<p style="text-align:justify"><span style="font-size:14px"><span style="background-color:white"><em><span style="color:black">A</span></em><span style="color:black">&nbsp;and&nbsp;<em>B</em>&nbsp;are&nbsp;<strong>mutually exclusive</strong>&nbsp;events if they cannot occur at the same time. This means that&nbsp;<em>A</em>&nbsp;and&nbsp;<em>B</em>&nbsp;do not share any outcomes and&nbsp;<em>P</em>(<em>A</em>&nbsp;AND&nbsp;<em>B</em>) = 0.</span></span></span></p>

<p style="text-align:justify">&nbsp;</p>

<p style="text-align:justify"><span style="font-size:14px"><span style="background-color:white"><span style="color:black">For example, suppose the sample space&nbsp;<em>S</em>&nbsp;= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Let&nbsp;<em>A</em>&nbsp;= {1, 2, 3, 4, 5},&nbsp;<em>B</em>&nbsp;= {4, 5, 6, 7, 8}, and&nbsp;<em>C</em>&nbsp;= {7, 9}.&nbsp;<em>A</em>&nbsp;AND&nbsp;<em>B</em>&nbsp;= {4, 5}.&nbsp;<em>P</em>(<em>A</em>&nbsp;AND&nbsp;<em>B</em>) =&nbsp;210210&nbsp;and is not equal to zero. Therefore,&nbsp;<em>A</em>&nbsp;and&nbsp;<em>B</em>&nbsp;are not mutually exclusive.</span></span></span></p>

<p style="text-align:justify"><br />
<span style="font-size:14px"><span style="background-color:white"><span style="color:black"><em>A</em>&nbsp;and&nbsp;<em>C</em>&nbsp;do not have any numbers in common so&nbsp;<em>P</em>(<em>A</em>&nbsp;AND&nbsp;<em>C</em>) = 0. Therefore,&nbsp;<em>A</em>&nbsp;and&nbsp;<em>C</em>&nbsp;are mutually exclusive.</span></span></span></p>

<p style="text-align:justify">&nbsp;</p>

<p style="text-align:justify"><span style="font-size:14px"><span style="background-color:white"><span style="color:black">If it is not known whether&nbsp;<em>A</em>&nbsp;and&nbsp;<em>B</em>&nbsp;are mutually exclusive,&nbsp;<strong>assume they are not until you can show otherwise</strong>. The following examples illustrate these definitions and terms.</span></span></span></p>

<p style="text-align:justify">&nbsp;</p>

<p style="text-align:justify"><span style="font-size:14px"><span style="color:#000000"><strong>EXAMPLE&nbsp;3.6</strong></span></span></p>

<p style="text-align:justify">&nbsp;</p>

<p style="text-align:justify"><span style="font-size:14px"><span style="color:black">Flip two fair coins. This is an experiment.</span></span></p>

<p style="text-align:justify">&nbsp;</p>

<p style="text-align:justify"><span style="font-size:14px"><span style="color:black">The sample space is {<em>HH</em>,&nbsp;<em>HT</em>,&nbsp;<em>TH</em>,&nbsp;<em>TT</em>}, where&nbsp;<em>T</em>&nbsp;= tails and&nbsp;<em>H</em>&nbsp;= heads. The outcomes are&nbsp;<em>HH</em>,&nbsp;<em>HT</em>,&nbsp;<em>TH</em>, and&nbsp;<em>TT</em>. The outcomes&nbsp;<em>HT</em>&nbsp;and&nbsp;<em>TH</em>&nbsp;are different. The&nbsp;<em>HT</em>&nbsp;means that the first coin showed heads and the second coin showed tails. The&nbsp;<em>TH</em>&nbsp;means that the first coin showed tails and the second coin showed heads.</span></span></p>

<p style="text-align:justify">&nbsp;</p>

<ul>
	<li style="text-align:justify">
	<p><span style="font-size:14px"><span style="color:black">Let&nbsp;<em>A</em>&nbsp;= the event of getting&nbsp;<strong>at most one tail</strong>. At most one tail means zero or one tail. Then&nbsp;<em>A</em>&nbsp;can be written as {<em>HH</em>,&nbsp;<em>HT</em>,&nbsp;<em>TH</em>}. The outcome&nbsp;<em>HH</em>&nbsp;shows zero tails.&nbsp;<em>HT</em>&nbsp;and&nbsp;<em>TH</em>&nbsp;each show one tail.</span></span></p>
	</li>
	<li style="text-align:justify">
	<p><span style="font-size:14px"><span style="color:black">Let&nbsp;<em>B</em>&nbsp;= the event of getting all tails.&nbsp;<em>B</em>&nbsp;can be written as {<em>TT</em>}.&nbsp;<em>B</em>&nbsp;is the&nbsp;<strong>complement event</strong>&nbsp;of&nbsp;<em>A</em>, so&nbsp;<em>B</em>&nbsp;=&nbsp;<em>A&prime;</em>. Also,&nbsp;<em>P</em>(<em>A</em>) +&nbsp;<em>P</em>(<em>B</em>) =&nbsp;<em>P</em>(<em>A</em>) +&nbsp;<em>P</em>(<em>A&prime;</em>) = 1.</span></span></p>
	</li>
	<li>
	<p style="text-align:justify"><span style="font-size:14px"><span style="color:black">The probabilities for&nbsp;<em>A</em>&nbsp;and for&nbsp;<em>B</em>&nbsp;are&nbsp;</span><span class="math-tex">\(P(A)=\frac{3}{4}\hspace{0.1cm}and\hspace{0.1cm}P(B)=\frac{1}{4}\)</span></span></p>
	</li>
	<li style="text-align:justify">
	<p><span style="font-size:14px"><span style="color:black">Let&nbsp;<em>C</em>&nbsp;= the event of getting all heads.&nbsp;<em>C</em>&nbsp;= {<em>HH</em>}. Since&nbsp;<em>B</em>&nbsp;= {<em>TT</em>},&nbsp;<em>P</em>(<em>B</em>&nbsp;AND&nbsp;<em>C</em>) = 0.&nbsp;<em>B</em>&nbsp;and&nbsp;<em>C</em>&nbsp;are mutually exclusive. (<em>B</em>&nbsp;and&nbsp;<em>C</em>&nbsp;have no members in common because you cannot have all tails and all heads at the same time.)</span></span></p>
	</li>
	<li>
	<p style="text-align:justify"><span style="font-size:14px"><span style="color:black">Let&nbsp;<em>D</em>&nbsp;= event of getting&nbsp;<strong>more than one</strong>&nbsp;tail.&nbsp;<em>D</em>&nbsp;= {<em>TT</em>}.&nbsp;<span class="math-tex">\(P(D)=\frac{1}{4}\)</span></span></span></p>
	</li>
	<li>
	<p style="text-align:justify"><span style="font-size:14px"><span style="color:black">Let&nbsp;<em>E</em>&nbsp;= event of getting a head on the first roll. This implies you can get either a head or tail on the second roll.&nbsp;<em>E</em>&nbsp;= {<em>HT</em>,&nbsp;<em>HH</em>}.&nbsp;<span class="math-tex">\(P(E)=\frac{2}{4}\)</span></span></span></p>
	</li>
	<li>
	<p style="text-align:justify"><span style="font-size:14px"><span style="color:black">Find the probability of getting&nbsp;<strong>at least one</strong>&nbsp;(one or two) tail in two flips. Let&nbsp;<em>F</em>&nbsp;= event of getting at least one tail in two flips.&nbsp;<em>F</em>&nbsp;= {<em>HT</em>,&nbsp;<em>TH</em>,&nbsp;<em>TT</em>}.&nbsp;<span class="math-tex">\(P(F)=\frac{3}{4}\)</span></span></span></p>
	</li>
</ul>

