1991 lines
55 KiB
XML
1991 lines
55 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, September 24.
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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
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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
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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
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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>,
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<m>B = \{2, 4, 9, 10, 12, 14, 19\}</m>, and
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<m>C = \{9, 10, 11, 14, 16, 17, 20\}</m>, which are all subsets of
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<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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<solution>
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<p>
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The set <m>A - (B \cap C)</m> consists of elements in the set
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<m>A</m> which are not in the overlap of <m>B</m> and <m>C</m>.
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The overlap is <m>B \cap C = \{9, 10, 14\}</m>, and of these
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elements, <m>9</m> and <m>10</m> are in <m>A</m>.
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So <m>A - (B \cap C) = \{1, 2, 3, 4, 5, 6, 7, 8\}</m>.
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</p>
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</solution>
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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>,
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<m>|A \cap B|</m>, <m>|B\cap C|</m>, <m>|A\cap C|</m>, and
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<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
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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>,
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<m>|A \cup B| = 13</m>, <m>|A \cap B| = 4</m>, <m>|B\cap C| = 3</m>,
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<m>|A\cap C| = 2</m>, <m>|A\cup B\cup C| = 17</m>.
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In particular, note that
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<m>|A\cup B| = 13 \neq 10 + 7 = |A| + |B|</m>, so it is not true in
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general that the size of the union of sets is the sum of the sizes
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of the individual sets.
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</p>
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</answer>
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<solution>
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<p>
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<m>|X|</m> counts the number of elements in a finite set <m>X</m>.
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We can quickly count the elements in sets <m>A, B, C</m> to see that
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<m>|A| = 10</m>, <m>|B| = 7</m>, and <m>|C| = 7</m>.
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</p>
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<p>
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The union of two sets includes all elements from either set, so:
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<md>
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<mrow> A \cup B \amp = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 19\} </mrow>
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<mrow> A \cup B \cup C \amp = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 14, 16, 17, 19, 20\} </mrow>
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</md>
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Counting the elements, we see <m>|A \cup B| = 13</m> and
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<m>|A \cup B \cup C| = 17</m>.
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</p>
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<p>
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The intersection of two sets is the overlap, consisting of elements
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which show up in both sets simultaneously, so:
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<md>
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<mrow> A \cap B \amp = \{2, 4, 9, 10\} </mrow>
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<mrow> B \cap C \amp = \{9, 10, 14\} </mrow>
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<mrow> A \cap C \amp = \{9, 10\} </mrow>
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</md>
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Counting the elements, we see <m>|A\cap B| = 4</m>,
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<m>|B \cap C| = 3</m>, and <m>|A\cap C| = 2</m>.
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</p>
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<p>
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In particular, note that
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<m>|A\cup B| = 13 \neq 10 + 7 = |A| + |B|</m>, so it is not true in
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general that the size of the union of sets is the sum of the sizes
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of the individual sets.
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</p>
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</solution>
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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>,
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<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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<solution>
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<p>
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<m>A^c</m> consists of elements of <m>\Omega</m> which are not in
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<m>A</m>, thus
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<m>A^c = \{11, 12, 13, 14, 15, 16, 17, 18, 19, 20\}</m>.
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Similarly, <m>(A\cup B)^c</m> consists of elements that are not in
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<m>A</m> nor in <m>B</m>.
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Therefore, <m>(A\cup B)^c = \{11, 13, 15, 16, 17, 18, 20\}</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 we have a 6-sided die that's weighted to roll a 6 half of the
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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
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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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</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> \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
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probabilities.
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</p>
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</solution>
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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?
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Four times?
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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:
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<md>
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<mrow> \Omega = \{ \amp HHH, HHT, HTH, THH, </mrow>
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<mrow> \amp HTT, THT, TTH, TTT \}</mrow>
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</md>
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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
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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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<m>|\Omega| = 6^{10}</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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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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</solution>
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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>
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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>
|
|
</p>
|
|
</answer>
|
|
</task>
|
|
|
|
<task>
|
|
<statement>
|
|
<p>
|
|
Make a probability distribution table for <m>\Omega</m> assuming the
|
|
die is fair.
|
|
</p>
|
|
</statement>
|
|
|
|
<answer>
|
|
<p>
|
|
We'll avoid an overly large table and note that, since the die is
|
|
fair, every outcome is equally likely.
