<p style="text-align:justify"><span style="font-size:14px"><span style="color:black">A&nbsp;<strong>tree diagram</strong>&nbsp;is a special type of graph used to determine the outcomes of an experiment. It consists of&nbsp;<em>branches</em>&nbsp;that are labeled with either frequencies or probabilities. Tree diagrams can make some probability problems easier to visualize and solve. The following example illustrates how to use a tree diagram:</span></span></p>

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

<p style="text-align:justify"><span style="font-size:14px"><strong>EXAMPLE&nbsp;3.24</strong></span></p>

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

<p style="text-align:justify"><span style="font-size:14px">In an urn, there are 11 balls. Three balls are red (<em>R</em>) and eight balls are blue (<em>B</em>). Draw two balls, one at a time,&nbsp;<strong>with replacement</strong>.&nbsp;<em>With replacement</em>&nbsp;means that you put the first ball back in the urn before you select the second ball. Therefore, you are selecting from exactly the same group each time, so each draw is independent. The tree diagram shows all the possible outcomes.</span></p>

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

<p style="text-align:justify"><span style="font-size:14px"><img alt="" src="https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16239179121623917912.jpg" style="height:438px; width:800px" /></span></p>

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

<p style="text-align:justify"><span style="font-size:14px"><strong>Figure&nbsp;3.10</strong>&nbsp;Total = 64 + 24 + 24 + 9 = 121.</span></p>

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

<p style="text-align:justify"><span style="font-size:14px">The first set of branches represents the first draw. There are 8 ways to draw a blue marble and 3 ways to draw a red one. The second set of branches represents the second draw. Regardless of the choice on the first draw, there are again eight ways to draw a blue marble and 3 ways to draw a red one. Read down each branch to see the total number of possible outcomes. For example, there are 8 ways to get a blue marble on the first draw, and eight ways to get one on the second draw, so there are 8 &times; 8 = 64 different ways to draw two blue marbles in succession. Each of the outcomes is distinct. In fact, we can list each red ball as&nbsp;<em>R</em>1,&nbsp;<em>R</em>2, and&nbsp;<em>R</em>3 and each blue ball as&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>B</em>6,&nbsp;<em>B</em>7, and&nbsp;<em>B</em>8. Then the nine&nbsp;<em>RR</em>&nbsp;outcomes can be written as follows:</span></p>

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

<p style="text-align:justify"><span style="font-size:14px"><em>R</em>1<em>R</em>1,&nbsp;<em>R</em>1<em>R</em>2,&nbsp;<em>R</em>1<em>R</em>3,&nbsp;<em>R</em>2<em>R</em>1,&nbsp;<em>R</em>2<em>R</em>2,&nbsp;<em>R</em>2<em>R</em>3,&nbsp;<em>R</em>3<em>R</em>1,&nbsp;<em>R</em>3<em>R</em>2,&nbsp;<em>R</em>3<em>R</em>3.</span></p>

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

<p style="text-align:justify"><span style="font-size:14px">The other outcomes are similar.</span></p>

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

<p style="text-align:justify"><span style="font-size:14px">There are a total of 11 balls in the urn. Draw two balls, one at a time, with replacement. There are 11(11) = 121 outcomes, the size of the&nbsp;<strong>sample space</strong>.</span><br />
&nbsp;</p>

<p style="text-align:justify"><span style="font-size:14px">a. List the 24&nbsp;<em>BR</em>&nbsp;outcomes:&nbsp;<em>B</em>1<em>R</em>1,&nbsp;<em>B</em>1<em>R</em>2,&nbsp;<em>B</em>1<em>R</em>3, . . .</span></p>

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

<p style="text-align:justify"><span style="font-size:14px"><span style="color:#000000"><strong>Solution&nbsp;3.24</strong></span></span></p>

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

<p style="text-align:justify"><span style="font-size:14px"><span style="color:#002060">a. </span><span style="color:black">We know that there will be 24 different possible outcomes because there are eight ways to draw blue and three ways to draw red. Make a systematic list of possible outcomes that consist of a blue marble on the first draw and a red marble on the second draw.</span></span><br />
<br />
&nbsp;</p>