<p style="text-align:justify">&nbsp;</p>

<p style="text-align:justify"><span style="font-size:14px"><span style="color:#000000"><strong>EXAMPLE&nbsp;3.7</strong></span></span></p>

<p style="text-align:justify">&nbsp;</p>

<p style="text-align:justify"><span style="font-size:14px">Flip two fair coins. Find the probabilities of the events.</span></p>

<p style="text-align:justify">&nbsp;</p>

<ol style="list-style-type:lower-alpha">
	<li>
	<p style="text-align:justify"><span style="font-size:14px">Let&nbsp;<em>F</em>&nbsp;= the event of getting at most one tail (zero or one tail).</span></p>
	</li>
	<li>
	<p style="text-align:justify"><span style="font-size:14px">Let&nbsp;<em>G</em>&nbsp;= the event of getting two faces that are the same.</span></p>
	</li>
	<li>
	<p style="text-align:justify"><span style="font-size:14px">Let&nbsp;<em>H</em>&nbsp;= the event of getting a head on the first flip followed by a head or tail on the second flip.</span></p>
	</li>
	<li>
	<p style="text-align:justify"><span style="font-size:14px">Are&nbsp;<em>F</em>&nbsp;and&nbsp;<em>G</em>&nbsp;mutually exclusive?</span></p>
	</li>
	<li>
	<p style="text-align:justify"><span style="font-size:14px">Let&nbsp;<em>J</em>&nbsp;= the event of getting all tails. Are&nbsp;<em>J</em>&nbsp;and&nbsp;<em>H</em>&nbsp;mutually exclusive?</span></p>
	</li>
</ol>

<p style="text-align:justify">&nbsp;</p>

<p style="text-align:justify"><span style="font-size:14px"><strong><span style="color:#002060">Solution&nbsp;3.7</span></strong></span></p>

<p style="text-align:justify">&nbsp;</p>

<p style="text-align:justify"><span style="font-size:14px"><span style="color:black">Look at the sample space in&nbsp;</span><a href="https://openstax.org/books/statistics/pages/3-2-independent-and-mutually-exclusive-events#element-931" style="color:blue; text-decoration:underline"><span style="color:#027eb5">Example 3.6</span></a><span style="color:#424242">.</span></span></p>

<p style="text-align:justify">&nbsp;</p>

<ol start="1" style="list-style-type:lower-alpha">
	<li>
	<p style="text-align:justify"><span style="font-size:14px"><span style="color:black">Zero (0) or one (1) tails occur when the outcomes&nbsp;<em>HH</em>,&nbsp;<em>TH</em>,&nbsp;<em>HT</em>&nbsp;show up.&nbsp;<span class="math-tex">\(P(F)=\frac{3}{4}\)</span></span></span></p>
	</li>
	<li>
	<p style="text-align:justify"><span style="font-size:14px"><span style="color:black">Two faces are the same if&nbsp;<em>HH</em>&nbsp;or&nbsp;<em>TT</em>&nbsp;show up.&nbsp;<span class="math-tex">\(P(G)=\frac{2}{4}\)</span></span></span></p>
	</li>
	<li>
	<p style="text-align:justify"><span style="font-size:14px"><span style="color:black">A head on the first flip followed by a head or tail on the second flip occurs when&nbsp;<em>HH</em>&nbsp;or&nbsp;<em>HT</em>&nbsp;show up.&nbsp;<span class="math-tex">\(P(H)=\frac{2}{4}\)</span></span></span></p>
	</li>
	<li style="text-align:justify">
	<p><span style="font-size:14px"><span style="color:black"><em>F</em>&nbsp;and&nbsp;<em>G</em>&nbsp;share&nbsp;<em>HH</em>&nbsp;so&nbsp;<em>P</em>(<em>F</em>&nbsp;AND&nbsp;<em>G</em>) is not equal to zero (0).&nbsp;<em>F</em>&nbsp;and&nbsp;<em>G</em>&nbsp;are not mutually exclusive.</span></span></p>
	</li>
	<li style="text-align:justify">
	<p><span style="font-size:14px"><span style="color:black">Getting all tails occurs when tails shows up on both coins (<em>TT</em>).&nbsp;<em>H</em>&rsquo;s outcomes are&nbsp;<em>HH</em>&nbsp;and&nbsp;<em>HT</em>.</span></span></p>
	</li>
</ol>

