{"id":134,"date":"2016-04-21T22:43:44","date_gmt":"2016-04-21T22:43:44","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/introstats1xmaster\/?post_type=chapter&#038;p=134"},"modified":"2021-07-06T18:37:22","modified_gmt":"2021-07-06T18:37:22","slug":"tree-and-venn-diagrams","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/frontrange-introstats1\/chapter\/tree-and-venn-diagrams\/","title":{"raw":"Tree and Venn Diagrams","rendered":"Tree and Venn Diagrams"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul id=\"list1523423\">\r\n \t<li>Construct and interpret Tree Diagrams<\/li>\r\n \t<li>Construct and interpret Venn Diagrams<\/li>\r\n<\/ul>\r\n<\/div>\r\nSometimes, when the probability problems are complex, it can be helpful to graph the situation. Tree diagrams and Venn diagrams are two tools that can be used to visualize and solve conditional probabilities.\r\n<h2 data-type=\"title\">Tree Diagrams<\/h2>\r\nA <strong>tree diagram<\/strong> is a special type of graph used to determine the outcomes of an experiment. It consists of \"branches\" 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.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nIn an urn, there are [latex]11[\/latex] balls. Three balls are red ([latex]R[\/latex]) and eight balls are blue ([latex]B[\/latex]). Draw two balls, one at a time, <strong>with replacement<\/strong>. \"With replacement\" means that you put the first ball back in the urn before you select the second ball. The tree diagram using frequencies that show all the possible outcomes follows.\r\n<figure id=\"element-325\" class=\"ui-has-child-figcaption\"><span id=\"id47069242\" data-type=\"media\" data-alt=\"This is a tree diagram with branches showing frequencies of each draw. The first branch shows two lines: 8B and 3R. The second branch has a set of two lines (8B and 3R) for each line of the first branch. Multiply along each line to find 64BB, 24BR, 24RB, and 9RR.\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/132\/2016\/04\/21214403\/fig-ch03_07_01N.jpg\" alt=\"This is a tree diagram with branches showing frequencies of each draw. The first branch shows two lines: 8B and 3R. The second branch has a set of two lines (8B and 3R) for each line of the first branch. Multiply along each line to find 64BB, 24BR, 24RB, and 9RR.\" width=\"400\" data-media-type=\"image\/jpg\" data-print-width=\"4in\" \/><\/span><figcaption>Total = [latex]64 + 24 + 24 + 9 = 121[\/latex]<\/figcaption><\/figure>\r\nThe first set of branches represents the first draw. The second set of branches represents the second draw. Each of the outcomes is distinct. In fact, we can list each red ball as [latex]R1[\/latex], [latex]R2[\/latex], and [latex]R3[\/latex] and each blue ball as [latex]B1[\/latex], [latex]B2[\/latex], [latex]B3[\/latex], [latex]B4[\/latex], [latex]B5[\/latex], [latex]B6[\/latex], [latex]B7[\/latex], and [latex]B8[\/latex]. Then the nine [latex]RR[\/latex] outcomes can be written as:\r\n[latex]R1R1;\\,\\, R1R2;\\,\\, R1R3;\\,\\, R2R1;\\,\\, R2R2;\\,\\, R2R3;\\,\\, R3R1;\\,\\, R3R2;\\,\\, R3R3[\/latex]\r\nThe other outcomes are similar.\r\n\r\nThere are a total of [latex]11[\/latex] balls in the urn. Draw two balls, one at a time, with replacement. There are [latex]11(11) = 121[\/latex] outcomes, the size of the <strong>sample space<\/strong>.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>List the [latex]24[\/latex] [latex]BR[\/latex] outcomes: [latex]B1R1[\/latex], [latex]B1R2[\/latex], [latex]B1R3[\/latex], ...<\/li>\r\n \t<li>Using the tree diagram, calculate [latex]P(RR)[\/latex].<\/li>\r\n \t<li>Using the tree diagram, calculate [latex]P(RB \\text{ OR } BR)[\/latex].<\/li>\r\n \t<li>Using the tree diagram, calculate [latex]P(R \\text{ on 1st draw AND } B \\text { on 2nd draw })[\/latex].<\/li>\r\n \t<li>Using the tree diagram, calculate [latex]P(R \\text{ on 2nd draw GIVEN } B \\text { on 1st draw })[\/latex].<\/li>\r\n \t<li>Using the tree diagram, calculate [latex]P(BB)[\/latex].<\/li>\r\n \t<li>Using the tree diagram, calculate [latex]P(B \\text{ on the 2nd draw given } R \\text { on the first draw })[\/latex].