<p style="text-align:justify"><span style="font-size:14px"><em>B</em>1<em>R</em>1,&nbsp;<em>B</em>1<em>R</em>2,&nbsp;<em>B</em>1<em>R</em>3<br />
<em>B</em>2<em>R</em>1,&nbsp;<em>B</em>2<em>R</em>2,&nbsp;<em>B</em>2<em>R</em>3<br />
<em>B</em>3<em>R</em>1,&nbsp;<em>B</em>3<em>R</em>2,&nbsp;<em>B</em>3<em>R</em>3<br />
<em>B</em>4<em>R</em>1,&nbsp;<em>B</em>4<em>R</em>2,&nbsp;<em>B</em>4<em>R</em>3<br />
<em>B</em>5<em>R</em>1,&nbsp;<em>B</em>5<em>R</em>2,&nbsp;<em>B</em>5<em>R</em>3<br />
<em>B</em>6<em>R</em>1,&nbsp;<em>B</em>6<em>R</em>2,&nbsp;<em>B</em>6<em>R</em>3<br />
<em>B</em>7<em>R</em>1,&nbsp;<em>B</em>7<em>R</em>2,&nbsp;<em>B</em>7<em>R</em>3<br />
<em>B</em>8<em>R</em>1,&nbsp;<em>B</em>8<em>R</em>2,&nbsp;<em>B</em>8<em>R</em>3</span><br />
&nbsp;</p>

<p style="text-align:justify"><span style="font-size:14px">b. Calculate&nbsp;<em>P</em>(<em>RR</em>).</span></p>

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

<p style="text-align:justify"><span style="font-size:14px"><span style="color:#000000"><strong>Solution&nbsp;2</strong></span></span></p>

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

<p style="text-align:justify"><span style="font-size:14px">b. You can use the tree diagram. There are nine ways to draw two reds and 121 possible outcomes. So,&nbsp;<span class="math-tex">\(P(RR)=\frac{9}{121}\)</span></span></p>

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

<p style="text-align:justify"><span style="font-size:14px">Each draw is independent, so you can also use the formula:&nbsp;<span class="math-tex">\(P(RR)=P(R)P(R)=(\frac{3}{11})(\frac{3}{11})=\frac{9}{121}\)</span></span></p>

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

<p style="text-align:justify"><span style="font-size:14px">c. Calculate&nbsp;<em>P</em>(<em>RB</em>&nbsp;OR&nbsp;<em>BR</em>).</span></p>

<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</strong></span></span></p>

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

<p style="text-align:justify"><span style="font-size:14px">c. The tree diagram shows that there are 24 ways to draw&nbsp;<em>RB</em>&nbsp;and 24 ways to draw&nbsp;<em>BR</em>. There are 121 possible outcomes, so&nbsp;<em>P</em>(<em>RB</em>&nbsp;or&nbsp;<em>BR</em>) <span class="math-tex">\(=\frac{24+24}{121}=\frac{48}{121}\)</span></span></p>

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

<p style="text-align:justify"><span style="font-size:14px">The events&nbsp;<em>RB</em>&nbsp;and&nbsp;<em>BR</em>&nbsp;are mutually exclusive, so&nbsp;<em>P</em>(<em>RB</em>&nbsp;OR&nbsp;<em>BR</em>) =&nbsp;<em>P</em>(<em>RB</em>) +&nbsp;<em>P</em>(<em>BR</em>) =&nbsp;<em>P</em>(<em>R</em>)<em>P</em>(<em>B</em>) +&nbsp;<em>P</em>(<em>B</em>)<em>P</em>(<em>R</em>)<span class="math-tex">\(=(\frac{3}{11})(\frac{8}{11})+(\frac{3}{11})(\frac{8}{11})=\frac{48}{121}\)</span></span></p>

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

<p style="text-align:justify"><span style="font-size:14px">d. Using the tree diagram, calculate&nbsp;<em>P</em>(<em>R</em>&nbsp;on 1st draw AND&nbsp;<em>B</em>&nbsp;on 2nd draw).</span></p>

