice as large as 2; a result of 6 is six times more likely than 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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</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 the die is not fair.
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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.
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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 an Even-biased 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>1/9</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>2/9</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>1/9</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>2/9</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>1/9</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>2/9</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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</exercise>
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<exercise>
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<statement>
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<p>
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A toxin molecule inside a cell has a 0.3 probability of leaving the cell during a 1-minute period.
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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.
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What is the probability of the molecule leaving the cell during the first 3 minutes?
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</p>
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</statement>
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<answer>
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<p>
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For short, write <m>\Pr(n)</m> to mean the probability of the toxin molecule leaving during the <m>n</m>th minute.
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Then <m>\Pr(n) = (0.7)^{n - 1} (0.3).</m>
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</p>
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<p>
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The probability of leaving during the first 3 minutes is <m>\Pr(1) + \Pr(2) + \Pr(3) = 0.657.</m>
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</p>
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</answer>
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</exercise>
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<!-- <exercise>
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<statement>
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<p>
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Each of 10 toxin molecules inside a cell has a 0.3 probability of leaving the cell during a 1-minute period.
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For each value of <m>n = 1, 2, 3, \dotsc</m>, and for each value of <m>0\leq k \leq n</m>, find the probability that exactly <m>k</m> toxin molecules remain in the cell after the <m>n</m>th minute.
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</p>
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</statement>
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</exercise> -->
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<!-- TODO write a solution --> <exercisegroup>
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<introduction>
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<p>
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In each of the following scenarios with given events <m>A</m> and <m>B</m>, alculate <m>\Pr(A), \Pr(B)</m>, <m>\Pr(A\cap B)</m>, <m>\Pr(A \mid B)</m>, and <m>\Pr(B \mid A)</m>.
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</p>
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</introduction>
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<exercise>
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<statement>
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<p>
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An experiment consists of rolling a fair die two times.
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Let <m>A</m> be the event that the sum is even, and let <m>B</m> be the event that the second roll is higher than the first.
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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> A = \{ \amp (1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), </mrow>
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<mrow> \amp (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (4, 6), </mrow>
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<mrow> \amp (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)\} </mrow>
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<mrow> B = \{ \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> A \cap B = \{ \amp (1, 3), (1, 5), (2, 4), (2, 6), (3, 5), (4, 6)\} </mrow>
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</md>
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So <m>\Pr(A) = \frac{18}{36} = \frac{1}{2}</m>, <m>\Pr(B) = \frac{15}{36} = \frac{5}{12}</m>, and <m>\Pr(A\cap B) = \frac{6}{36} = \frac{1}{6}</m>.
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Finally:
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<md>
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<mrow> \Pr(A \mid B) \amp = \frac{\Pr(A\cap B)}{\Pr(B)} = \frac{6/36}{15/36} = \frac{6}{15} = \frac{2}{5} </mrow>
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<mrow> \Pr(B \mid A) \amp = \frac{\Pr(B\cap A)}{\Pr(A)} = \frac{6/36}{18/36} = \frac{6}{18} = \frac{1}{3} </mrow>
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</md>
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</p>
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</answer>
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</exercise>
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<exercise>
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<statement>
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<p>
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An experiment consists of flipping a fair coin three times.
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Let <m>A</m> be the event that the first and second flips match.
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Let <m>B</m> be the event that there are at least two heads.
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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> A \amp = \{ HHH, HHT, TTH, TTT \} </mrow>
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<mrow> B \amp = \{ HHH, HHT, HTH, THH \} </mrow>
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<mrow> A\cap B \amp = \{HHH, HHT\} </mrow>
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</md>
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So <m>\Pr(A) = \frac{4}{8} = \frac{1}{2}</m>, <m>\Pr(B) = \frac{4}{8} = \frac{1}{2}</m>, and <m>\Pr(A\cap B) = \frac{2}{8} = \frac{1}{4}</m>.