<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"124076\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"124076\"]\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li style=\"list-style-type: none;\">\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]B1R1; B1R2; B1R3; B2R1; B2R2; B2R3; B3R1; B3R2; B3R3; B4R1; B4R2; B4R3; B5R1; B5R2; B5R3; B6R1; [\/latex]<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n[latex] B6R2; B6R3; B7R1; B7R2; B7R3; B8R1; B8R2; B8R3 [\/latex]\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li style=\"list-style-type: none;\">\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]P(RR) = \\frac{3}{11}\\frac{3}{11} = \\frac{9}{121}[\/latex]<\/li>\r\n \t<li>[latex]P(RB \\text{ OR } BR) = \\frac{3}{11}\\frac{8}{11} + \\frac{8}{11}\\frac{3}{11} = \\frac{48}{121}[\/latex]<\/li>\r\n \t<li>[latex]P(R \\text{ on 1st draw AND } B \\text{ on 2nd draw}) = P(RB) = \\frac{3}{11}\\frac{8}{11} = \\frac{24}{121}[\/latex]<\/li>\r\n \t<li>[latex]P(R \\text{ on 2nd draw GIVEN } B \\text{ on 1st draw}) = P(R \\text{ on 2nd }|B \\text{ on 1st }) = \\frac{24}{88} = \\frac{3}{11}.[\/latex] This problem is a conditional one. The sample space has been reduced to those outcomes that already have a blue on the first draw. There are [latex]24 + 64 = 88[\/latex] possible outcomes ([latex]24 BR \\text{ and } 64 BB[\/latex]). Twenty-four of the [latex]88[\/latex] possible outcomes are [latex]BR[\/latex]. [latex]\\frac{24}{88} = \\frac{3}{11}[\/latex].<\/li>\r\n \t<li>[latex]P(BB) = \\frac{64}{121}[\/latex]<\/li>\r\n \t<li>[latex]P(B \\text{ on 2nd draw }|R \\text{ on 1st draw }) = \\frac{8}{11}[\/latex] There are [latex]9 + 24[\/latex] outcomes that have [latex]R[\/latex] on the first draw ([latex]9 RR \\text{ and } 24 RB[\/latex]). The sample space is then [latex]9 + 24 = 33[\/latex]. [latex]24[\/latex] of the [latex]33[\/latex] outcomes have [latex]B[\/latex] on the second draw. The probability is then [latex]\\frac{24}{33}[\/latex] which reduces to\u00a0[latex]\\frac{8}{11}[\/latex].<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nAn 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. <strong>\"Without replacement\"<\/strong> 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, [latex](\\frac{3}{11})(\\frac{2}{10})=(\\frac{6}{110})[\/latex].\r\n<figure id=\"element-325a\" class=\"ui-has-child-figcaption\"><span id=\"id47078287\" data-type=\"media\" data-alt=\"This is a tree diagram with branches showing probabilities of each draw. The first branch shows 2 lines: B 8\/11 and R 3\/11. The second branch has a set of 2 lines for each first branch line. Below B 8\/11 are B 7\/10 and R 3\/10. Below R 3\/11 are B 8\/10 and R 2\/10. Multiply along each line to find BB 56\/110, BR 24\/110, RB 24\/110, and RR 6\/110.\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/132\/2016\/04\/21214405\/fig-ch03_07_02.jpg\" alt=\"This is a tree diagram with branches showing probabilities of each draw. The first branch shows 2 lines: B 8\/11 and R 3\/11. The second branch has a set of 2 lines for each first branch line. Below B 8\/11 are B 7\/10 and R 3\/10. Below R 3\/11 are B 8\/10 and R 2\/10. Multiply along each line to find BB 56\/110, BR 24\/110, RB 24\/110, and RR 6\/110.\" width=\"400\" data-media-type=\"image\/jpg\" \/><\/span><figcaption><center><\/center>Total = [latex]\\displaystyle\\frac{{56+24+24+6}}{{110}}=\\frac{{110}}{{110}}=1[\/latex]<\/figcaption><\/figure>\r\n<div class=\"textbox shaded\">\r\n<h3>Note<\/h3>\r\nIf you draw a red on the first draw from the three red possibilities, there are two red marbles left to draw on the second draw. You do not put back or replace the first marble after you have drawn it. You draw <strong>without replacement<\/strong>, so that on the second draw there are ten marbles left in the urn.\r\n\r\n<\/div>\r\nCalculate the following probabilities using the tree diagram.\r\n\r\na. [latex]P(RR)[\/latex] = ________\r\n\r\nb. Fill in the blanks:\r\n[latex]P(RB \\text{ OR } BR = (\\frac{3}{11})(\\frac{8}{10}) + (\\rule{1cm}{0.15mm})(\\rule{1cm}{0.15mm}) = \\frac{48}{110}[\/latex]\r\n\r\nc. [latex]P(R \\text{ on 2nd|}B \\text{ on 1st}) =[\/latex]\r\n\r\nd. Fill in the blanks.