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

<p style="text-align:justify"><span style="font-size:14px"><span style="color:#000000"><strong>Solution&nbsp;4</strong></span></span></p>

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

<p style="text-align:justify"><span style="font-size:14px">d. Follow the path on the tree. There are three ways to get a red marble on the first draw and eight ways to get a blue on the second draw. There are 3 &times; 8 = 24 ways to draw red then blue, so&nbsp;<em>P</em>(<em>RB</em>) =&nbsp;24121.</span></p>

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

<p style="text-align:justify"><span style="font-size:14px">Can you think of another way to find this probability?&nbsp;<em>P</em>(<em>R</em>&nbsp;on 1st draw AND&nbsp;<em>B</em>&nbsp;on 2nd draw)&nbsp;<span class="math-tex">\(=P(RB)=(\frac{3}{11})(\frac{8}{11})=\frac{24}{121}\)</span></span></p>

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

<p style="text-align:justify"><span style="font-size:14px">e. Using the tree diagram, calculate&nbsp;<em>P</em>(<em>R</em>&nbsp;on 2nd draw GIVEN&nbsp;<em>B</em>&nbsp;on 1st draw).</span></p>

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

<p style="text-align:justify"><span style="font-size:14px"><span style="color:#000000"><strong>Solution&nbsp;5</strong></span></span></p>

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

<p style="text-align:justify"><span style="font-size:14px">e. Given that a blue marble is selected first, we need only follow the left set of branches on the tree diagram. In this case, there are three ways to obtain red on the second draw and 11 possible outcomes.</span></p>

<p style="text-align:justify"><span style="font-size:14px"><img alt="" src="https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16239184651623918465.png" style="height:533px; width:800px" /></span></p>

<p style="text-align:justify"><span style="font-size:14px"><strong><span style="color:#424242">Figure&nbsp;3.11</span></strong></span></p>

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

<p style="text-align:justify"><span style="font-size:14px">P(R on 2nd draw GIVEN B on 1st)=P(R on 2nd |B on 1st)<span class="math-tex">\(=\frac{3}{11}\)</span></span></p>

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

<p style="text-align:justify"><span style="font-size:14px">You can also use the formula</span></p>

<p style="text-align:justify"><span style="font-size:14px">P(R on 2nd |B on 1st)</span></p>

<p style="text-align:justify"><span style="font-size:14px"><span class="math-tex">\(=\frac{P(R\hspace{0.1cm}on\hspace{0.1cm}2nd\hspace{0.1cm}AND\hspace{0.1cm}B\hspace{0.1cm}on\hspace{0.1cm}1st)}{P(B\hspace{0.1cm}on\hspace{0.1cm}1st)}=\frac{\frac{24}{121}}{\frac{64+24}{121}}=\frac{24}{88}=\frac{3}{11}\)</span></span></p>

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

<p style="text-align:justify"><span style="font-size:14px">f. Using the tree diagram, calculate&nbsp;<em>P</em>(<em>BB</em>).</span></p>

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

<p style="text-align:justify"><span style="font-size:14px"><span style="color:#000000"><strong>Solution&nbsp;6</strong></span></span></p>

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

<p style="text-align:justify"><span style="font-size:14px">f.&nbsp;<span class="math-tex">\(P(BB)=\frac{64}{121}\)</span></span></p>

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

<p style="text-align:justify"><span style="font-size:14px">g. Using the tree diagram, calculate&nbsp;<em><span style="color:#424242">P</span></em>(<em><span style="color:#424242">B</span></em>&nbsp;on the 2nd draw GIVEN&nbsp;<em><span style="color:#424242">R</span></em>&nbsp;on the first draw).</span></p>

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

<p style="text-align:justify"><span style="font-size:14px"><span style="color:#000000"><strong>Solution&nbsp;7</strong></span></span></p>

