Independent Events solutions

source/sec-Independent-Events.ptx

@@ -2,21 +2,26 @@
   <title>Independent Events</title>
 
   <p>
-    The idea of conditional probability is that knowledge of one event can change our understanding of the probability of another.
+    The idea of conditional probability is that knowledge of one event can
+    change our understanding of the probability of another.
     But it's also important to understand when this is not the case.
   </p>
 
   <definition xml:id="def-independent-events">
     <statement>
       <p>
-        Events <m>A, B</m> are <term>independent</term> if <m>\Pr(A \mid B) = \Pr(A)</m>.
+        Events <m>A, B</m> are <term>independent</term> if
+        <m>\Pr(A \mid B) = \Pr(A)</m>.
       </p>
     </statement>
   </definition>
 
   <p>
-    The equation <m>\Pr(A\mid B) = \Pr(A)</m> very concretely says: knowledge that the event <m>B</m> has occurred does not change our understanding of <m>\Pr(A)</m>.
-    Assuming <m>\Pr(A) \neq 0</m>, this is equivalent to <m>\Pr(A\cap B) = \Pr(A)\Pr(B)</m>, since:
+    The equation <m>\Pr(A\mid B) = \Pr(A)</m> very concretely says: knowledge
+    that the event <m>B</m> has occurred does not change our understanding of
+    <m>\Pr(A)</m>.
+    Assuming <m>\Pr(A) \neq 0</m>, this is equivalent to
+    <m>\Pr(A\cap B) = \Pr(A)\Pr(B)</m>, since:
     <md>
       <mrow> \Pr(A\mid B) \amp = \Pr(A) </mrow>
       <mrow> \frac{\Pr(A\cap B)}{\Pr(B)} \amp = \Pr(A) </mrow>
@@ -28,12 +33,17 @@
   <example xml:id="example-rolls-independent">
     <statement>
       <p>
-        In <xref ref="example-rolls-conditional"/>, we considered an experiment in which we rolled a fair 6-sided die twice.
-        We defined events <m>A</m> that the sum of the rolls is at least 10 and <m>B</m> that the first roll is a 6, and we found that <m>\Pr(A \mid B) \neq \Pr(A)</m>, so <m>A</m> and <m>B</m> are not independent.
+        In <xref ref="example-rolls-conditional"/>, we considered an experiment
+        in which we rolled a fair 6-sided die twice.
+        We defined events <m>A</m> that the sum of the rolls is at least 10 and
+        <m>B</m> that the first roll is a 6, and we found that
+        <m>\Pr(A \mid B) \neq \Pr(A)</m>, so <m>A</m> and <m>B</m> are not
+        independent.
       </p>
 
       <p>
-        Now consider the event <m>C</m> that the sum of the rolls is 7, which sounds very similar to the event <m>A</m>.
+        Now consider the event <m>C</m> that the sum of the rolls is 7, which
+        sounds very similar to the event <m>A</m>.
         We have:
         <md>
           <mrow> C \amp = \{(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)\} </mrow>
@@ -42,22 +52,30 @@
           <mrow> \Pr(B\cap C) \amp = \frac{1}{36} </mrow>
           <mrow> \text{therefore: } \Pr(C\mid B) \amp = \frac{\Pr(C\cap B)}{\Pr(B)} = \frac{1/36}{1/6} = \frac{1}{6} = \Pr(C) </mrow>
         </md>
-        So, even though the descriptions of events <m>A</m> and <m>C</m> are very similar, the event <m>C</m> is independent with <m>B</m>, while <m>A</m> is not.
+        So, even though the descriptions of events <m>A</m> and <m>C</m> are
+        very similar, the event <m>C</m> is independent with <m>B</m>, while
+        <m>A</m> is not.
       </p>
     </statement>
   </example>
-
   <exercises xml:id="exercises-Independent-Events">
     <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.
+          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>
@@ -70,14 +88,16 @@
             <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>.
+          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>
+      </solution>
     </exercise>
 
     <exercise>
@@ -91,27 +111,37 @@
       </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>.
+          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>
+      </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.
+          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>
 
@@ -156,6 +186,49 @@
             </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
@@ -164,7 +237,7 @@
           </md>
           so <m>A</m> and <m>B</m> are independent.
         </p>
-    </answer>
+      </solution>
     </exercise>
   </exercises>
 </section>