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Finally:
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<md>
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<mrow> \Pr(A \mid B) \amp = \frac{\Pr(A\cap B)}{\Pr(B)} = \frac{2/8}{4/8} = \frac{2}{4} = \frac{1}{2} </mrow>
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<mrow> \Pr(B \mid A) \amp = \frac{\Pr(B\cap A)}{\Pr(A)} = \frac{2/8}{4/8} = \frac{2}{4} = \frac{1}{2} </mrow>
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</md>
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</p>
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</answer>
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</exercise>
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</exercisegroup>
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<exercise>
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<introduction>
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<p>
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A diagnostic test is developed to detect a disease present in 3.2% of the population.
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For a patient who has the disease, the test will accurately give a positive result 65% of the time.
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When the patient does not have the disease, the test will accurately give a negative result 99.9% of the time.
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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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For a patient who receives a positive test, what is the probability they have the disease?
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</p>
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</statement>
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<answer>
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<p>
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Let <m>P</m> be the event of testing positive and <m>D</m> the event of having the disease.
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Then the prevalence <m>\Pr(D)</m> is given as 3.2%, or 0.032.
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The sensitivity is <m>\Pr(P\mid D) = 0.65</m>, and the specificity is <m>\Pr(P^c\mid D^c) = 0.999</m>.
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So, according to Bayes' Theorem:
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<md>
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<mrow> \Pr(D\mid P) \amp = \frac{\Pr(P\mid D)\Pr(D)}{\Pr(P\mid D)\Pr(D) + (1 - \Pr(P^c\mid D^c))\Pr(D^c)} </mrow>
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<mrow> \amp = \frac{(0.65)(0.032)}{(0.65)(0.032) + (1 - 0.999)(1 - 0.032)} </mrow>
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<mrow> \amp \approx 0.96 </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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For a patient who receives a negative test, what is the probability they do not have the disease?
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</p>
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</statement>
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<answer>
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|
<p>
|
|
<md>
|
|
<mrow> \Pr(D^c\mid P^c) \amp = \frac{\Pr(P^c\mid D^c)\Pr(D^c)}{\Pr(P^c\mid D^c)\Pr(D^c) + (1 - \Pr(P\mid D))\Pr(D)} </mrow>
|
|
<mrow> \amp = \frac{(0.999)(1 - 0.032)}{(0.999)(1 - 0.032) + (1 - 0.65)(0.032)} </mrow>
|
|
<mrow> \amp \approx 0.99 </mrow>
|
|
</md>
|
|
</p>
|
|
</answer>
|
|
</task>
|
|
</exercise>
|
|
|
|
<exercise>
|
|
<statement>
|
|
<p>
|
|
An experiment consists of rolling a fair die two times.
|
|
Let <m>A</m> be the event that the sum is even, and let <m>B</m> be the event that the second roll is higher than the first.
|
|
Are <m>A</m> and <m>B</m> independent?
|
|
</p>
|
|
</statement>
|
|
|
|
<answer>
|
|
<p>
|
|
<md>
|
|
<mrow> A = \{ \amp (1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), </mrow>
|
|
<mrow> \amp (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (4, 6), </mrow>
|
|
<mrow> \amp (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)\} </mrow>
|
|
<mrow> B = \{ \amp (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), </mrow>
|
|
<mrow> \amp (2, 3), (2, 4), (2, 5), (2, 6), </mrow>
|
|
<mrow> \amp (3, 4), (3, 5), (3, 6), </mrow>
|
|
<mrow> \amp (4, 5), (4, 6), </mrow>
|
|
<mrow> \amp (5, 6)\} </mrow>
|
|
<mrow> A \cap B = \{ \amp (1, 3), (1, 5), (2, 4), (2, 6), (3, 5), (4, 6)\} </mrow>
|
|
</md>
|
|
So <m>\Pr(A) = \frac{18}{36} = \frac{1}{2}</m>, <m>\Pr(B) = \frac{15}{36} = \frac{5}{12}</m>, and <m>\Pr(A\cap B) = \frac{6}{36} = \frac{1}{6}</m>.