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

<p style="text-align:justify"><span style="font-size:14px"><em><span style="color:black">P</span></em><span style="color:black">(<em>B</em>&nbsp;on&nbsp;2nd&nbsp;</span>draw|<em><span style="color:#424242">R</span></em>&nbsp;on&nbsp;1st&nbsp;draw) <span class="math-tex">\(=\frac{8}{11}\)</span></span></p>

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

<p style="text-align:justify"><span style="font-size:14px">There are 9 + 24 outcomes that have&nbsp;<em><span style="color:#424242">R</span></em>&nbsp;on the first draw (9&nbsp;<em><span style="color:#424242">RR</span></em>&nbsp;and 24&nbsp;<em><span style="color:#424242">RB</span></em>). The sample space is then 9 + 24 = 33. Twenty-four of the 33 outcomes have&nbsp;<em><span style="color:#424242">B</span></em>&nbsp;on the second draw. The probability is then&nbsp;<span class="math-tex">\(\frac{24}{33}\)</span></span></p>

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

<p style="text-align:justify"><span style="font-size:14px"><strong>EXAMPLE&nbsp;3.25</strong></span></p>

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

<p style="text-align:justify"><span style="font-size:14px">An urn has three red marbles and eight blue marbles in it. Draw two marbles, one at a time, this time without replacement, from the urn.&nbsp;<strong><em>Without replacement</em></strong>&nbsp;means that you do not put the first ball back before you select the second marble. Following is a tree diagram for this situation. The branches are labeled with probabilities instead of frequencies. The numbers at the ends of the branches are calculated by multiplying the numbers on the two corresponding branches, for example,&nbsp;<em>P</em>(<em>RR</em>) <span class="math-tex">\(=(\frac{3}{11})(\frac{2}{10})=\frac{6}{110}\)</span></span></p>

<p style="text-align:justify"><span style="font-size:14px"><img alt="" src="https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16239192451623919245.png" style="height:533px; width:800px" /><strong><span style="color:#424242">Figure&nbsp;3.13</span></strong><span style="color:#424242">&nbsp;Total </span></span></p>

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

<p style="text-align:justify"><span style="font-size:14px"><span class="math-tex">\(=\frac{56+24+24+6}{110}=\frac{110}{110}=1\)</span></span></p>

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

<p style="text-align:justify"><span style="font-size:14px">Calculate the following probabilities using the tree diagram:</span><br />
<br />
&nbsp;</p>

<p style="text-align:justify"><span style="font-size:14px">a.&nbsp;<em>P</em>(<em>RR</em>) = ________</span></p>

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

<p style="text-align:justify"><span style="font-size:14px">Calculate the following probabilities using the tree diagram:</span><br />
&nbsp;</p>

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

<p style="text-align:justify"><span style="font-size:14px">a.&nbsp;<em><span style="color:#424242">P</span></em>(<em><span style="color:#424242">RR</span></em>) = ________</span></p>

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

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

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

<p style="text-align:justify"><span style="font-size:14px">a.&nbsp;<em><span style="color:#424242">P</span></em>(<em><span style="color:#424242">RR</span></em>) <span class="math-tex">\(=(\frac{3}{11})(\frac{2}{10})=\frac{6}{110}\)</span></span></p>

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

<p style="text-align:justify"><span style="font-size:14px">b. Fill in the blanks.</span></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">RB</span></em>&nbsp;OR&nbsp;<em><span style="color:#424242">BR</span></em>) <span class="math-tex">\(=(\frac{3}{11})(\frac{8}{10})\)</span>&nbsp;<span style="color:#424242">+&nbsp;(________)(________)&nbsp;</span><span class="math-tex">\(=\frac{48}{110}\)</span></span></p>