|
|
Therefore, <m>\Pr(x) = \frac{1}{36}</m> for every
|
|
<m>x\in \Omega</m>.
|
|
</p>
|
|
</answer>
|
|
</task>
|
|
|
|
<task>
|
|
<statement>
|
|
<p>
|
|
Let <m>A</m> be the event that the second roll is higher than the
|
|
first, and let <m>B</m> be the event that the first roll is even.
|
|
Find <m>\Pr(A), \Pr(B)</m>, and <m>\Pr(A \cap B)</m>.
|
|
</p>
|
|
</statement>
|
|
|
|
<answer>
|
|
<p>
|
|
<m>\Pr(A) = \frac{5}{12}, \Pr(B) = \frac{1}{2}, \Pr(A \cap B) = \frac{1}{6}</m>.
|
|
</p>
|
|
</answer>
|
|
|
|
<solution>
|
|
<p>
|
|
<md>
|
|
<mrow> A = \{ \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> B = \{ \amp (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), </mrow>
|
|
<mrow> \amp (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), </mrow>
|
|
<mrow> \amp (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)\} </mrow>
|
|
<mrow> A \cap B = \{ \amp (2, 3), (2, 4), (2, 5), (2, 6), </mrow>
|
|
<mrow> \amp (4, 5), (4, 6)\} </mrow>
|
|
</md>
|
|
Therefore
|
|
<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>
|
|
</p>
|
|
</solution>
|
|
</task>
|
|
</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, the probabilities scale by the same amount as the face
|
|
values.
|
|
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.
|
|
Write a probability distribution table for this die.
|
|
</p>
|
|
</statement>
|
|
|
|
<answer>
|
|
<table>
|
|
<title>Probability Distribution for a Linearly Scaled 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>1/21</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>
|
|
</answer>
|
|
|
|
<solution>
|
|
<p>
|
|
Let <m>\Pr(1) = x</m>.
|
|
Then <m>\Pr(2) = 2 \Pr(1) = 2x</m>, and <m>\Pr(3) = 3\Pr(1) = 3x</m>,
|
|
and so on.
|
|
So the total probability in the space is:
|
|
<md>
|
|
<mrow> Pr(\Omega) \amp = \Pr(1) + \Pr(2) + \Pr(3) + \Pr(4) + \Pr(5) + \Pr(6) </mrow>
|
|
<mrow> \amp = x + 2x + 3x + 4x + 5x + 6x </mrow>
|
|
<mrow> \amp = 21x </mrow>
|
|
</md>
|
|
Since the total probability must add up to 1, we have <m>1 = 21x</m>,
|
|
so <m>x = \frac{1}{21}</m>, and we can calculate the rest of the
|
|
probabilities from there.
|
|
</p>
|
|
|
|
<table>
|
|
<title>Probability Distribution for a Linearly Scaled 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>1/21</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>
|
|
<exercisegroup>
|
|
<introduction>
|
|
<p>
|
|
In each of the following scenarios with given events <m>A</m> and
|
|
<m>B</m>, calculate <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>.
|
|
</p>
|
|
</introduction>
|
|
|
|
<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.
|
|
</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} </mrow>
|
|
<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>
|
|
</md>
|
|
</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.
|
|
</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} </mrow>
|
|
<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>
|
|
</md>
|
|
</p>
|
|
</answer>
|
|
</exercise>
|
|
</exercisegroup>
|
|
|
|
<exercise>
|
|
<introduction>
|
|
<p>
|
|
A diagnostic test is developed to detect a disease present in 3.2% of
|
|
the population.
|
|
For a patient who has the disease, the test will accurately give a
|
|
positive result 65% of the time.
|
|
When the patient does not have the disease, the test will accurately
|
|
give a negative result 99.9% of the time.
|
|
</p>
|
|
</introduction>
|
|
|
|
<task>
|
|
<statement>
|
|
<p>
|
|
For a patient who receives a positive test, what is the probability
|
|
they have the disease?
|
|
</p>
|
|
</statement>
|
|
|
|
<answer>
|
|
<p>
|
|
<m>\approx 0.96</m>.