|
|
Finally:
|
|
<md>
|
|
<mrow> \Pr(A \mid B) \amp = \frac{\Pr(A\cap B)}{\Pr(B)} = \frac{6/36}{15/36} = \frac{6}{15} = \frac{2}{5} \neq \Pr(A), </mrow>
|
|
</md>
|
|
so <m>A</m> and <m>B</m> are not independent.
|
|
</p>
|
|
</answer>
|
|
</exercise>
|
|
|
|
<exercise>
|
|
<statement>
|
|
<p>
|
|
An experiment consists of flipping a fair coin three times.
|
|
Let <m>A</m> be the event that the first and second flips match.
|
|
Let <m>B</m> be the event that there are at least two heads.
|
|
Are <m>A</m> and <m>B</m> independent?
|
|
</p>
|
|
</statement>
|
|
|
|
<answer>
|
|
<p>
|
|
<md>
|
|
<mrow> A \amp = \{ HHH, HHT, TTH, TTT \} </mrow>
|
|
<mrow> B \amp = \{ HHH, HHT, HTH, THH \} </mrow>
|
|
<mrow> A\cap B \amp = \{HHH, HHT\} </mrow>
|
|
</md>
|
|
So <m>\Pr(A) = \frac{4}{8} = \frac{1}{2}</m>, <m>\Pr(B) = \frac{4}{8} = \frac{1}{2}</m>, and <m>\Pr(A\cap B) = \frac{2}{8} = \frac{1}{4}</m>.
|
|
Finally:
|
|
<md>
|
|
<mrow> \Pr(A \mid B) \amp = \frac{\Pr(A\cap B)}{\Pr(B)} = \frac{2/8}{4/8} = \frac{2}{4} = \frac{1}{2} = \Pr(A), </mrow>
|
|
</md>
|
|
so <m>A</m> and <m>B</m> are independent.
|
|
</p>
|
|
</answer>
|
|
</exercise>
|
|
|
|
<exercise>
|
|
<statement>
|
|
<p>
|
|
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>.
|
|
Create a probability distribution for <m>\Omega</m> so that <m>A, B</m> are independent.
|
|
</p>
|
|
</statement>
|
|
|
|
<answer>
|
|
<table>
|
|
<title>Example Distribution</title>
|
|
|
|
<tabular halign="center">
|
|
<row bottom="minor">
|
|
<cell><m>x</m></cell>
|
|
<cell><m>\Pr(x)</m></cell>
|
|
</row>
|
|
|
|
<row>
|
|
<cell>1</cell>
|
|
<cell>0.1</cell>
|
|
</row>
|
|
|
|
<row>
|
|
<cell>2</cell>
|
|
<cell>0.2</cell>
|
|
</row>
|
|
|
|
<row>
|
|
<cell>3</cell>
|
|
<cell>0.2</cell>
|
|
</row>
|
|
|
|
<row>
|
|
<cell>4</cell>
|
|
<cell>0.1</cell>
|
|
</row>
|
|
|
|
<row>
|
|
<cell>5</cell>
|
|
<cell>0.1</cell>
|
|
</row>
|
|
|
|
<row>
|
|
<cell>6</cell>
|
|
<cell>0.3</cell>
|
|
</row>
|
|
</tabular>
|
|
</table>
|
|
|
|
<p>
|
|
Now <m>\Pr(A) = 0.5</m>, <m>\Pr(B) = 0.4</m>, and
|
|
<md>
|
|
<mrow>\Pr(A\cap B) = 0.2 = (0.5)(0.4) = \Pr(A)\Pr(B),</mrow>
|
|
</md>
|
|
so <m>A</m> and <m>B</m> are independent.
|
|
</p>
|
|
</answer>
|
|
</exercise>
|
|
|
|
<exercise>
|
|
<statement>
|
|
<p>
|
|
An experiment consists of flipping a biased coin 20 times.