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

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

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

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

<p style="text-align:justify"><span style="font-size:14px">b.&nbsp;<em><span style="color:#424242">P</span></em>(<em><span style="color:#424242">RB</span></em>&nbsp;OR&nbsp;<em><span style="color:#424242">BR</span></em>) =&nbsp;<em><span style="color:#424242">P</span></em>(<em><span style="color:#424242">RB</span></em>) +&nbsp;<em><span style="color:#424242">P</span></em>(<em><span style="color:#424242">BR</span></em>) =&nbsp;<em><span style="color:#424242">P</span></em>(<em><span style="color:#424242">R</span></em>&nbsp;on 1st)&nbsp;<em><span style="color:#424242">P</span></em>(<em><span style="color:#424242">B</span></em>&nbsp;on 2nd) +&nbsp;<em><span style="color:#424242">P</span></em>(<em><span style="color:#424242">B</span></em>&nbsp;on 1st)&nbsp;<em><span style="color:#424242">P</span></em>(<em><span style="color:#424242">R</span></em>&nbsp;on 2nd) <span class="math-tex">\(=(\frac{3}{11})(\frac{8}{10})+(\frac{8}{11})(\frac{3}{10})=\frac{48}{110}\)</span></span></p>

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

<p style="text-align:justify"><span style="font-size:14px">c. Because this is a conditional probability, we restrict the sample space to consider only those outcomes that have a blue marble in the first draw. Look at the second level of the tree to see that&nbsp;<em><span style="color:#424242">P</span></em>(<em><span style="color:#424242">R</span></em>&nbsp;on 2nd|<em><span style="color:#424242">B</span></em>&nbsp;on 1st) <span class="math-tex">\(=\frac{3}{10}\)</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>Solution&nbsp;c</strong></span></span></p>

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

<p style="text-align:justify"><span style="font-size:14px">c.&nbsp;<em><span style="color:#424242">P</span></em>(<em><span style="color:#424242">R</span></em>&nbsp;on 2nd|<em><span style="color:#424242">B</span></em>&nbsp;on 1st)<span class="math-tex">\(=\frac{3}{10}\)</span></span></p>

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

<p style="text-align:justify"><span style="font-size:14px">d. Fill in the blanks.</span></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">R</span></em>&nbsp;on 1st AND&nbsp;<em><span style="color:#424242">B</span></em>&nbsp;on 2nd) =&nbsp;<em><span style="color:#424242">P</span></em>(<em><span style="color:#424242">RB</span></em>) = (________)(________)<span class="math-tex">\(=\frac{24}{100}\)</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>Solution&nbsp;d</strong></span></span></p>

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

<p style="text-align:justify"><span style="font-size:14px">d.&nbsp;<em><span style="color:#424242">P</span></em>(<em><span style="color:#424242">R</span></em>&nbsp;on 1st AND&nbsp;<em><span style="color:#424242">B</span></em>&nbsp;on 2nd) =&nbsp;<em><span style="color:#424242">P</span></em>(<em><span style="color:#424242">RB</span></em>) &nbsp;<span class="math-tex">\(=(\frac{3}{11})(\frac{8}{10})=\frac{24}{100}\)</span></span></p>

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

<p style="text-align:justify"><span style="font-size:14px">e. Find&nbsp;<em><span style="color:#424242">P</span></em>(<em><span style="color:#424242">BB</span></em>).</span></p>

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

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

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

<p style="text-align:justify"><span style="font-size:14px">e.&nbsp;<span class="math-tex">\(P(BB)=(\frac{8}{11})(\frac{7}{10})\)</span></span></p>

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

<p style="text-align:justify"><span style="font-size:14px">f. Find&nbsp;<em>P</em>(<em>B</em>&nbsp;on 2nd|<em>R</em>&nbsp;on 1st).</span></p>

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

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

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

<p style="text-align:justify"><span style="font-size:14px">Using the tree diagram,&nbsp;<em>P</em>(<em>B</em>&nbsp;on 2nd|<em>R</em>&nbsp;on 1st) =&nbsp;<em>P</em>(<em>R</em>|<em>B</em>) <span class="math-tex">\(=\frac{8}{10}\)</span></span></p>

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

<p style="text-align:justify"><span style="font-size:14px">If we are using probabilities, we can label the tree in the following general way:</span></p>