\r\n\r\n[latex]P(R \\text{ on 1st AND } B \\text{ on 2nd }) = P(RB) = (\\rule{1cm}{0.15mm})(\\rule{1cm}{0.15mm})= \\frac{24}{110}[\/latex]\r\n\r\ne. Find [latex]P(BB)[\/latex].\r\n\r\nf. Find [latex]P(B \\text{ on 2nd|}R \\text{ on 1st})[\/latex].\r\n\r\n[reveal-answer q=\"124075\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"124075\"]\r\n\r\na. [latex]P(RR) = (\\frac{3}{11})(\\frac{2}{10}) = \\frac{6}{110} = \\frac{3}{55}[\/latex]\r\n\r\nb. [latex]P(RB \\text{ OR } BR = (\\frac{3}{11})(\\frac{8}{10}) + (\\frac{8}{11})(\\frac{3}{10}) = \\frac{48}{110} = \\frac{24}{55}[\/latex]\r\n\r\nc. [latex]P(R \\text{ on 2nd|}B \\text{ on 1st }) =\\frac{3}{10}[\/latex]\r\n\r\nd. [latex]P(R \\text{ on 1st AND } B \\text{ on 2nd }) = P(RB) = (\\frac{3}{11})(\\frac{8}{10})= \\frac{24}{110} = \\frac{12}{55}[\/latex]\r\n\r\ne. [latex]P(BB) = (\\frac{8}{11})(\\frac{7}{10}) = \\frac{56}{110} = \\frac{28}{55}[\/latex]\r\n\r\nf. Using the tree diagram, [latex]P(B \\text{ on 2nd|}R \\text{ on 1st }) = P(R|B) = \\frac{8}{10} = \\frac{4}{5}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\nIf we are using probabilities, we can label the tree in the following general way.\r\n\r\n<span data-type=\"media\" data-alt=\"This is a tree diagram for a two-step experiment. The first branch shows first outcome: P(B) and P(R). The second branch has a set of 2 lines for each line of the first branch: the probability of B given B = P(BB), the probability of R given B = P(RB), the probability of B given R = P(BR), and the probability of R given R = P(RR).\" data-display=\"block\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/132\/2016\/04\/21214407\/fig-ch03_07_03N.jpg\" alt=\"This is a tree diagram for a two-step experiment. The first branch shows first outcome: P(B) and P(R). The second branch has a set of 2 lines for each line of the first branch: the probability of B given B = P(BB), the probability of R given B = P(RB), the probability of B given R = P(BR), and the probability of R given R = P(RR).\" width=\"400\" data-media-type=\"image\/jpg\" \/><\/span>\r\n<ul>\r\n \t<li>[latex]P(RR)[\/latex] here means [latex]P(R \\text{ on 2nd|}R \\text{ on 1st})[\/latex]<\/li>\r\n \t<li>[latex]P(BR)[\/latex] here means [latex]P(B \\text{ on 2nd|}R \\text{ on 1st})[\/latex]<\/li>\r\n \t<li>[latex]P(RB)[\/latex] here means [latex]P(R \\text{ on 2nd|}B \\text{ on 1st})[\/latex]<\/li>\r\n \t<li>[latex]P(BB)[\/latex] here means [latex]P(B \\text{ on 2nd|}B \\text{ on 1st})[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\nhttps:\/\/youtu.be\/Zxvc6iPKdec\r\n\r\n<section class=\"ui-body\">\r\n<h2 data-type=\"title\">Venn Diagram<\/h2>\r\nA <strong>Venn diagram<\/strong> is a picture that represents the outcomes of an experiment. It generally consists of a box that represents the sample space S together with circles or ovals. The circles or ovals represent events.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSuppose an experiment has the outcomes [latex]1, 2, 3, ... , 12[\/latex] where each outcome has an equal chance of occurring. Let event [latex]A = \\{1, 2, 3, 4, 5, 6\\}[\/latex] and event [latex]B = \\{6, 7, 8, 9\\}[\/latex]. Then [latex] A \\text{ AND } B = \\{6\\}[\/latex] and [latex]A \\text{ OR }B = \\{1, 2, 3, 4, 5, 6, 7, 8, 9\\}[\/latex]. The Venn diagram is as follows:\r\n<figure id=\"eip-idm17287840\"><span id=\"id18119489\" data-type=\"media\" data-alt=\"A Venn diagram. An oval representing set A contains the values 1, 2, 3, 4, 5, and 6. An oval representing set B also contains the 6, along with 7, 8, and 9. The values 10, 11, and 12 are present but not contained in either set.\" data-display=\"block\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/132\/2016\/04\/21214409\/fig-ch03_06_01.jpg\" alt=\"A Venn diagram. An oval representing set A contains the values 1, 2, 3, 4, 5, and 6. An oval representing set B also contains the 6, along with 7, 8, and 9. The values 10, 11, and 12 are present but not contained in either set.\" width=\"380\" data-media-type=\"image\/png\" \/><\/span><\/figure>\r\n&nbsp;\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFlip two fair coins. Let [latex]A[\/latex] = tails on the first coin. Let [latex]B[\/latex] = tails on the second coin. Then [latex]A = \\{TT, TH\\}[\/latex] and [latex]B = \\{TT, HT\\}[\/latex]. Therefore,[latex]A \\text{ AND } B = \\{TT\\}[\/latex]. [latex]A \\text{ OR } B = \\{TH, TT, HT\\}[\/latex].