<p style="text-align:justify">&nbsp;</p>

<p style="text-align:justify"><span style="font-size:14px"><em><span style="color:black">J</span></em><span style="color:black">&nbsp;and&nbsp;<em>H</em>&nbsp;have nothing in common so&nbsp;<em>P</em>(<em>J</em>&nbsp;AND&nbsp;<em>H</em>) = 0.&nbsp;<em>J</em>&nbsp;and&nbsp;<em>H</em>&nbsp;are mutually exclusive.</span></span></p>

<p style="text-align:justify">&nbsp;</p>

<p style="text-align:justify"><span style="font-size:14px"><strong>EXAMPLE&nbsp;3.8</strong></span></p>

<p style="text-align:justify">&nbsp;</p>

<p style="text-align:justify"><span style="font-size:14px"><span style="color:black">Roll one fair, six-sided die. The sample space is {1, 2, 3, 4, 5, 6}. Let event&nbsp;<em>A</em>&nbsp;= a face is odd. Then&nbsp;<em>A</em>&nbsp;= {1, 3, 5}. Let event&nbsp;<em>B</em>&nbsp;= a face is even. Then&nbsp;<em>B</em>&nbsp;= {2, 4, 6}.</span></span></p>

<p style="text-align:justify">&nbsp;</p>

<p style="text-align:justify"><span style="font-size:14px"><span style="color:black">Find the complement of&nbsp;<em>A</em>,&nbsp;<em>A&prime;</em>. The complement of&nbsp;<em>A</em>,&nbsp;<em>A&prime;</em>, is&nbsp;<em>B</em>&nbsp;because&nbsp;<em>A</em>&nbsp;and&nbsp;<em>B</em>&nbsp;together make up the sample space.&nbsp;<em>P</em>(<em>A</em>) +&nbsp;<em>P</em>(<em>B</em>) =&nbsp;<em>P</em>(<em>A</em>) +&nbsp;<em>P</em>(<em>A&prime;</em>) = 1. Also,&nbsp;</span><span class="math-tex">\(P(A)=\frac{3}{6}\hspace{0.1cm}and\hspace{0.1cm}P(B)=\frac{3}{6}\)</span></span></p>

<p style="text-align:justify">&nbsp;</p>

<ul>
	<li style="text-align:justify">
	<p><span style="font-size:14px"><span style="color:black">Let event&nbsp;<em>C</em>&nbsp;= odd faces larger than two. Then&nbsp;<em>C</em>&nbsp;= {3, 5}. Let event&nbsp;<em>D</em>&nbsp;= all even faces smaller than five. Then&nbsp;<em>D</em>&nbsp;= {2, 4}.&nbsp;<em>P</em>(<em>C</em>&nbsp;AND&nbsp;<em>D</em>) = 0 because you cannot have an odd and even face at the same time. Therefore,&nbsp;<em>C</em>&nbsp;and&nbsp;<em>D</em>&nbsp;are mutually exclusive events.</span></span></p>
	</li>
	<li>
	<p style="text-align:justify"><span style="font-size:14px"><span style="color:black">Let event&nbsp;<em>E</em>&nbsp;= all faces less than five.&nbsp;<em>E</em>&nbsp;= {1, 2, 3, 4}.</span></span></p>
	</li>
</ul>

<p style="text-align:justify">&nbsp;</p>

<p style="text-align:justify"><span style="font-size:14px"><span style="color:black">Are&nbsp;<em>C</em>&nbsp;and&nbsp;<em>E</em>&nbsp;mutually exclusive events? Answer yes or no. Why or why not?</span></span></p>

<p style="text-align:justify">&nbsp;</p>

<p style="text-align:justify"><span style="font-size:14px"><strong>Solution&nbsp;3.8</strong></span></p>

<p style="text-align:justify">&nbsp;</p>

<ul>
	<li>
	<p style="text-align:justify"><span style="font-size:14px"><span style="color:black">No.&nbsp;<em>C</em>&nbsp;= {3, 5} and&nbsp;<em>E</em>&nbsp;= {1, 2, 3, 4}.&nbsp;<em>P</em>(<em>C</em>&nbsp;AND&nbsp;<em>E</em>) </span><span class="math-tex">\(=\frac{1}{6}\)</span></span></p>
	</li>
	<li style="text-align:justify">
	<p><span style="font-size:14px"><span style="color:black">&nbsp;&nbsp;To be mutually exclusive,&nbsp;<em>P</em>(<em>C</em>&nbsp;AND&nbsp;<em>E</em>) must be zero.</span></span></p>
	</li>
	<li>
	<p style="text-align:justify"><span style="font-size:14px"><span style="color:black">Find&nbsp;<em>P</em>(<em>C</em>|<em>A</em>). This is a conditional probability. Recall that event&nbsp;<em>C</em>&nbsp;is {3, 5} and event&nbsp;<em>A</em>&nbsp;is {1, 3, 5}. To find&nbsp;<em>P</em>(<em>C</em>|<em>A</em>), find the probability of&nbsp;<em>C</em>&nbsp;using the sample space&nbsp;<em>A</em>. You have reduced the sample space from the original sample space {1, 2, 3, 4, 5, 6} to {1, 3, 5}. So,&nbsp;<em>P</em>(<em>C</em>|<em>A</em>)=<span class="math-tex">\(\frac{2}{3}\)</span></span></span></p>
	</li>
</ul>

