Probability solutions

source/sec-Probability.ptx

@@ -2,13 +2,17 @@
   <title>Definition of Probability</title>
 
   <p>
-      Now that we have the language to refer to outcomes and events of an experiment, we want to start quantifying how likely those outcomes/events are to occur.
+    Now that we have the language to refer to outcomes and events of an
+    experiment, we want to start quantifying how likely those outcomes/events
+    are to occur.
   </p>
 
   <definition xml:id="def-probability-distribution">
     <statement>
       <p>
-          A <term>probability distribution</term> on a sample space <m>\Omega</m> assigns probabilities to every event, satisfying the following conditions:
+        A <term>probability distribution</term> on a sample space <m>\Omega</m>
+        assigns probabilities to every event, satisfying the following
+        conditions:
         <ol>
           <li>
             <p>
@@ -24,7 +28,8 @@
 
           <li>
             <p>
-                If <m>A \cap B = \emptyset</m>, then <m>\Pr(A\cup B) = \Pr(A) + \Pr(B)</m>.
+              If <m>A \cap B = \emptyset</m>, then
+              <m>\Pr(A\cup B) = \Pr(A) + \Pr(B)</m>.
             </p>
           </li>
         </ol>
@@ -33,8 +38,11 @@
   </definition>
 
   <p>
-      For small probability spaces (i.e., with finitely many outcomes in the sample space), we'll usually assign probabilities to each individual outcome, and perhaps list them in a table.
+    For small probability spaces (i.e., with finitely many outcomes in the
-      Then, to find the probability of any event, simply add together the probabilities of each outcome in that event.
+    sample space), we'll usually assign probabilities to each individual
+    outcome, and perhaps list them in a table.
+    Then, to find the probability of any event, simply add together the
+    probabilities of each outcome in that event.
   </p>
 
   <example>
@@ -87,7 +95,8 @@
       </table>
 
       <p>
-          One possible event is <m>A = \{2, 4, 6\}</m>, i.e., the event that the result of the roll is even.
+        One possible event is <m>A = \{2, 4, 6\}</m>, i.e., the event that the
+        result of the roll is even.
         The probability of <m>A</m> is:
         <md>
           <mrow> \Pr(A) = \Pr(2) + \Pr(4) + \Pr(6) = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2} </mrow>
@@ -99,16 +108,20 @@
   <definition xml:id="def-discrete-uniform-distribution">
     <statement>
       <p>
-          Let <m>\Omega = \{x_1, x_2, \dotsc, x_n\}</m> be a sample space and <m>A\subset \Omega</m> an event.
+        Let <m>\Omega = \{x_1, x_2, \dotsc, x_n\}</m> be a sample space and
-          We refer to the distribution in which <m>\Pr(x_i) = \frac{1}{n}</m> for all <m>i</m> as the <term>uniform distribution</term>.
+        <m>A\subset \Omega</m> an event.
+        We refer to the distribution in which <m>\Pr(x_i) = \frac{1}{n}</m> for
+        all <m>i</m> as the <term>uniform distribution</term>.
         In this case, it follows that <m>\Pr(A) = \frac{|A|}{|\Omega|}</m>.
       </p>
     </statement>
   </definition>
 
   <p>
-      When we talk about a fair coin flip or a fair die roll, the word "fair" is indicating a uniform distribution.
+    When we talk about a fair coin flip or a fair die roll, the word "fair" is
-      However, don't make the mistake of assuming that all distributions are uniform by default.
+    indicating a uniform distribution.
+    However, don't make the mistake of assuming that all distributions are
+    uniform by default.
   </p>
 
   <example>
@@ -121,27 +134,32 @@
 
     <answer>
       <p>
-          Without assuming the distribution is fair (i.e., that each value <m>1, 2, \dotsc, 10</m>) has probability <m>1/10</m> of occurring), we don't have enough information to answer this question.
+        Without assuming the distribution is fair (i.e., that each value
+        <m>1, 2, \dotsc, 10</m> ) has probability <m>1/10</m> of occurring), we
+        don't have enough information to answer this question.
       </p>
 
       <p>
-          In fact, the situation is even more vague than that: the sample space itself is unclear.
+        In fact, the situation is even more vague than that: the sample space
-          Are we only allowed to pick integer values? What about fractions like <m>7/2</m>? What about irrational numbers like <m>\pi</m>?
+        itself is unclear.
+        Are we only allowed to pick integer values?
+        What about fractions like <m>7/2</m>?
+        What about irrational numbers like <m>\pi</m>?
       </p>
     </answer>
   </example>
-
   <exercises xml:id="exercises-Probability">
     <exercise>
       <statement>
         <p>
-          Consider the sample space <m>\Omega = \{1, 2, 3, 4, 5, 6, 7, 8\}</m> with probability distribution below.
+          Consider the sample space <m>\Omega = \{1, 2, 3, 4, 5, 6, 7, 8\}</m>
-          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>.
+          with probability distribution below.
+          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>.
         </p>
 
         <table>
-          <title></title>
+          <title/>
-
           <tabular halign="center">
             <row bottom="minor">
               <cell><m>x</m></cell>
@@ -196,11 +214,41 @@
         </table>
       </statement>
 