\r\n\r\nThe sample space when you flip two fair coins is [latex]X = \\{HH, HT, TH, TT\\}[\/latex]. The outcome [latex]HH[\/latex] is in NEITHER [latex]A[\/latex] NOR [latex]B[\/latex]. The Venn diagram is as follows:\r\n<figure id=\"eip-idm154602320\"><span id=\"id18154607\" data-type=\"media\" data-alt=\"This is a venn diagram. An oval representing set A contains Tails + Heads and Tails + Tails. An oval representing set B also contains Tails + Tails, along with Heads + Tails. The universe S contains Heads + Heads, but this value is not contained in either set A or B.\" data-display=\"block\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/132\/2016\/04\/21214411\/fig-ch03_06_02.jpg\" alt=\"This is a venn diagram. An oval representing set A contains Tails + Heads and Tails + Tails. An oval representing set B also contains Tails + Tails, along with Heads + Tails. The universe S contains Heads + Heads, but this value is not contained in either set A or B.\" width=\"400\" data-media-type=\"image\/jpg\" \/><\/span><\/figure>\r\n<\/div>\r\n&nbsp;\r\n\r\nhttps:\/\/youtu.be\/MassxXy8iko\r\n<h2 data-type=\"glossary-title\">Glossary<\/h2>\r\n<dl id=\"treediagram\" class=\"definition\">\r\n \t<dt>Tree Diagram<\/dt>\r\n \t<dd id=\"id18749941\">the useful visual representation of a sample space and events in the form of a \u201ctree\u201d with branches marked by possible outcomes together with associated probabilities (frequencies, relative frequencies)<\/dd>\r\n<\/dl>\r\n<dl id=\"vendiagram\" class=\"definition\">\r\n \t<dt>Venn Diagram<\/dt>\r\n \t<dd id=\"id18154967\">the visual representation of a sample space and events in the form of circles or ovals showing their intersections<\/dd>\r\n<\/dl>\r\n<div data-type=\"newline\" data-count=\"2\"><\/div>\r\n<\/section>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul id=\"list1523423\">\n<li>Construct and interpret Tree Diagrams<\/li>\n<li>Construct and interpret Venn Diagrams<\/li>\n<\/ul>\n<\/div>\n<p>Sometimes, when the probability problems are complex, it can be helpful to graph the situation. Tree diagrams and Venn diagrams are two tools that can be used to visualize and solve conditional probabilities.<\/p>\n<h2 data-type=\"title\">Tree Diagrams<\/h2>\n<p>A <strong>tree diagram<\/strong> is a special type of graph used to determine the outcomes of an experiment. It consists of &#8220;branches&#8221; 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.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>In an urn, there are [latex]11[\/latex] balls. Three balls are red ([latex]R[\/latex]) and eight balls are blue ([latex]B[\/latex]). Draw two balls, one at a time, <strong>with replacement<\/strong>. &#8220;With replacement&#8221; means that you put the first ball back in the urn before you select the second ball. The tree diagram using frequencies that show all the possible outcomes follows.<\/p>\n<figure id=\"element-325\" class=\"ui-has-child-figcaption\"><span id=\"id47069242\" data-type=\"media\" data-alt=\"This is a tree diagram with branches showing frequencies of each draw. The first branch shows two lines: 8B and 3R. The second branch has a set of two lines (8B and 3R) for each line of the first branch. Multiply along each line to find 64BB, 24BR, 24RB, and 9RR.\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/132\/2016\/04\/21214403\/fig-ch03_07_01N.jpg\" alt=\"This is a tree diagram with branches showing frequencies of each draw. The first branch shows two lines: 8B and 3R. The second branch has a set of two lines (8B and 3R) for each line of the first branch. Multiply along each line to find 64BB, 24BR, 24RB, and 9RR.