|
|
If the coin comes up heads with probability <m>p = 0.3</m>, find the probability of seeing 5 heads.
|
|
Find the probability of seeing up to (and including) 3 heads.
|
|
</p>
|
|
</statement>
|
|
|
|
<solution>
|
|
<p>
|
|
Let <m>S</m> be the number of heads.
|
|
Then <m>S \sim \Bin(20, 0.3)</m>, so:
|
|
<md>
|
|
<mrow> \Pr(S = 5) \amp = {20 \choose 5} (0.3)^5 (0.7)^{20 - 5} </mrow>
|
|
<mrow> \amp = \frac{20!}{(5!)(15!)} (0.3)^5 (0.7)^{15} </mrow>
|
|
<mrow> \amp = \frac{20 \times 19 \times 18 \times 17 \times 16}{5 \times 4 \times 3 \times 2 \times 1} (0.3)^5 (0.7)^{15} </mrow>
|
|
<mrow> \amp = (19 \times 3 \times 17 \times 16) (0.3)^5 (0.7)^{15} </mrow>
|
|
<mrow> \amp \approx 0.179 </mrow>
|
|
</md>
|
|
</p>
|
|
</solution>
|
|
</exercise>
|
|
|
|
<exercise>
|
|
<introduction>
|
|
<p>
|
|
An experiment consists of flipping a coin repeatedly until we first see heads.
|
|
</p>
|
|
</introduction>
|
|
|
|
|
|
<task>
|
|
<statement>
|
|
<p>
|
|
If the coin comes up heads with probability 0.4, what is the probability we'll see our first heads within three flips? What about precisely on the third flip?
|
|
</p>
|
|
</statement>
|
|
|
|
<solution>
|
|
<p>
|
|
Let <m>T</m> be the number of flips until we see heads.
|
|
Then <m>T</m> is geometric with parameter <m>p = 0.4</m>, so:
|
|
<md>
|
|
<mrow> \Pr(T = k) \amp = (1-0.4)^{k-1}(0.4) = 0.6^{k-1} \cdot 0.4 </mrow>
|
|
</md>
|
|
</p>
|
|
</solution>
|
|
</task>
|
|
|
|
|
|
<task>
|
|
<statement>
|
|
<p>
|
|
Which flip has the highest chance of being the first flip to come up heads?
|
|
</p>
|
|
</statement>
|
|
</task>
|
|
</exercise>
|
|
|
|
<exercise>
|
|
<statement>
|
|
<p>
|
|
A particular store has an average of 20 customers each hour.
|
|
During a 4-hour afternoon shift, what is the probability of serving 80 customers.
|
|
</p>
|
|
</statement>
|
|
</exercise>
|
|
|
|
<exercise>
|
|
<introduction>
|
|
<p>
|
|
A continuous random variable <m>X</m> taking values in <m>[1, 4]</m> has p.d.f.
|
|
<m>f(x) = k(x - \sqrt{x})</m> for some constant <m>k</m>.
|
|
</p>
|
|
</introduction>
|
|
|
|
|
|
<task>
|
|
<statement>
|
|
<p>
|
|
What is the value of <m>k</m>?
|
|
</p>
|
|
</statement>
|
|
|
|
<answer>
|
|
<p>
|
|
<m>k = \frac{6}{17}.</m>
|
|
</p>
|
|
</answer>
|
|
</task>
|
|
|
|
|
|
<task>
|
|
<statement>
|
|
<p>
|
|
Find <m>\Pr(2 \leq X \leq 3).</m>
|
|
</p>
|
|
</statement>
|
|
|
|
<answer>
|
|
<p>
|
|
<m>\Pr(2 \leq X \leq 3) \approx 0.325</m>
|
|
</p>
|
|
</answer>
|
|
</task>
|
|
</exercise>
|
|
|
|
<exercise>
|
|
<statement>
|
|
<p>
|
|
A continuous random variable <m>X</m> taking values in <m>[1, 2]</m> has p.d.f.