<p style="text-align:justify">&nbsp;</p>

<p style="text-align:justify"><span style="font-size:14px"><strong><span style="color:#002060">EXAMPLE&nbsp;3.9</span></strong></span></p>

<p style="text-align:justify">&nbsp;</p>

<p style="text-align:justify"><span style="font-size:14px"><span style="color:black">Let event&nbsp;<em>G</em>&nbsp;= taking a math class. Let event&nbsp;<em>H</em>&nbsp;= taking a science class. Then,&nbsp;<em>G</em>&nbsp;AND&nbsp;<em>H</em>&nbsp;= taking a math class and a science class. Suppose&nbsp;<em>P</em>(<em>G</em>) = .6,&nbsp;<em>P</em>(<em>H</em>) = .5, and&nbsp;<em>P</em>(<em>G</em>&nbsp;AND&nbsp;<em>H</em>) = .3. Are&nbsp;<em>G</em>&nbsp;and&nbsp;<em>H</em>&nbsp;independent?</span></span></p>

<p style="text-align:justify"><span style="font-size:14px"><span style="color:black">If&nbsp;<em>G</em>&nbsp;and&nbsp;<em>H</em>&nbsp;are independent, then you must show&nbsp;<strong>ONE</strong>&nbsp;of the following:</span></span></p>

<p style="text-align:justify">&nbsp;</p>

<ul>
	<li style="text-align:justify">
	<p><span style="font-size:14px"><span style="color:black"><em>P</em>(<em>G</em>|<em>H</em>) =&nbsp;<em>P</em>(<em>G</em>)</span></span></p>
	</li>
	<li style="text-align:justify">
	<p><span style="font-size:14px"><span style="color:black"><em>P</em>(<em>H</em>|<em>G</em>) =&nbsp;<em>P</em>(<em>H</em>)</span></span></p>
	</li>
	<li style="text-align:justify">
	<p><span style="font-size:14px"><span style="color:black"><em>P</em>(<em>G</em>&nbsp;AND&nbsp;<em>H</em>) =&nbsp;<em>P</em>(<em>G</em>)<em>P</em>(<em>H</em>)</span></span></p>
	</li>
</ul>

<p style="text-align:justify">&nbsp;</p>

<p style="text-align:justify"><span style="font-size:14px"><span style="color:black">a. Show that&nbsp;<em>P</em>(<em>G</em>|<em>H</em>) =&nbsp;<em>P</em>(<em>G</em>).</span></span></p>

<p style="text-align:justify">&nbsp;</p>

<p style="text-align:justify"><strong><span style="font-size:14px"><span style="color:#002060">Solution&nbsp;3.9</span></span></strong></p>

<p style="text-align:justify">&nbsp;</p>

<p style="text-align:justify"><span style="font-size:14px"><span style="color:black"><span class="math-tex">\(P(G\H)=\frac{P(G\hspace{0.1cm}AND\hspace{0.1cm}H)}{P(H)}=\frac{3}{5}=6=P(G)\)</span></span></span></p>

<p style="text-align:justify">&nbsp;</p>

<p style="text-align:justify"><span style="font-size:14px"><span style="color:black">. Show&nbsp;<em>P</em>(<em>G</em>&nbsp;AND&nbsp;<em>H</em>) =&nbsp;<em>P</em>(<em>G</em>)<em>P</em>(<em>H</em>).</span></span></p>

<p style="text-align:justify">&nbsp;</p>

<p style="text-align:justify"><strong><span style="font-size:14px"><span style="color:#002060">Solution&nbsp;3.9</span></span></strong></p>

<p style="text-align:justify">&nbsp;</p>

<p style="text-align:justify"><span style="font-size:14px"><em><span style="color:#424242">P</span></em>(<em><span style="color:#424242">G</span></em>)<em><span style="color:#424242">P</span></em>(<em><span style="color:#424242">H</span></em>) = (.6)(.5) = .3 =&nbsp;<em><span style="color:#424242">P</span></em>(<em><span style="color:#424242">G</span></em>&nbsp;AND&nbsp;<em><span style="color:#424242">H</span></em>)</span></p>

<p style="text-align:justify">&nbsp;</p>

<p style="text-align:justify"><span style="font-size:14px"><span style="color:black">Since&nbsp;<em>G</em>&nbsp;and&nbsp;<em>H</em>&nbsp;are independent, knowing that a person is taking a science class does not change the chance that he or she is taking a math class. If the two events had not been independent, that is, they are dependent, then knowing that a person is taking a science class would change the chance he or she is taking math. For practice, show that&nbsp;<em>P</em>(<em>H</em>|<em>G</em>) =&nbsp;<em>P</em>(<em>H</em>) to show that&nbsp;<em>G</em>&nbsp;and&nbsp;<em>H</em>&nbsp;are independent events.</span></span></p>

<p style="text-align:justify">&nbsp;</p>

<p style="text-align:justify"><span style="font-size:14px"><span style="color:#000000"><strong>EXAMPLE&nbsp;3.10</strong></span></span></p>

<p style="text-align:justify">&nbsp;</p>

<p style="text-align:justify"><span style="font-size:14px"><span style="color:black">Let event&nbsp;<em>C</em>&nbsp;= taking an English class. Let event&nbsp;<em>D</em>&nbsp;= taking a speech class.</span></span></p>

<p style="text-align:justify">&nbsp;</p>

<p style="text-align:justify"><span style="font-size:14px"><span style="color:black">Suppose&nbsp;<em>P</em>(<em>C</em>) = .75,&nbsp;<em>P</em>(<em>D</em>) = .3,&nbsp;<em>P</em>(<em>C</em>|<em>D</em>) = .75 and&nbsp;<em>P</em>(<em>C</em>&nbsp;AND&nbsp;<em>D</em>) = .225.</span></span></p>