\" width=\"400\" data-media-type=\"image\/jpg\" data-print-width=\"4in\" \/><\/span><figcaption>Total = [latex]64 + 24 + 24 + 9 = 121[\/latex]<\/figcaption><\/figure>\n<p>The first set of branches represents the first draw. The second set of branches represents the second draw. Each of the outcomes is distinct. In fact, we can list each red ball as [latex]R1[\/latex], [latex]R2[\/latex], and [latex]R3[\/latex] and each blue ball as [latex]B1[\/latex], [latex]B2[\/latex], [latex]B3[\/latex], [latex]B4[\/latex], [latex]B5[\/latex], [latex]B6[\/latex], [latex]B7[\/latex], and [latex]B8[\/latex]. Then the nine [latex]RR[\/latex] outcomes can be written as:<br \/>\n[latex]R1R1;\\,\\, R1R2;\\,\\, R1R3;\\,\\, R2R1;\\,\\, R2R2;\\,\\, R2R3;\\,\\, R3R1;\\,\\, R3R2;\\,\\, R3R3[\/latex]<br \/>\nThe other outcomes are similar.<\/p>\n<p>There are a total of [latex]11[\/latex] balls in the urn. Draw two balls, one at a time, with replacement. There are [latex]11(11) = 121[\/latex] outcomes, the size of the <strong>sample space<\/strong>.<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>List the [latex]24[\/latex] [latex]BR[\/latex] outcomes: [latex]B1R1[\/latex], [latex]B1R2[\/latex], [latex]B1R3[\/latex], &#8230;<\/li>\n<li>Using the tree diagram, calculate [latex]P(RR)[\/latex].<\/li>\n<li>Using the tree diagram, calculate [latex]P(RB \\text{ OR } BR)[\/latex].<\/li>\n<li>Using the tree diagram, calculate [latex]P(R \\text{ on 1st draw AND } B \\text { on 2nd draw })[\/latex].<\/li>\n<li>Using the tree diagram, calculate [latex]P(R \\text{ on 2nd draw GIVEN } B \\text { on 1st draw })[\/latex].<\/li>\n<li>Using the tree diagram, calculate [latex]P(BB)[\/latex].<\/li>\n<li>Using the tree diagram, calculate [latex]P(B \\text{ on the 2nd draw given } R \\text { on the first draw })[\/latex].<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q124076\">Show Solution<\/span><\/p>\n<div id=\"q124076\" class=\"hidden-answer\" style=\"display: none\">\n<ol style=\"list-style-type: lower-alpha;\">\n<li style=\"list-style-type: none;\">\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]B1R1; B1R2; B1R3; B2R1; B2R2; B2R3; B3R1; B3R2; B3R3; B4R1; B4R2; B4R3; B5R1; B5R2; B5R3; B6R1;[\/latex]<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<p>[latex]B6R2; B6R3; B7R1; B7R2; B7R3; B8R1; B8R2; B8R3[\/latex]<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li style=\"list-style-type: none;\">\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]P(RR) = \\frac{3}{11}\\frac{3}{11} = \\frac{9}{121}[\/latex]<\/li>\n<li>[latex]P(RB \\text{ OR } BR) = \\frac{3}{11}\\frac{8}{11} + \\frac{8}{11}\\frac{3}{11} = \\frac{48}{121}[\/latex]<\/li>\n<li>[latex]P(R \\text{ on 1st draw AND } B \\text{ on 2nd draw}) = P(RB) = \\frac{3}{11}\\frac{8}{11} = \\frac{24}{121}[\/latex]<\/li>\n<li>[latex]P(R \\text{ on 2nd draw GIVEN } B \\text{ on 1st draw}) = P(R \\text{ on 2nd }|B \\text{ on 1st }) = \\frac{24}{88} = \\frac{3}{11}.[\/latex] This problem is a conditional one. The sample space has been reduced to those outcomes that already have a blue on the first draw. There are [latex]24 + 64 = 88[\/latex] possible outcomes ([latex]24 BR \\text{ and } 64 BB[\/latex]). Twenty-four of the [latex]88[\/latex] possible outcomes are [latex]BR[\/latex]. [latex]\\frac{24}{88} = \\frac{3}{11}[\/latex].<\/li>\n<li>[latex]P(BB) = \\frac{64}{121}[\/latex]<\/li>\n<li>[latex]P(B \\text{ on 2nd draw }|R \\text{ on 1st draw }) = \\frac{8}{11}[\/latex] There are [latex]9 + 24[\/latex] outcomes that have [latex]R[\/latex] on the first draw ([latex]9 RR \\text{ and } 24 RB[\/latex]). The sample space is then [latex]9 + 24 = 33[\/latex]. [latex]24[\/latex] of the [latex]33[\/latex] outcomes have [latex]B[\/latex] on the second draw. The probability is then [latex]\\frac{24}{33}[\/latex] which reduces to\u00a0[latex]\\frac{8}{11}[\/latex].<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>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. <strong>&#8220;Without replacement&#8221;<\/strong> 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, [latex](\\frac{3}{11})(\\frac{2}{10})=(\\frac{6}{110})[\/latex].<\/p>\n<figure id=\"element-325a\" class=\"ui-has-child-figcaption\"><span id=\"id47078287\" data-type=\"media\" data-alt=\"This is a tree diagram with branches showing probabilities of each draw. The first branch shows 2 lines: B 8\/11 and R 3\/11. The second branch has a set of 2 lines for each first branch line. Below B 8\/11 are B 7\/10 and R 3\/10. Below R 3\/11 are B 8\/10 and R 2\/10. Multiply along each line to find BB 56\/110, BR 24\/110, RB 24\/110, and RR 6\/110.