|
|
</p>
|
|
</answer>
|
|
|
|
<solution>
|
|
<p>
|
|
Let <m>P</m> be the event of testing positive and <m>D</m> the event
|
|
of having the disease.
|
|
Then the prevalence <m>\Pr(D)</m> is given as 3.2%, or 0.032.
|
|
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>.
|
|
So, according to Bayes' Theorem:
|
|
<md>
|
|
<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>
|
|
<mrow> \amp = \frac{(0.65)(0.032)}{(0.65)(0.032) + (1 - 0.999)(1 - 0.032)} </mrow>
|
|
<mrow> \amp \approx 0.96 </mrow>
|
|
</md>
|
|
</p>
|
|
</solution>
|
|
</task>
|
|
|
|
<task>
|
|
<statement>
|
|
<p>
|
|
For a patient who receives a negative test, what is the probability
|
|
they do not have the disease?
|
|
</p>
|
|
</statement>
|
|
|
|
<answer>
|
|
<p>
|
|
<m>\approx 0.99</m>.
|
|
</p>
|
|
</answer>
|
|
|
|
<solution>
|
|
<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>
|
|
</solution>
|
|
</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>
|
|
<m>A</m> and <m>B</m> are not independent.
|
|
</p>
|
|
</answer>
|
|
|
|
<solution>
|
|
<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>
|
|
</solution>
|
|
</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>
|
|
<m>A</m> and <m>B</m> are independent.
|
|
</p>
|
|
</answer>
|
|
|
|
<solution>
|
|
<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>
|
|
</solution>
|
|
</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>
|
|
</answer>
|
|
|
|
<solution>
|
|
<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>
|
|
</solution>
|
|
</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>
|
|
|
|
<solution>
|
|
<p>
|
|
<m>\Pr(T = k) = 0.6^{k-1} \cdot 0.4</m>, so every additional flip
|
|
multiplies the probability by 0.6.
|
|
Therefore, the highest value for <m>\Pr(T = k)</m> occurs when
|
|
<m>k = 1</m>, in which case <m>\Pr(T = 1) = 0.4.</m>
|
|
</p>
|
|
</solution>
|
|
</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>
|
|
|
|
<solution>
|
|
<p>
|
|
Let <m>N</m> be the number of customers seen during the afternoon
|
|
shift.
|
|
Since the store averages 20 customers per hour, it will average 80 per
|
|
4-hours.
|
|
So <m>N \sim \Poiss(20, 4)</m> will have distribution
|
|
<m>\Pr(N = k) = \frac{80^{k}}{k!} e^{-80}</m>.
|
|
Therefore:
|
|
<md>
|
|
<mrow> \Pr(N = 80) = \frac{80^{80}}{80!} e^{-80} \approx 0.045. </mrow>
|
|
</md>
|
|
</p>
|
|
</solution>
|
|
</exercise>
|
|
|
|
<exercise xml:id="exercise-continuous-RV-find-k">
|
|
<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>
|
|
Let <m>T \sim \Exp(3)</m>.
|
|
Find <m>\Pr(1 \leq T \leq 3)</m> and <m>\Pr(T \geq 0.5)</m>.
|
|
</p>
|
|
</statement>
|
|
|
|
<answer>
|
|
<p>
|
|
<m>\Pr(1 \leq T \leq 3) \approx 0.050</m>.
|
|
<m>\Pr(T \geq 0.5) \approx 0.223</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 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 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/>
|
|
<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>
|
|
<statement>
|
|
<p>
|
|
Let <m>X</m> and <m>Y</m> be random variables each taking the values
|
|
1, 2, 3, 4, 5.
|
|
Write different distribution tables for <m>X</m> and <m>Y</m> so that
|
|
they have the same expected value.