<p style="text-align:justify">&nbsp;</p>

<p style="text-align:justify"><span style="font-size:14px"><span style="color:black">Justify your answers to the following questions numerically.</span></span></p>

<p style="text-align:justify">&nbsp;</p>

<ol style="list-style-type:lower-alpha">
	<li>
	<p style="text-align:justify"><span style="font-size:14px"><span style="color:black">Are&nbsp;<em>C</em>&nbsp;and&nbsp;<em>D</em>&nbsp;independent?</span></span></p>
	</li>
	<li>
	<p style="text-align:justify"><span style="font-size:14px"><span style="color:black">Are&nbsp;<em>C</em>&nbsp;and&nbsp;<em>D</em>&nbsp;mutually exclusive?</span></span></p>
	</li>
	<li>
	<p style="text-align:justify"><span style="font-size:14px"><span style="color:black">What is&nbsp;<em>P</em>(<em>D</em>|<em>C</em>)?</span></span></p>
	</li>
</ol>

<p style="text-align:justify">&nbsp;</p>

<p style="text-align:justify"><span style="font-size:14px"><span style="color:#000000"><strong>Solution&nbsp;3.10</strong></span></span></p>

<p style="text-align:justify">&nbsp;</p>

<ol style="list-style-type:lower-alpha">
	<li>
	<p style="text-align:justify"><span style="font-size:14px"><span style="color:black">Yes, because&nbsp;<em>P</em>(<em>C</em>|<em>D</em>) = .75 =&nbsp;<em>P</em>(<em>C</em>).</span></span></p>
	</li>
	<li>
	<p style="text-align:justify"><span style="font-size:14px"><span style="color:black">No, because&nbsp;<em>P</em>(<em>C</em>&nbsp;AND&nbsp;<em>D</em>) is not equal to zero.</span></span></p>
	</li>
	<li>
	<p style="text-align:justify"><span style="font-size:14px"><em><span style="color:black">P</span></em><span style="color:black">(</span><em><span style="color:black">D</span></em><span style="color:black">|</span><em><span style="color:black">C</span></em><span style="color:black">) </span><span class="math-tex">\(=\frac{P(C\hspace{0.1cm}AND\hspace{0.1cm}D)}{P(C)}=\frac{0.225}{0.750}=.3\)</span></span></p>
	</li>
</ol>

<p style="text-align:justify">&nbsp;</p>

<p style="text-align:justify"><span style="font-size:14px"><strong>EXAMPLE&nbsp;3.11</strong></span></p>

<p style="text-align:justify">&nbsp;</p>

<ol style="list-style-type:lower-alpha">
	<li style="text-align:justify">
	<p><span style="font-size:14px"><span style="color:black">In a box there are three red cards and five blue cards. The red cards are marked with the numbers 1, 2, and 3, and the blue cards are marked with the numbers 1, 2, 3, 4, and 5. The cards are well-shuffled. You reach into the box (you cannot see into it) and draw one card.</span></span></p>
	</li>
</ol>

<p style="text-align:justify">&nbsp;</p>

<p style="margin-left:48px; text-align:justify"><span style="font-size:14px"><span style="color:black">Let&nbsp;<em>R</em>&nbsp;= red card is drawn,&nbsp;<em>B</em>&nbsp;= blue card is drawn,&nbsp;<em>E</em>&nbsp;= even-numbered card is drawn.</span></span></p>

<p style="margin-left:48px; text-align:justify">&nbsp;</p>

<p style="margin-left:48px; text-align:justify"><span style="font-size:14px"><span style="color:black">The sample space&nbsp;<em>S</em>&nbsp;=&nbsp;<em>R</em>1,&nbsp;<em>R</em>2,&nbsp;<em>R</em>3,&nbsp;<em>B</em>1,&nbsp;<em>B</em>2,&nbsp;<em>B</em>3,&nbsp;<em>B</em>4,&nbsp;<em>B</em>5.&nbsp;<em>S</em>&nbsp;has eight outcomes.</span></span></p>