\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/132\/2016\/04\/21214405\/fig-ch03_07_02.jpg\" alt=\"This is a tree diagram with branches showing probabilities of each draw. The first branch shows 2 lines: B 8\/11 and R 3\/11. The second branch has a set of 2 lines for each first branch line. Below B 8\/11 are B 7\/10 and R 3\/10. Below R 3\/11 are B 8\/10 and R 2\/10. Multiply along each line to find BB 56\/110, BR 24\/110, RB 24\/110, and RR 6\/110.\" width=\"400\" data-media-type=\"image\/jpg\" \/><\/span><figcaption><\/figcaption><\/figure>\n<\/div>\n<p>Total = [latex]\\displaystyle\\frac{{56+24+24+6}}{{110}}=\\frac{{110}}{{110}}=1[\/latex]<\/p>\n<div class=\"textbox shaded\">\n<h3>Note<\/h3>\n<p>If you draw a red on the first draw from the three red possibilities, there are two red marbles left to draw on the second draw. You do not put back or replace the first marble after you have drawn it. You draw <strong>without replacement<\/strong>, so that on the second draw there are ten marbles left in the urn.<\/p>\n<\/div>\n<p>Calculate the following probabilities using the tree diagram.<\/p>\n<p>a. [latex]P(RR)[\/latex] = ________<\/p>\n<p>b. Fill in the blanks:<br \/>\n[latex]P(RB \\text{ OR } BR = (\\frac{3}{11})(\\frac{8}{10}) + (\\rule{1cm}{0.15mm})(\\rule{1cm}{0.15mm}) = \\frac{48}{110}[\/latex]<\/p>\n<p>c. [latex]P(R \\text{ on 2nd|}B \\text{ on 1st}) =[\/latex]<\/p>\n<p>d. Fill in the blanks.<\/p>\n<p>[latex]P(R \\text{ on 1st AND } B \\text{ on 2nd }) = P(RB) = (\\rule{1cm}{0.15mm})(\\rule{1cm}{0.15mm})= \\frac{24}{110}[\/latex]<\/p>\n<p>e. Find [latex]P(BB)[\/latex].<\/p>\n<p>f. Find [latex]P(B \\text{ on 2nd|}R \\text{ on 1st})[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q124075\">Show Solution<\/span><\/p>\n<div id=\"q124075\" class=\"hidden-answer\" style=\"display: none\">\n<p>a. [latex]P(RR) = (\\frac{3}{11})(\\frac{2}{10}) = \\frac{6}{110} = \\frac{3}{55}[\/latex]<\/p>\n<p>b. [latex]P(RB \\text{ OR } BR = (\\frac{3}{11})(\\frac{8}{10}) + (\\frac{8}{11})(\\frac{3}{10}) = \\frac{48}{110} = \\frac{24}{55}[\/latex]<\/p>\n<p>c. [latex]P(R \\text{ on 2nd|}B \\text{ on 1st }) =\\frac{3}{10}[\/latex]<\/p>\n<p>d. [latex]P(R \\text{ on 1st AND } B \\text{ on 2nd }) = P(RB) = (\\frac{3}{11})(\\frac{8}{10})= \\frac{24}{110} = \\frac{12}{55}[\/latex]<\/p>\n<p>e. [latex]P(BB) = (\\frac{8}{11})(\\frac{7}{10}) = \\frac{56}{110} = \\frac{28}{55}[\/latex]<\/p>\n<p>f. Using the tree diagram, [latex]P(B \\text{ on 2nd|}R \\text{ on 1st }) = P(R|B) = \\frac{8}{10} = \\frac{4}{5}[\/latex]<\/p>\n<\/div>\n<\/div>\n<p>If we are using probabilities, we can label the tree in the following general way.<\/p>\n<p><span data-type=\"media\" data-alt=\"This is a tree diagram for a two-step experiment. The first branch shows first outcome: P(B) and P(R). The second branch has a set of 2 lines for each line of the first branch: the probability of B given B = P(BB), the probability of R given B = P(RB), the probability of B given R = P(BR), and the probability of R given R = P(RR).\" data-display=\"block\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/132\/2016\/04\/21214407\/fig-ch03_07_03N.jpg\" alt=\"This is a tree diagram for a two-step experiment. The first branch shows first outcome: P(B) and P(R). The second branch has a set of 2 lines for each line of the first branch: the probability of B given B = P(BB), the probability of R given B = P(RB), the probability of B given R = P(BR), and the probability of R given R = P(RR).\" width=\"400\" data-media-type=\"image\/jpg\" \/><\/span><\/p>\n<ul>\n<li>[latex]P(RR)[\/latex] here means [latex]P(R \\text{ on 2nd|}R \\text{ on 1st})[\/latex]<\/li>\n<li>[latex]P(BR)[\/latex] here means [latex]P(B \\text{ on 2nd|}R \\text{ on 1st})[\/latex]<\/li>\n<li>[latex]P(RB)[\/latex] here means [latex]P(R \\text{ on 2nd|}B \\text{ on 1st})[\/latex]<\/li>\n<li>[latex]P(BB)[\/latex] here means [latex]P(B \\text{ on 2nd|}B \\text{ on 1st})[\/latex]<\/li>\n<\/ul>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Count outcomes using tree diagram | Statistics and probability | 7th grade | Khan Academy\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/Zxvc6iPKdec?