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

<p style="text-align:justify"><span style="font-size:14px"><img alt="" src="https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16239204011623920401.jpg" style="height:297px; width:800px" /></span></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">R</span></em>|<em><span style="color:#424242">R</span></em>) here means&nbsp;<em><span style="color:#424242">P</span></em>(<em><span style="color:#424242">R</span></em>&nbsp;on 2nd|<em><span style="color:#424242">R</span></em>&nbsp;on 1st)</span></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">B</span></em>|<em><span style="color:#424242">R</span></em>) here means&nbsp;<em><span style="color:#424242">P</span></em>(<em><span style="color:#424242">B</span></em>&nbsp;on 2nd|<em><span style="color:#424242">R</span></em>&nbsp;on 1st)</span></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">R</span></em>|<em><span style="color:#424242">B</span></em>) here means&nbsp;<em><span style="color:#424242">P</span></em>(<em><span style="color:#424242">R</span></em>&nbsp;on 2nd|<em><span style="color:#424242">B</span></em>&nbsp;on 1st)</span></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">B</span></em>|<em><span style="color:#424242">B</span></em>) here means&nbsp;<em><span style="color:#424242">P</span></em>(<em><span style="color:#424242">B</span></em>&nbsp;on 2nd|<em><span style="color:#424242">B</span></em>&nbsp;on 1st)</span></p>

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

<p style="text-align:justify"><span style="font-size:14px"><span style="color:white">Example 3.27</span></span></p>

<p style="text-align:justify"><span style="font-size:14px"><span style="color:black">Suppose an experiment has the outcomes 1, 2, 3, . . . , 12 where each outcome has an equal chance of occurring.</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 </span><em><span style="color:black">A </span></em><span style="color:black">= {1, 2, 3, 4, 5, 6} and event </span><em><span style="color:black">B </span></em><span style="color:black">= {6, 7, 8, 9}. Then </span><em><span style="color:black">A </span></em><span style="color:black">AND </span><em><span style="color:black">B </span></em><span style="color:black">= {6} and </span><em><span style="color:black">A </span></em><span style="color:black">OR </span><em><span style="color:black">B </span></em><span style="color:black">= {1, 2, 3, 4, 5, 6, 7,8, 9}. The Venn diagram is as follows:</span></span></p>

<p style="text-align:justify">&nbsp;</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;3.26</strong></span></span></p>

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

<p style="text-align:justify"><span style="font-size:14px">A litter of kittens available for adoption at the Humane Society has four tabby kittens and five black kittens. A family comes in and randomly selects two kittens (without replacement) for adoption.</span></p>

<p style="text-align:justify"><span style="font-size:14px"><img alt="" src="https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16239204851623920485.png" style="height:533px; width:800px" /><span style="color:black">Which shows the </span></span></p>

<p style="text-align:justify"><span style="font-size:14px"><span style="color:black">probability that both kittens are tabby?</span><br />
<br />
<span class="math-tex">\(a(\frac{1}{2})(\frac{1}{2})\hspace{0.1cm}b(\frac{4}{9})(\frac{4}{9})\hspace{0.1cm}c(\frac{4}{9})(\frac{3}{8})\hspace{0.1cm}d(\frac{4}{9})(\frac{5}{9})\)</span></span></p>

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

<p style="text-align:justify"><span style="font-size:14px"><span style="color:black">b.What is the probability that one kitten of each coloring is selected?</span></span></p>

<p style="text-align:justify"><span style="font-size:14px"><span style="color:black"><img alt="" src="https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16239206881623920688.png" style="height:47px; width:400px" /></span></span></p>

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

<p style="text-align:justify"><span style="font-size:14px"><span style="color:black">What is the probability that a tabby is chosen as the second kitten when a black kitten was chosen as the first?</span></span></p>

<ol style="list-style-type:lower-alpha">
	<li>
	<p style="text-align:justify"><span style="font-size:14px"><span style="color:black">What is the probability of choosing two kittens of the same color?</span></span></p>
	</li>
</ol>

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

<p style="text-align:justify"><strong><span style="font-size:14px"><span style="color:black">Solution&nbsp;1</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"><img alt="" src="https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16239207671623920767.png" style="height:66px; width:400px" /></span></span></p>

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

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

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