<p style="margin-left:48px; text-align:justify">&nbsp;</p>

<ul>
	<li style="text-align:justify">
	<p><span style="font-size:14px"><span style="color:black"><span class="math-tex">\(P(R)=\frac{3}{8}. P(B)=\frac{5}{8}\)</span></span>&nbsp;&nbsp;<em><span style="color:black">P</span></em><span style="color:black">(<em>R</em>&nbsp;AND&nbsp;<em>B</em>) = 0. You cannot draw one card that is both red and blue.</span></span></p>
	</li>
	<li style="text-align:justify">
	<p><span style="font-size:14px"><span style="color:black"><span class="math-tex">\(P(E)=\frac{3}{8}. \)</span>There are three even-numbered cards,&nbsp;<em>R</em>2,&nbsp;<em>B</em>2, and&nbsp;<em>B</em>4.</span></span></p>
	</li>
	<li style="text-align:justify">
	<p><span style="font-size:14px"><em><span style="color:black">P</span></em><span style="color:black">(<em>E</em>|<em>B</em>&nbsp;</span>)<span class="math-tex">\(=\frac{2}{5}\)</span>&nbsp;<span style="color:black">. There are five blue cards:&nbsp;<em>B</em>1,&nbsp;<em>B</em>2,&nbsp;<em>B</em>3,&nbsp;<em>B</em>4, and&nbsp;<em>B</em>5. Out of the blue cards, there are two even cards;&nbsp;<em>B</em>2 and&nbsp;<em>B</em>4.</span></span></p>
	</li>
	<li style="text-align:justify">
	<p><span style="font-size:14px"><em><span style="color:black">P</span></em><span style="color:black">(<em>B</em>|<em>E</em>)&nbsp;<span class="math-tex">\(=\frac{2}{3}\)</span></span>&nbsp;</span></p>
	</li>
	<li>
	<p style="text-align:justify"><span style="font-size:14px"><span style="color:black">. There are three even-numbered cards:&nbsp;<em>R</em>2,&nbsp;<em>B</em>2, and&nbsp;<em>B</em>4. Out of the even-numbered cards, two are blue;&nbsp;<em>B</em>2 and&nbsp;<em>B</em>4.</span></span></p>
	</li>
	<li>
	<p style="text-align:justify"><span style="font-size:14px"><span style="color:black">The events&nbsp;<em>R</em>&nbsp;and&nbsp;<em>B</em>&nbsp;are mutually exclusive because&nbsp;<em>P</em>(<em>R</em>&nbsp;AND&nbsp;<em>B</em>) = 0.</span></span></p>
	</li>
	<li style="text-align:justify">
	<p><span style="font-size:14px"><span style="color:black">Let&nbsp;<em>G</em>&nbsp;= card with a number greater than 3.&nbsp;<em>G</em>&nbsp;= {<em>B</em>4,&nbsp;<em>B</em>5}.&nbsp;<em>P</em>(<em>G</em>)&nbsp;<span class="math-tex">\(=\frac{2}{8}\)</span>&nbsp;Let&nbsp;<em>H</em>&nbsp;= blue card numbered between one and four, inclusive.&nbsp;<em>H</em>&nbsp;= {<em>B</em>1,&nbsp;<em>B</em>2,&nbsp;<em>B</em>3,&nbsp;<em>B</em>4}.&nbsp;<em>P</em>(<em>G</em>|<em>H</em>)=&nbsp;<span class="math-tex">\(\frac{1}{4}\)</span>&nbsp;. The only card in&nbsp;<em>H</em>&nbsp;that has a number greater than three is&nbsp;<em>B</em>4. Since&nbsp;<span class="math-tex">\(=\frac{2}{8}=\frac{1}{4}\)</span>&nbsp;</span><em><span style="color:black">P</span></em><span style="color:black">(<em>G</em>) =&nbsp;<em>P</em>(<em>G</em>|<em>H</em>), which means that&nbsp;<em>G</em>&nbsp;and&nbsp;<em>H</em>&nbsp;are independent.</span></span></p>
	</li>
</ul>

<p style="margin-left:48px; text-align:justify"><span style="font-size:14px"><span style="color:black">??????????????</span></span></p>

<p style="text-align:justify"><span style="font-size:14px"><span style="color:#000000"><strong>EXAMPLE&nbsp;3.12</strong></span></span></p>

<p style="text-align:justify">&nbsp;</p>

<p style="text-align:justify"><span style="font-size:14px"><span style="color:black">In a particular class, 60 percent of the students are female. Fifty percent of all students in the class have long hair. Forty-five percent of the students are female and have long hair. Of the female students, 75 percent have long hair. Let&nbsp;<em>F</em>&nbsp;be the event that a student is female. Let&nbsp;<em>L</em>&nbsp;be the event that a student has long hair. One student is picked randomly. Are the events of being female and having long hair independent?</span></span></p>

<p style="text-align:justify">&nbsp;</p>

<p style="text-align:justify"><span style="font-size:14px"><span style="color:black">The following probabilities are given in this example:</span></span></p>

<p style="text-align:justify">&nbsp;</p>

<ul>
	<li>
	<p style="text-align:justify"><span style="font-size:14px"><span style="color:black"><em>P</em>(<em>F</em>) = 0.60;&nbsp;<em>P</em>(<em>L</em>) = 0.50</span></span></p>
	</li>
	<li>
	<p style="text-align:justify"><span style="font-size:14px"><span style="color:black"><em>P</em>(<em>F</em>&nbsp;AND&nbsp;<em>L</em>) = 0.45</span></span></p>
	</li>
	<li>
	<p style="text-align:justify"><span style="font-size:14px"><span style="color:black"><em>P</em>(<em>L</em>|<em>F</em>) = 0.75</span></span></p>
	</li>
</ul>

<p style="text-align:justify"><span style="font-size:14px"><span style="color:black">??????????????</span></span></p>

<p style="text-align:justify"><span style="font-size:14px"><strong>Solution 3.12</strong></span></p>