|
|
</p>
|
|
</statement>
|
|
|
|
<answer>
|
|
<table>
|
|
<title>Example Distributions</title>
|
|
|
|
<tabular>
|
|
<row bottom="minor">
|
|
<cell><m>a</m></cell>
|
|
<cell><m>\Pr(X = a)</m></cell>
|
|
<cell><m>\Pr(Y = a)</m></cell>
|
|
</row>
|
|
|
|
<row>
|
|
<cell>1</cell>
|
|
<cell>0.2</cell>
|
|
<cell>0.1</cell>
|
|
</row>
|
|
|
|
<row>
|
|
<cell>2</cell>
|
|
<cell>0.2</cell>
|
|
<cell>0.1</cell>
|
|
</row>
|
|
|
|
<row>
|
|
<cell>3</cell>
|
|
<cell>0.2</cell>
|
|
<cell>0.6</cell>
|
|
</row>
|
|
|
|
<row>
|
|
<cell>4</cell>
|
|
<cell>0.2</cell>
|
|
<cell>0.1</cell>
|
|
</row>
|
|
|
|
<row>
|
|
<cell>5</cell>
|
|
<cell>0.2</cell>
|
|
<cell>0.1</cell>
|
|
</row>
|
|
</tabular>
|
|
</table>
|
|
</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, 1]</m>
|
|
has p.d.f.
|
|
<m>f(x) = \frac{3x^2}{2}</m>.
|
|
What is <m>\E(X)</m>?
|
|
</p>
|
|
</statement>
|
|
|
|
<answer>
|
|
<p>
|
|
0.
|
|
</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, 4]</m>
|
|
has p.d.f.
|
|
<m>f(x) = k(x - \sqrt{x})</m> for some constant <m>k</m>.
|
|
In a previous problem ( <xref ref="exercise-continuous-RV-find-k"/> ),
|
|
you found the value of <m>k</m>.
|
|
Now, find <m>\E(X)</m>.
|
|
</p>
|
|
</statement>
|
|
|
|
<answer>
|
|
<p>
|
|
<m>\frac{258}{85} \approx 3.04.</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.
|
|
<m>\displaystyle{f(x) = \frac{1}{2}\left(\frac{1}{x^2} + x\right)}</m>.
|
|
Find <m>\E(X)</m>.
|
|
</p>
|
|
</statement>
|
|
|
|
<answer>
|
|
<p>
|
|
<m>\frac{1}{2}\left( \ln(2) + \frac{7}{3}\right) \approx 1.513</m>.
|
|
</p>
|
|
</answer>
|
|
</exercise>
|
|
|
|
<exercise>
|
|
<statement>
|
|
<p>
|
|
Consider a random variable <m>X</m> with probability distribution
|
|
below.
|
|
Find <m>\Var(X)</m>.
|
|
</p>
|
|
|
|
<table>
|
|
<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>
|
|
<m>5.76</m>.
|
|
</p>
|
|
</answer>
|
|
</exercise>
|
|
|
|
<exercise>
|
|
<statement>
|
|
<p>
|
|
Let <m>X</m> be a random variable taking the values 1, 2, 3, 4, 5.
|
|
Write a distribution table for <m>X</m>, then use your table to write
|
|
a distribution for <m>X^2</m>.
|
|
Then, find <m>\Var(X)</m>.
|
|
</p>
|
|
</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 = 0.3</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>
|
|
|
|
<answer>
|
|
<p>
|
|
When <m>n = 100</m> and <m>p = 0.3</m>, <m>\Var(N) = 21</m>.
|
|
When <m>n = 80</m> and <m>p = 0.6</m>, <m>\Var(N) = 19.2</m>.
|
|
When <m>n = 200</m> and <m>p = 0.5</m>, <m>\Var(N) = 50</m>.
|
|
</p>
|
|
</answer>
|
|
</exercise>
|
|
|
|
<exercise>
|
|
<statement>
|
|
<p>
|
|
If <m>\E(X) = 3</m>, <m>\Var(X) = 2</m>, what is <m>\E(X^2)</m>?
|
|
</p>
|
|
</statement>
|
|
|
|
<answer>
|
|
<p>
|
|
<m>11</m>.
|
|
</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>\Var(X)</m>?
|
|
</p>
|
|
</statement>
|
|
|
|
<answer>
|
|
<p>
|
|
<m>1/18</m>.
|
|
</p>
|
|
</answer>
|
|
</exercise>
|
|
|
|
<exercise>
|
|
<statement>
|
|
<p>
|
|
A continuous random variable <m>X</m> taking values in <m>[-1, 1]</m>
|
|
has p.d.f.
|
|
<m>f(x) = \frac{3x^2}{2}</m>.
|
|
What is <m>\Var(X)</m>?
|
|
</p>
|
|
</statement>
|
|
|
|
<answer>
|
|
<p>
|
|
<m>0.6</m>.