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<section class=\"ui-body\">\n<h2 data-type=\"title\">Venn Diagram<\/h2>\n<p>A <strong>Venn diagram<\/strong> is a picture that represents the outcomes of an experiment. It generally consists of a box that represents the sample space S together with circles or ovals. The circles or ovals represent events.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Suppose an experiment has the outcomes [latex]1, 2, 3, ... , 12[\/latex] where each outcome has an equal chance of occurring. Let event [latex]A = \\{1, 2, 3, 4, 5, 6\\}[\/latex] and event [latex]B = \\{6, 7, 8, 9\\}[\/latex]. Then [latex]A \\text{ AND } B = \\{6\\}[\/latex] and [latex]A \\text{ OR }B = \\{1, 2, 3, 4, 5, 6, 7, 8, 9\\}[\/latex]. The Venn diagram is as follows:<\/p>\n<figure id=\"eip-idm17287840\"><span id=\"id18119489\" data-type=\"media\" data-alt=\"A Venn diagram. An oval representing set A contains the values 1, 2, 3, 4, 5, and 6. An oval representing set B also contains the 6, along with 7, 8, and 9. The values 10, 11, and 12 are present but not contained in either set.\" data-display=\"block\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/132\/2016\/04\/21214409\/fig-ch03_06_01.jpg\" alt=\"A Venn diagram. An oval representing set A contains the values 1, 2, 3, 4, 5, and 6. An oval representing set B also contains the 6, along with 7, 8, and 9. The values 10, 11, and 12 are present but not contained in either set.\" width=\"380\" data-media-type=\"image\/png\" \/><\/span><\/figure>\n<p>&nbsp;<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Flip two fair coins. Let [latex]A[\/latex] = tails on the first coin. Let [latex]B[\/latex] = tails on the second coin. Then [latex]A = \\{TT, TH\\}[\/latex] and [latex]B = \\{TT, HT\\}[\/latex]. Therefore,[latex]A \\text{ AND } B = \\{TT\\}[\/latex]. [latex]A \\text{ OR } B = \\{TH, TT, HT\\}[\/latex].<\/p>\n<p>The sample space when you flip two fair coins is [latex]X = \\{HH, HT, TH, TT\\}[\/latex]. The outcome [latex]HH[\/latex] is in NEITHER [latex]A[\/latex] NOR [latex]B[\/latex]. The Venn diagram is as follows:<\/p>\n<figure id=\"eip-idm154602320\"><span id=\"id18154607\" data-type=\"media\" data-alt=\"This is a venn diagram. An oval representing set A contains Tails + Heads and Tails + Tails. An oval representing set B also contains Tails + Tails, along with Heads + Tails. The universe S contains Heads + Heads, but this value is not contained in either set A or B.\" data-display=\"block\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/132\/2016\/04\/21214411\/fig-ch03_06_02.jpg\" alt=\"This is a venn diagram. An oval representing set A contains Tails + Heads and Tails + Tails. An oval representing set B also contains Tails + Tails, along with Heads + Tails. The universe S contains Heads + Heads, but this value is not contained in either set A or B.\" width=\"400\" data-media-type=\"image\/jpg\" \/><\/span><\/figure>\n<\/div>\n<p>&nbsp;<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Solving Problems with Venn Diagrams\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/MassxXy8iko?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2 data-type=\"glossary-title\">Glossary<\/h2>\n<dl id=\"treediagram\" class=\"definition\">\n<dt>Tree Diagram<\/dt>\n<dd id=\"id18749941\">the useful visual representation of a sample space and events in the form of a \u201ctree\u201d with branches marked by possible outcomes together with associated probabilities (frequencies, relative frequencies)<\/dd>\n<\/dl>\n<dl id=\"vendiagram\" class=\"definition\">\n<dt>Venn Diagram<\/dt>\n<dd id=\"id18154967\">the visual representation of a sample space and events in the form of circles or ovals showing their intersections<\/dd>\n<\/dl>\n<div data-type=\"newline\" data-count=\"2\"><\/div>\n<\/section>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-134\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Tree and Venn Diagrams. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/30189442-6998-4686-ac05-ed152b91b9de@17.44\">http:\/\/cnx.org\/contents\/30189442-6998-4686-ac05-ed152b91b9de@17.44<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/about\/pdm\">Public Domain: No Known Copyright<\/a><\/em><\/li><li>Introductory Statistics . <strong>Authored by<\/strong>: Barbara Illowski, Susan Dean. <strong>Provided by<\/strong>: Open Stax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/30189442-6998-4686-ac05-ed152b91b9de@17.44\">http:\/\/cnx.org\/contents\/30189442-6998-4686-ac05-ed152b91b9de@17.44<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at http:\/\/cnx.org\/contents\/30189442-6998-4686-ac05-ed152b91b9de@17.44<\/li><li>Count outcomes using tree diagram. <strong>Provided by<\/strong>: Khan Acadamy. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/www.khanacademy.org\/math\/cc-seventh-grade-math\/cc-7th-probability-statistics\/cc-7th-compound-events\/v\/tree-diagram-to-count-outcomes\">https:\/\/www.khanacademy.org\/math\/cc-seventh-grade-math\/cc-7th-probability-statistics\/cc-7th-compound-events\/v\/tree-diagram-to-count-outcomes<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">All rights reserved content<\/div><ul class=\"citation-list\"><li>Solving Problems with Venn Diagrams. <strong>Authored by<\/strong>: Mathispower4u. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/MassxXy8iko\">https:\/\/youtu.be\/MassxXy8iko<\/a>. <strong>License<\/strong>: <em>All Rights Reserved<\/em>. <strong>License Terms<\/strong>: Standard YouTube License<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":21,"menu_order":7,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Tree and Venn Diagrams\",\"author\":\"\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/30189442-6998-4686-ac05-ed152b91b9de@17.44\",\"project\":\"\",\"license\":\"pd\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Introductory Statistics \",\"author\":\"Barbara Illowski, Susan Dean\",\"organization\":\"Open Stax\",\"url\":\"http:\/\/cnx.org\/contents\/30189442-6998-4686-ac05-ed152b91b9de@17.44\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at http:\/\/cnx.org\/contents\/30189442-6998-4686-ac05-ed152b91b9de@17.44\"},{\"type\":\"cc\",\"description\":\"Count outcomes using tree diagram\",\"author\":\"\",\"organization\":\"Khan Acadamy\",\"url\":\"https:\/\/www.khanacademy.org\/math\/cc-seventh-grade-math\/cc-7th-probability-statistics\/cc-7th-compound-events\/v\/tree-diagram-to-count-outcomes\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"\"},{\"type\":\"copyrighted_video\",\"description\":\"Solving Problems with Venn Diagrams\",\"author\":\"Mathispower4u\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/MassxXy8iko\",\"project\":\"\",\"license\":\"arr\",\"license_terms\":\"Standard YouTube License\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-134","chapter","type-chapter","status-publish","hentry"],"part":122,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/frontrange-introstats1\/wp-json\/pressbooks\/v2\/chapters\/134","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/frontrange-introstats1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/frontrange-introstats1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/frontrange-introstats1\/wp-json\/wp\/v2\/users\/21"}],"version-history":[{"count":16,"href":"https:\/\/courses.lumenlearning.com\/frontrange-introstats1\/wp-json\/pressbooks\/v2\/chapters\/134\/revisions"}],"predecessor-version":[{"id":2361,"href":"https:\/\/courses.lumenlearning.com\/frontrange-introstats1\/wp-json\/pressbooks\/v2\/chapters\/134\/revisions\/2361"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/frontrange-introstats1\/wp-json\/pressbooks\/v2\/parts\/122"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/frontrange-introstats1\/wp-json\/pressbooks\/v2\/chapters\/134\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/frontrange-introstats1\/wp-json\/wp\/v2\/media?parent=134"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/frontrange-introstats1\/wp-json\/pressbooks\/v2\/chapter-type?post=134"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/frontrange-introstats1\/wp-json\/wp\/v2\/contributor?post=134"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/frontrange-introstats1\/wp-json\/wp\/v2\/license?post=134"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}