+      <hint>
+        <p>
+          Remember that, to calculate the probability of an event, you should
+          add up the probabilities of each outcome in the event.
+        </p>
+      </hint>
+
       <answer>
         <p>
           <m>\Pr(A) = 0.45, \Pr(B) = 0.4, \Pr(A \cup B) = 0.6, \Pr(A \cap B) = 0.25.</m>
         </p>
       </answer>
+
+      <solution>
+        <p>
+          <md>
+            <mrow> \Pr(A) \amp = \Pr(1) + \Pr(3) + \Pr(7) + \Pr(8) </mrow>
+            <mrow> \amp = 0.1 + 0.2 + 0.05 + 0.1 </mrow>
+            <mrow> \amp = 0.45 </mrow>
+            <mrow> \Pr(B) \amp = \Pr(2) + \Pr(3) + \Pr(6) + \Pr(7) </mrow>
+            <mrow> \amp = 0.05 + 0.2 + 0.1 + 0.05 </mrow>
+            <mrow> \amp = 0.4 </mrow>
+          </md>
+          The other events are <m>A\cup B = \{1, 2, 3, 6, 7, 8\}</m> and
+          <m>A\cap B = \{3, 7\}</m>, so:
+          <md>
+            <mrow> \Pr(A \cup B) \amp = \Pr(1) + \Pr(2) + \Pr(3) + \Pr(6) + \Pr(7) + \Pr(8) </mrow>
+            <mrow> \amp = 0.1 + 0.05 + 0.2 + 0.1 + 0.05 + 0.1 </mrow>
+            <mrow> \amp = 0.6 </mrow>
+            <mrow> \Pr(A \cap B) \amp = \Pr(3) + \Pr(7) </mrow>
+            <mrow> \amp = 0.2 + 0.05 </mrow>
+            <mrow> \amp = 0.25 </mrow>
+          </md>
+        </p>
+      </solution>
     </exercise>
 
     <exercise>
@@ -208,11 +256,11 @@
         <p>
           Suppose we flip a coin two times.
           Answer the questions below.
-          What about three flips? What about four flips?
+          What about three flips?
+          What about four flips?
         </p>
       </introduction>
 
-
       <task>
         <statement>
           <p>
@@ -227,11 +275,11 @@
         </answer>
       </task>
 
-
       <task>
         <statement>
           <p>
-            Make a probability distribution table for <m>\Omega</m> assuming the coin is fair.
+            Make a probability distribution table for <m>\Omega</m> assuming the
+            coin is fair.
           </p>
         </statement>
 
@@ -269,11 +317,11 @@
         </answer>
       </task>
 
-
       <task>
         <statement>
           <p>
-            Make a probability distribution table assuming the coin comes up heads with probability 0.3.
+            Make a probability distribution table assuming the coin comes up
+            heads with probability 0.3.
           </p>
         </statement>
 
@@ -315,12 +363,11 @@
     <exercise>
       <introduction>
         <p>
-          Suppose we roll a die two times.
+          Suppose we roll a fair 6-sided die two times.
           Answer the questions below.
         </p>
       </introduction>
 
-
       <task>
         <statement>
           <p>
@@ -342,32 +389,40 @@
         </answer>
       </task>
 
-
       <task>
         <statement>
           <p>
-            Make a probability distribution table for <m>\Omega</m> assuming the die is fair.
+            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.
+            We'll avoid an overly large table and note that, since the die is
-            Therefore, <m>\Pr(x) = \frac{1}{36}</m> for every <m>x\in \Omega</m>.
+            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.
+            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>
@@ -381,18 +436,23 @@
               <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>
+            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>
-        </answer>
+        </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.
+          Suppose a die has the values <m>1, 2, 3, 4, 5, 6</m> on the faces, but
-          Instead, the probabilities scale by the same amount as the face values.
+          the die is not fair.
-          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.
+          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>
@@ -439,13 +499,74 @@
           </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.
+          Suppose a die has the values <m>1, 2, 3, 4, 5, 6</m> on the faces, but
-          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.
+          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>
@@ -492,36 +613,112 @@
           </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.
+          A toxin molecule inside a cell has a 0.3 probability of leaving the
-          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.
+          cell during a 1-minute period.
-          What is the probability of the molecule leaving the cell during the first 3 minutes?
+          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>
-          For short, write <m>\Pr(n)</m> to mean the probability of the toxin molecule leaving during the <m>n</m>th minute.
+          Write <m>T</m> for the minute that the toxin molecule leaves the cell.
-          Then <m>\Pr(n) = (0.7)^{n - 1} (0.3).</m>
+          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>
-          The probability of leaving during the first 3 minutes is <m>\Pr(1) + \Pr(2) + \Pr(3) = 0.657.</m>
+          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>
-      </answer>
+
-    </exercise>
-    <!--
-    <exercise>
-      <statement>
         <p>
-          Each of 10 toxin molecules inside a cell has a 0.3 probability of leaving the cell during a 1-minute period.
+          Then, the probaiblity of the molecule leaving within the first three minutes will be:
-          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.
+          <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>
-      </statement>
+      </solution>
     </exercise>
-    needs binomial coefficients --> </exercises>
+  </exercises>
-  </section>
+</section>