|
|
<m>\displaystyle{f(x) = \frac{1}{2}\left(\frac{1}{x^2} + x\right)}</m>.
|
|
Find the c.d.f.
|
|
<m>F(x)</m>.
|
|
Use your c.d.f.
|
|
to find <m>\Pr\left(1 \leq X \leq \frac{3}{2}\right)</m>.
|
|
</p>
|
|
</statement>
|
|
|
|
<answer>
|
|
<p>
|
|
<m>\frac{1}{2}\left(\frac{x^2}{2} - \frac{1}{x}\right) + \frac{1}{4}</m>. <m>\Pr\left(1 \leq X \leq \frac{3}{2}\right) \approx 0.479.</m>
|
|
</p>
|
|
</answer>
|
|
</exercise>
|
|
|
|
<exercise>
|
|
<statement>
|
|
<p>
|
|
A continuous random variable <m>X</m> taking values in <m>[2, 3]</m> has c.d.f.
|
|
<m>F(x) = \frac{x^3}{3} - x^2 + 4</m>.
|
|
Find the p.d.f.
|
|
<m>f(x)</m>.
|
|
</p>
|
|
</statement>
|
|
|
|
<answer>
|
|
<p>
|
|
<m>f(x) = x^2 - 2x.</m>
|
|
</p>
|
|
</answer>
|
|
</exercise>
|
|
|
|
<exercise>
|
|
<statement>
|
|
<p>
|
|
Consider <m>X, Y</m> with the joint distribution table below.
|
|
Are <m>X, Y</m> independent?
|
|
</p>
|
|
|
|
<table>
|
|
<title>Joint distribution for <m>X, Y</m></title>
|
|
|
|
<tabular halign="center">
|
|
<row bottom="minor">
|
|
<cell right="minor"></cell>
|
|
<cell right="minor"><m>X = 0</m></cell>
|
|
<cell><m>X = 1</m></cell>
|
|
</row>
|
|
|
|
<row bottom="minor">
|
|
<cell right="minor"><m>Y = 0</m></cell>
|
|
<cell right="minor">0.2</cell>
|
|
<cell>0.3</cell>
|
|
</row>
|
|
|
|
<row>
|
|
<cell right="minor"><m>Y = 1</m></cell>
|
|
<cell right="minor">0.4</cell>
|
|
<cell>0.1</cell>
|
|
</row>
|
|
</tabular>
|
|
</table>
|
|
</statement>
|
|
|
|
<answer>
|
|
<p>
|
|
No.
|
|
For example, <m>\Pr(X = 0, Y = 0) \neq \Pr(X = 0)\Pr(Y = 0).</m>
|
|
</p>
|
|
</answer>
|
|
</exercise>
|
|
|
|
<exercise>
|
|
<statement>
|
|
<p>
|
|
Suppose <m>X, Y</m> have the distributions:
|
|
<md>
|
|
<mrow> \Pr(X = 0) \amp = 0.1 \amp \Pr(Y = 0) \amp = 0.4 </mrow>
|
|
<mrow> \Pr(X = 1) \amp = 0.4 \amp \Pr(Y = 1) \amp = 0.6 </mrow>
|
|
<mrow> \Pr(X = 2) \amp = 0.5 </mrow>
|
|
</md>
|
|
Assuming <m>X, Y</m> are independent, write a joint distribution table.