<p style="text-align:justify">&nbsp;</p>

<p style="text-align:justify"><span style="font-size:14px"><span style="background-color:#ededed"><span style="color:black">Check whether&nbsp;</span></span><em><span style="background-color:#ededed"><span style="color:#424242">P</span></span></em><span style="background-color:#ededed"><span style="color:black">(</span></span><em><span style="background-color:#ededed"><span style="color:#424242">F</span></span></em><span style="background-color:#ededed"><span style="color:black">&nbsp;AND&nbsp;</span></span><em><span style="background-color:#ededed"><span style="color:#424242">L</span></span></em><span style="background-color:#ededed"><span style="color:black">) =&nbsp;</span></span><em><span style="background-color:#ededed"><span style="color:#424242">P</span></span></em><span style="background-color:#ededed"><span style="color:black">(</span></span><em><span style="background-color:#ededed"><span style="color:#424242">F</span></span></em><span style="background-color:#ededed"><span style="color:black">)</span></span><em><span style="background-color:#ededed"><span style="color:#424242">P</span></span></em><span style="background-color:#ededed"><span style="color:black">(</span></span><em><span style="background-color:#ededed"><span style="color:#424242">L</span></span></em><span style="background-color:#ededed"><span style="color:black">). We are given that&nbsp;</span></span><em><span style="background-color:#ededed"><span style="color:#424242">P</span></span></em><span style="background-color:#ededed"><span style="color:black">(</span></span><em><span style="background-color:#ededed"><span style="color:#424242">F</span></span></em><span style="background-color:#ededed"><span style="color:black">&nbsp;AND&nbsp;</span></span><em><span style="background-color:#ededed"><span style="color:#424242">L</span></span></em><span style="background-color:#ededed"><span style="color:black">) = 0.45, but&nbsp;</span></span><em><span style="background-color:#ededed"><span style="color:#424242">P</span></span></em><span style="background-color:#ededed"><span style="color:black">(</span></span><em><span style="background-color:#ededed"><span style="color:#424242">F</span></span></em><span style="background-color:#ededed"><span style="color:black">)</span></span><em><span style="background-color:#ededed"><span style="color:#424242">P</span></span></em><span style="background-color:#ededed"><span style="color:black">(</span></span><em><span style="background-color:#ededed"><span style="color:#424242">L</span></span></em><span style="background-color:#ededed"><span style="color:black">) = (.60)(.50) = .30. The events of being female and having long hair are not independent because&nbsp;</span></span><em><span style="background-color:#ededed"><span style="color:#424242">P</span></span></em><span style="background-color:#ededed"><span style="color:black">(</span></span><em><span style="background-color:#ededed"><span style="color:#424242">F</span></span></em><span style="background-color:#ededed"><span style="color:black">&nbsp;AND&nbsp;</span></span><em><span style="background-color:#ededed"><span style="color:#424242">L</span></span></em><span style="background-color:#ededed"><span style="color:black">) does not equal&nbsp;</span></span><em><span style="background-color:#ededed"><span style="color:#424242">P</span></span></em><span style="background-color:#ededed"><span style="color:black">(</span></span><em><span style="background-color:#ededed"><span style="color:#424242">F</span></span></em><span style="background-color:#ededed"><span style="color:black">)</span></span><em><span style="background-color:#ededed"><span style="color:#424242">P</span></span></em><span style="background-color:#ededed"><span style="color:black">(</span></span><em><span style="background-color:#ededed"><span style="color:#424242">L</span></span></em><span style="background-color:#ededed"><span style="color:black">).</span></span></span></p>

<p style="text-align:justify">&nbsp;</p>

<p style="text-align:justify"><span style="font-size:14px"><strong>Solution 3.12</strong></span></p>

<p style="text-align:justify">&nbsp;</p>

<p style="text-align:justify"><span style="font-size:14px"><span style="background-color:#ededed"><span style="color:black">Check whether&nbsp;</span></span><em><span style="background-color:#ededed"><span style="color:#424242">P</span></span></em><span style="background-color:#ededed"><span style="color:black">(</span></span><em><span style="background-color:#ededed"><span style="color:#424242">L</span></span></em><span style="background-color:#ededed"><span style="color:black">|</span></span><em><span style="background-color:#ededed"><span style="color:#424242">F</span></span></em><span style="background-color:#ededed"><span style="color:black">) equals&nbsp;</span></span><em><span style="background-color:#ededed"><span style="color:#424242">P</span></span></em><span style="background-color:#ededed"><span style="color:black">(</span></span><em><span style="background-color:#ededed"><span style="color:#424242">L</span></span></em><span style="background-color:#ededed"><span style="color:black">). We are given that&nbsp;</span></span><em><span style="background-color:#ededed"><span style="color:#424242">P</span></span></em><span style="background-color:#ededed"><span style="color:black">(</span></span><em><span style="background-color:#ededed"><span style="color:#424242">L</span></span></em><span style="background-color:#ededed"><span style="color:black">|</span></span><em><span style="background-color:#ededed"><span style="color:#424242">F</span></span></em><span style="background-color:#ededed"><span style="color:black">) = .75, but&nbsp;</span></span><em><span style="background-color:#ededed"><span style="color:#424242">P</span></span></em><span style="background-color:#ededed"><span style="color:black">(</span></span><em><span style="background-color:#ededed"><span style="color:#424242">L</span></span></em><span style="background-color:#ededed"><span style="color:black">) = .50; they are not equal. The events of being female and having long hair are not independent.</span></span></span></p>

<p style="text-align:justify"><br />
<br />
<strong><span style="font-size:14px">Interpretation of Results</span></strong></p>

<p style="text-align:justify">&nbsp;</p>

<p style="text-align:justify"><span style="font-size:14px">The events of being female and having long hair are not independent; knowing that a student is female changes the probability that a student has long hair.</span></p>

<p style="text-align:justify">&nbsp;</p>

<p style="text-align:justify"><span style="font-size:14px"><span style="color:#000000">EXAMPLE&nbsp;3.13</span></span></p>