|
|
</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>\Var(X)</m>?
|
|
</p>
|
|
</statement>
|
|
|
|
<answer>
|
|
<p>
|
|
<m>4 - \left(\frac{4}{3}\ln(4)\right)^2 \approx 0.583</m>.
|
|
</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) = k(x - \sqrt{x})</m> for some constant <m>k</m>.
|
|
In a previous problem ( <xref ref="exercise-continuous-RV-find-k"/> ),
|
|
you found the value of <m>k</m>.
|
|
Now, find <m>\Var(X)</m>.
|
|
</p>
|
|
</statement>
|
|
|
|
<answer>
|
|
<p>
|
|
<m>\frac{2307}{238} - \left(\frac{258}{85}\right)^2 \approx 0.48</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.
|
|
<m>\displaystyle{f(x) = \frac{1}{2}\left(\frac{1}{x^2} + x\right)}</m>.
|
|
Find <m>\Var(X)</m>.
|
|
</p>
|
|
</statement>
|
|
|
|
<answer>
|
|
<p>
|
|
<m>\frac{29}{24} - \frac{1}{2}\ln(2) \approx 0.862</m>.
|
|
</p>
|
|
</answer>
|
|
</exercise>
|
|
|
|
<exercise>
|
|
<statement>
|
|
<p>
|
|
Suppose a coin has an unknown probability of coming up heads.
|
|
We perform the experiment in five independent trials, during which it
|
|
takes 4, 5, 4, 3, and 6 flips to see our first heads in each trial.
|
|
What is the maximum likelihood estimation for the probability of the
|
|
coin coming up heads on a flip?
|
|
</p>
|
|
</statement>
|
|
|
|
<answer>
|
|
<p>
|
|
<m>5/22 \approx 0.227</m>.
|
|
</p>
|
|
</answer>
|
|
</exercise>
|
|
|
|
<exercise>
|
|
<statement>
|
|
<p>
|
|
A particular store owner wants to approximate the average hourly rate
|
|
at which customers come into the store.
|
|
They observe 80 customers enter during a particular 4-hour shift.
|
|
What is the maximum likelihood estimation for the hourly customer
|
|
rate?
|
|
</p>
|
|
</statement>
|
|
|
|
<answer>
|
|
<p>
|
|
<m>20</m>.
|
|
</p>
|
|
</answer>
|
|
</exercise>
|
|
|
|
<exercise>
|
|
<statement>
|
|
<p>
|
|
A radioactive material emits particles at an unknown probabilistic
|
|
rate <m>\lambda</m> particles per minute.
|
|
We observe particles emitted at times 1.1, 1.7, 1.3, 2.2, 1.9, and 1.8
|
|
minutes.
|
|
Write the likelihood function <m>\mathcal{L}(\lambda)</m> based on
|
|
this data.
|
|
What is the maximum likelihood estimation for <m>\lambda</m>?
|
|
</p>
|
|
</statement>
|
|
|
|
<answer>
|
|
<p>
|
|
<m>\mathcal{L}(\lambda) = \left( \lambda e^{-1.1\lambda} \right) \left( \lambda e^{-1.7\lambda} \right) \left( \lambda e^{-1.3\lambda} \right) \left( \lambda e^{-2.2\lambda} \right) \left( \lambda e^{-1.9\lambda} \right) \left( \lambda e^{-1.8\lambda} \right)</m>.
|
|
The MLE is <m>0.6</m>.
|
|
</p>
|
|
</answer>
|
|
</exercise>
|
|
|
|
<exercise>
|
|
<statement>
|
|
<p>
|
|
Suppose a parameter <m>\theta</m> takes values in <m>[0, 1]</m> with
|
|
likelihood function
|
|
<m>\mathcal{L}(\theta) = \sqrt{\theta} - \theta^2</m>.
|
|
Find the maximum likelihood estimation of <m>\theta</m>.
|
|
</p>
|
|
</statement>
|
|
|
|
<answer>
|
|
<p>
|
|
<m>\left(\frac{1}{4}\right)^{2/3} \approx 0. 37</m>.
|
|
</p>
|
|
</answer>
|
|
</exercise>
|
|
</exercises>
|
|
</section> |