|
|
</p>
|
|
</statement>
|
|
|
|
<answer>
|
|
<table>
|
|
<title>Joint Distribution</title>
|
|
|
|
<tabular halign="center">
|
|
<row bottom="minor">
|
|
<cell right="minor"></cell>
|
|
<cell right="minor"><m>X = 0</m></cell>
|
|
<cell right="minor"><m>X = 1</m></cell>
|
|
<cell><m>X = 2</m></cell>
|
|
</row>
|
|
|
|
<row bottom="minor">
|
|
<cell right="minor"><m>Y = 0</m></cell>
|
|
<cell right="minor">0.04</cell>
|
|
<cell right="minor">0.16</cell>
|
|
<cell>0.2</cell>
|
|
</row>
|
|
|
|
<row>
|
|
<cell right="minor"><m>Y = 1</m></cell>
|
|
<cell right="minor">0.06</cell>
|
|
<cell right="minor">0.24</cell>
|
|
<cell>0.3</cell>
|
|
</row>
|
|
</tabular>
|
|
</table>
|
|
</answer>
|
|
</exercise>
|
|
|
|
<exercise>
|
|
<statement>
|
|
<p>
|
|
Consider a random variable <m>X</m> with probability distribution below.
|
|
Find <m>\E(X)</m>.
|
|
</p>
|
|
|
|
<table>
|
|
<title></title>
|
|
|
|
<tabular halign="center">
|
|
<row header="yes" bottom="minor">
|
|
<cell><m>x</m></cell>
|
|
<cell><m>\Pr(X = x)</m></cell>
|
|
</row>
|
|
|
|
<row>
|
|
<cell>1</cell>
|
|
<cell>0.1</cell>
|
|
</row>
|
|
|
|
<row>
|
|
<cell>2</cell>
|
|
<cell>0.05</cell>
|
|
</row>
|
|
|
|
<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>
|
|
</tabular>
|
|
</table>
|
|
</statement>
|
|
|
|
<answer>
|
|
<p>
|
|
4.8.
|
|
</p>
|
|
</answer>
|
|
</exercise>
|
|
|
|
<exercise>
|
|
<introduction>
|
|
<p>
|
|
Suppose we flip a coin <m>n = 100</m> times, and let <m>N</m> count the number of heads.
|
|
</p>
|
|
</introduction>
|
|
|
|
|
|
<task>
|
|
<statement>
|
|
<p>
|
|
If the coin comes up heads on a flip with probability <m>p = 0.4</m>, what is <m>\E(N)</m>?
|
|
</p>
|
|
</statement>
|
|
|
|
<answer>
|
|
<p>
|
|
40.
|
|
</p>
|
|
</answer>
|
|
</task>
|
|
|
|
|
|
<task>
|
|
<statement>
|
|
<p>
|
|
What if <m>n = 80</m> and <m>p = 0.6</m>?
|
|
</p>
|
|
</statement>
|
|
|
|
<answer>
|
|
<p>
|
|
48.
|
|
</p>
|
|
</answer>
|
|
</task>
|
|
|
|
|
|
<task>
|
|
<statement>
|
|
<p>
|
|
What if <m>n = 200</m> and <m>p = 0.5</m>?
|
|
</p>
|
|
</statement>
|
|
|
|
<answer>
|
|
<p>
|
|
100.
|
|
</p>
|
|
</answer>
|
|
</task>
|
|
</exercise>
|
|
|
|
<exercise>
|
|
<statement>
|
|
<p>
|
|
If <m>\E(X) = 3</m>, <m>\E(Y) = -2</m>, and <m>\E(Z) = 1</m>, what is <m>\E(4X + 5Y - Z + 3)</m>?
|
|
</p>
|
|
</statement>
|
|
|
|
<answer>
|
|
<p>
|
|
4.
|
|
</p>
|
|
</answer>
|
|
</exercise>
|
|
|
|
<exercise>
|
|
<statement>
|
|
<p>
|
|
A continuous random variable <m>X</m> taking values in <m>[0, 1]</m> has p.d.f.
|
|
<m>f(x) = 2x</m>.
|
|
What is <m>\E(X)</m>?
|
|
</p>
|
|
</statement>
|
|
|
|
<answer>
|
|
<p>
|
|
2/3.
|
|
</p>
|
|
</answer>
|
|
</exercise>
|
|
|
|
<exercise>
|
|
<statement>
|
|
<p>
|
|
A continuous random variable <m>X</m> taking values in <m>[1, 4]</m> has p.d.f.