<p style="text-align:justify">&nbsp;</p>

<ol style="list-style-type:lower-alpha">
	<li>
	<p style="text-align:justify"><span style="font-size:14px"><span style="color:black">Toss one fair coin (the coin has two sides,&nbsp;<em>H</em>&nbsp;and&nbsp;<em>T</em>). The outcomes are ________. Count the outcomes. There are ________ outcomes.</span></span></p>
	</li>
	<li>
	<p style="text-align:justify"><span style="font-size:14px"><span style="color:black">Toss one fair, six-sided die (the die has 1, 2, 3, 4, 5, or 6 dots on a side). The outcomes are ________. Count the outcomes. There are ________ outcomes.</span></span></p>
	</li>
	<li>
	<p style="text-align:justify"><span style="font-size:14px"><span style="color:black">Multiply the two numbers of outcomes. The answer is ________.</span></span></p>
	</li>
	<li>
	<p style="text-align:justify"><span style="font-size:14px"><span style="color:black">If you flip one fair coin and follow it with the toss of one fair, six-sided die, the answer in Part c is the number of outcomes (size of the sample space). List the outcomes. Hint&mdash;Two of the outcomes are&nbsp;<em>H</em>1 and&nbsp;<em>T</em>6.</span></span></p>
	</li>
	<li>
	<p style="text-align:justify"><span style="font-size:14px"><span style="color:black">Event&nbsp;<em>A</em>&nbsp;= heads (<em>H</em>) on the coin followed by an even number (2, 4, 6) on the die.<br />
	<em>A</em>&nbsp;= {________}. Find&nbsp;<em>P</em>(<em>A</em>).</span></span></p>
	</li>
	<li>
	<p style="text-align:justify"><span style="font-size:14px"><span style="color:black">Event&nbsp;<em>B</em>&nbsp;= heads on the coin followed by a three on the die.&nbsp;<em>B</em>&nbsp;= {________}. Find&nbsp;<em>P</em>(<em>B</em>).</span></span></p>
	</li>
	<li>
	<p style="text-align:justify"><span style="font-size:14px"><span style="color:black">Are&nbsp;<em>A</em>&nbsp;and&nbsp;<em>B</em>&nbsp;mutually exclusive? Hint&mdash;What is&nbsp;<em>P</em>(<em>A</em>&nbsp;AND&nbsp;<em>B</em>)? If&nbsp;<em>P</em>(<em>A</em>&nbsp;AND&nbsp;<em>B</em>) = 0, then&nbsp;<em>A</em>&nbsp;and&nbsp;<em>B</em>&nbsp;are mutually exclusive.</span></span></p>
	</li>
	<li>
	<p style="text-align:justify"><span style="font-size:14px"><span style="color:black">Are&nbsp;<em>A</em>&nbsp;and&nbsp;<em>B</em>&nbsp;independent? Hint&mdash;Is&nbsp;<em>P</em>(<em>A</em>&nbsp;AND&nbsp;<em>B</em>) =&nbsp;<em>P</em>(<em>A</em>)<em>P</em>(<em>B</em>)? If&nbsp;<em>P</em>(<em>A</em>&nbsp;AND&nbsp;<em>B</em>) =&nbsp;<em>P</em>(<em>A</em>)<em>P</em>(<em>B</em>), then&nbsp;<em>A</em>&nbsp;and&nbsp;<em>B</em>&nbsp;are independent. If not, then they are dependent.</span></span></p>
	</li>
</ol>

<p style="text-align:justify"><span style="font-size:14px"><span style="color:black">???????</span></span></p>

<p style="text-align:justify"><span style="font-size:14px"><span style="color:#000000"><strong>Solution&nbsp;3.13</strong></span></span></p>

<p style="text-align:justify"><span style="font-size:14px"><span style="color:#000000"><strong>???????</strong></span></span></p>

<ol style="list-style-type:lower-alpha">
	<li>
	<p style="text-align:justify"><span style="font-size:14px"><em><span style="color:black">H</span></em><span style="color:black">&nbsp;and&nbsp;<em>T</em>; 2</span></span></p>
	</li>
	<li>
	<p style="text-align:justify"><span style="font-size:14px"><span style="color:black">1, 2, 3, 4, 5, 6; 6</span></span></p>
	</li>
	<li>
	<p style="text-align:justify"><span style="font-size:14px"><span style="color:black">2(6) = 12</span></span></p>
	</li>
	<li style="text-align:justify">
	<p><span style="font-size:14px"><span style="color:black">Make a systematic list of possible outcomes. Start by listing all possible outcomes when the coin shows tails (<em>T</em>). Then list the outcomes that are possible when the coin shows heads (<em>H</em>):&nbsp;<em>T</em>1,&nbsp;<em>T</em>2,&nbsp;<em>T</em>3,&nbsp;<em>T</em>4,&nbsp;<em>T</em>5,&nbsp;<em>T</em>6,&nbsp;<em>H</em>1,&nbsp;<em>H</em>2,&nbsp;<em>H</em>3,&nbsp;<em>H</em>4,&nbsp;<em>H</em>5,&nbsp;<em>H</em>6</span></span></p>
	</li>
	<li style="text-align:justify">
	<p><span style="font-size:14px"><em><span style="color:black">A</span></em><span style="color:black">&nbsp;= {<em>H</em>2,&nbsp;<em>H</em>4,&nbsp;<em>H</em>6};&nbsp;</span><span class="math-tex">\(P(A)=\frac{NUMBER\hspace{0.1cm}OF\hspace{0.1cm}OUTCOMES\hspace{0.1cm}IN\hspace{0.1cm}A}{NUMBER\hspace{0.1cm}OF\hspace{0.1cm}POSSIBLE\hspace{0.1cm}OUTCOMES}=\frac{3}{12}\)</span></span></p>
	</li>
	<li style="text-align:justify">
	<p><span style="font-size:14px"><em><span style="color:black">B</span></em><span style="color:black">&nbsp;= {<em>H</em>3};&nbsp;<em>P</em>(<em>B</em>) </span><span class="math-tex">\(=\frac{1}{2}\)</span></span></p>
	</li>
	<li>
	<p style="text-align:justify"><span style="font-size:14px"><span style="color:black">Yes, because&nbsp;<em>P</em>(<em>A</em>&nbsp;AND&nbsp;<em>B</em>) = 0</span></span></p>
	</li>
	<li style="text-align:justify">
	<p><span style="font-size:14px"><em><span style="color:black">P</span></em><span style="color:black">(<em>A</em>&nbsp;AND&nbsp;<em>B</em>) = 0.&nbsp;<em>P</em>(<em>A</em>)<em>P</em>(<em>B</em>)=<span class="math-tex">\((\frac{3}{12})(\frac{1}{2})\)</span>&nbsp;</span><em><span style="color:black">P</span></em><span style="color:black">(<em>A</em>&nbsp;AND&nbsp;<em>B</em>) does not equal&nbsp;<em>P</em>(<em>A</em>)<em>P</em>(<em>B</em>), so&nbsp;<em>A</em>&nbsp;and&nbsp;<em>B</em>&nbsp;are dependent.</span></span></p>
	</li>
</ol>

<p>&nbsp;</p>