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Independent Events solutions

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<title>Independent Events</title> <title>Independent Events</title>
<p> <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. But it's also important to understand when this is not the case.
</p> </p>
<definition xml:id="def-independent-events"> <definition xml:id="def-independent-events">
<statement> <statement>
<p> <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> </p>
</statement> </statement>
</definition> </definition>
<p> <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>. The equation <m>\Pr(A\mid B) = \Pr(A)</m> very concretely says: knowledge
Assuming <m>\Pr(A) \neq 0</m>, this is equivalent to <m>\Pr(A\cap B) = \Pr(A)\Pr(B)</m>, since: 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> <md>
<mrow> \Pr(A\mid B) \amp = \Pr(A) </mrow> <mrow> \Pr(A\mid B) \amp = \Pr(A) </mrow>
<mrow> \frac{\Pr(A\cap B)}{\Pr(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"> <example xml:id="example-rolls-independent">
<statement> <statement>
<p> <p>
In <xref ref="example-rolls-conditional"/>, we considered an experiment in which we rolled a fair 6-sided die twice. In <xref ref="example-rolls-conditional"/>, we considered an experiment
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 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>
<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: We have:
<md> <md>
<mrow> C \amp = \{(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)\} </mrow> <mrow> C \amp = \{(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)\} </mrow>
@@ -42,129 +52,192 @@
<mrow> \Pr(B\cap C) \amp = \frac{1}{36} </mrow> <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> <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> </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> </p>
</statement> </statement>
</example> </example>
<exercises xml:id="exercises-Independent-Events"> <exercises xml:id="exercises-Independent-Events">
<exercise> <exercise>
<statement> <statement>
<p> <p>
An experiment consists of rolling a fair die two times. 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
Are <m>A</m> and <m>B</m> independent? the event that the second roll is higher than the first.
</p> Are <m>A</m> and <m>B</m> independent?
</statement> </p>
</statement>
<answer> <answer>
<p> <p>
<md> <m>A</m> and <m>B</m> are not independent.
<mrow> A = \{ \amp (1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), </mrow> </p>
<mrow> \amp (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (4, 6), </mrow> </answer>
<mrow> \amp (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)\} </mrow>
<mrow> B = \{ \amp (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), </mrow>
<mrow> \amp (2, 3), (2, 4), (2, 5), (2, 6), </mrow>
<mrow> \amp (3, 4), (3, 5), (3, 6), </mrow>
<mrow> \amp (4, 5), (4, 6), </mrow>
<mrow> \amp (5, 6)\} </mrow>
<mrow> A \cap B = \{ \amp (1, 3), (1, 5), (2, 4), (2, 6), (3, 5), (4, 6)\} </mrow>
</md>
So <m>\Pr(A) = \frac{18}{36} = \frac{1}{2}</m>, <m>\Pr(B) = \frac{15}{36} = \frac{5}{12}</m>, and <m>\Pr(A\cap B) = \frac{6}{36} = \frac{1}{6}</m>.
Finally:
<md>
<mrow> \Pr(A \mid B) \amp = \frac{\Pr(A\cap B)}{\Pr(B)} = \frac{6/36}{15/36} = \frac{6}{15} = \frac{2}{5} \neq \Pr(A), </mrow>
</md>
so <m>A</m> and <m>B</m> are not independent.
</p>
</answer>
</exercise>
<exercise> <solution>
<statement> <p>
<p> <md>
An experiment consists of flipping a fair coin three times. <mrow> A = \{ \amp (1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), </mrow>
Let <m>A</m> be the event that the first and second flips match. <mrow> \amp (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (4, 6), </mrow>
Let <m>B</m> be the event that there are at least two heads. <mrow> \amp (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)\} </mrow>
Are <m>A</m> and <m>B</m> independent? <mrow> B = \{ \amp (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), </mrow>
</p> <mrow> \amp (2, 3), (2, 4), (2, 5), (2, 6), </mrow>
</statement> <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>
<answer> <exercise>
<p> <statement>
<md> <p>
<mrow> A \amp = \{ HHH, HHT, TTH, TTT \} </mrow> An experiment consists of flipping a fair coin three times.
<mrow> B \amp = \{ HHH, HHT, HTH, THH \} </mrow> Let <m>A</m> be the event that the first and second flips match.
<mrow> A\cap B \amp = \{HHH, HHT\} </mrow> Let <m>B</m> be the event that there are at least two heads.
</md> Are <m>A</m> and <m>B</m> independent?
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>. </p>
Finally: </statement>
<md>
<mrow> \Pr(A \mid B) \amp = \frac{\Pr(A\cap B)}{\Pr(B)} = \frac{2/8}{4/8} = \frac{2}{4} = \frac{1}{2} = \Pr(A), </mrow>
</md>
so <m>A</m> and <m>B</m> are independent.
</p>
</answer>
</exercise>
<exercise> <answer>
<statement> <p>
<p> <m>A</m> and <m>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>. </p>
Create a probability distribution for <m>\Omega</m> so that <m>A, B</m> are independent. </answer>
</p>
</statement>
<answer> <solution>
<table> <p>
<title>Example Distribution</title> <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>
<tabular halign="center"> <exercise>
<row bottom="minor"> <statement>
<cell><m>x</m></cell> <p>
<cell><m>\Pr(x)</m></cell> Let <m>A = \{1, 2, 3\}</m> and <m>B = \{3, 4, 5\}</m> be events in the
</row> 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>
<row> <answer>
<cell>1</cell> <table>
<cell>0.1</cell> <title>Example Distribution</title>
</row>
<row> <tabular halign="center">
<cell>2</cell> <row bottom="minor">
<cell>0.2</cell> <cell><m>x</m></cell>
</row> <cell><m>\Pr(x)</m></cell>
</row>
<row> <row>
<cell>3</cell> <cell>1</cell>
<cell>0.2</cell> <cell>0.1</cell>
</row> </row>
<row> <row>
<cell>4</cell> <cell>2</cell>
<cell>0.1</cell> <cell>0.2</cell>
</row> </row>
<row> <row>
<cell>5</cell> <cell>3</cell>
<cell>0.1</cell> <cell>0.2</cell>
</row> </row>
<row> <row>
<cell>6</cell> <cell>4</cell>
<cell>0.3</cell> <cell>0.1</cell>
</row> </row>
</tabular>
</table>
<p> <row>
Now <m>\Pr(A) = 0.5</m>, <m>\Pr(B) = 0.4</m>, and <cell>5</cell>
<md> <cell>0.1</cell>
<mrow>\Pr(A\cap B) = 0.2 = (0.5)(0.4) = \Pr(A)\Pr(B),</mrow> </row>
</md>
so <m>A</m> and <m>B</m> are independent. <row>
</p> <cell>6</cell>
</answer> <cell>0.3</cell>
</exercise> </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>
</exercises> </exercises>
</section> </section>