|
|
<m>f(x) = \frac{4}{3x^2}</m>.
|
|
What is <m>\E(X)</m>?
|
|
</p>
|
|
</statement>
|
|
|
|
<answer>
|
|
<p>
|
|
<m>\frac{4}{3}\ln(4) \approx 1.85.</m>
|
|
</p>
|
|
</answer>
|
|
</exercise>
|
|
|
|
<exercise>
|
|
<statement>
|
|
<p>
|
|
A continuous random variable <m>X</m> taking values in <m>[1, 2]</m> has p.d.f.
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<m>\displaystyle{f(x) = \frac{1}{2}\left(\frac{1}{x^2} + x\right)}</m>.
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Find <m>\E(X)</m>.
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</p>
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</statement>
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|
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<answer>
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<p>
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<m>\frac{1}{2}\left( \ln(2) + \frac{7}{3}\right) \approx 1.513.</m>
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</p>
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</answer>
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</exercise>
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|
|
|
<exercise>
|
|
<statement>
|
|
<p>
|
|
Consider a random variable <m>X</m> with probability distribution below.
|
|
Find <m>\Var(X)</m>.
|
|
</p>
|
|
|
|
<table>
|
|
<title></title>
|
|
|
|
<tabular halign="center">
|
|
<row header="yes" bottom="minor">
|
|
<cell><m>x</m></cell>
|
|
<cell><m>\Pr(X = x)</m></cell>
|
|
</row>
|
|
|
|
<row>
|
|
<cell>1</cell>
|
|
<cell>0.1</cell>
|
|
</row>
|
|
|
|
<row>
|
|
<cell>2</cell>
|
|
<cell>0.05</cell>
|
|
</row>
|
|
|
|
<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>
|
|
</tabular>
|
|
</table>
|
|
</statement>
|
|
</exercise>
|
|
|
|
<exercise>
|
|
<statement>
|
|
<p>
|
|
Suppose we flip a coin <m>n = 100</m> times, and let <m>N</m> count the number of heads.
|
|
If the coin comes up heads on a flip with probability <m>p = 0.4</m>, what is <m>\Var(N)</m>? What if <m>n = 80</m> and <m>p = 0.6</m>? What if <m>n = 200</m> and <m>p = 0.5</m>?
|
|
</p>
|
|
</statement>
|
|
</exercise>
|
|
|
|
<exercise>
|
|
<statement>
|
|
<p>
|
|
If <m>\E(X) = 3</m>, <m>\Var(X) = 2</m>, what is <m>\E(X^2)</m>?
|
|
</p>
|
|
</statement>
|
|
</exercise>
|
|
|
|
<exercise>
|
|
<statement>
|
|
<p>
|
|
A continuous random variable <m>X</m> taking values in <m>[0, 1]</m> has p.d.f.
|
|
<m>f(x) = 2x</m>.
|
|
What is <m>\Var(X)</m>?
|
|
</p>
|
|
</statement>
|
|
</exercise>
|
|
|
|
<exercise>
|
|
<statement>
|
|
<p>
|
|
A continuous random variable <m>X</m> taking values in <m>[1, 4]</m> has p.d.f.
|
|
<m>f(x) = \frac{4}{3x^2}</m>.
|
|
What is <m>\Var(X)</m>?
|
|
</p>
|
|
</statement>
|
|
</exercise>
|
|
|
|
<exercise>
|
|
<statement>
|
|
<p>
|
|
A continuous random variable <m>X</m> taking values in <m>[1, 2]</m> has p.d.f.
|
|
<m>\displaystyle{f(x) = \frac{1}{2}\left(\frac{1}{x^2} + x\right)}</m>.
|
|
Find <m>\Var(X)</m>.
|
|
</p>
|
|
</statement>
|
|
</exercise>
|
|
</exercises>
|
|
</section> |