{"id":3172,"date":"2023-09-18T00:15:38","date_gmt":"2023-09-18T00:15:38","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/introstatstest\/?post_type=chapter&#038;p=3172"},"modified":"2025-03-25T14:24:11","modified_gmt":"2025-03-25T14:24:11","slug":"conditional-probabilities-learn-it-3-update","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/introstatstest\/chapter\/conditional-probabilities-learn-it-3-update\/","title":{"raw":"Probability with Tree Diagrams: Learn It 2","rendered":"Probability with Tree Diagrams: Learn It 2"},"content":{"raw":"<section class=\"textbox learningGoals\">\r\n<ul>\r\n\t<li>Learn how to construct and interpret a tree diagram to represent sequential events or decisions<\/li>\r\n\t<li>Calculate conditional probabilities using tree diagrams, considering both dependent and independent events<\/li>\r\n\t<li>Apply tree diagrams to solve problems involving probability of multiple events, such as probability of compound events and conditional probability<\/li>\r\n<\/ul>\r\n<\/section>\r\n<p>Tree diagrams are particularly useful when investigating the probability of compound events. By creating a probability tree, we can visualize the sequence of events and determine the probability by multiplying the appropriate path along corresponding branches in the tree diagram. This method is beneficial when solving for two events that are either independent or dependent.\u00a0<\/p>\r\n<section class=\"textbox keyTakeaway\">\r\n<h3>multiplication rule of probability<\/h3>\r\n<p>Tree diagrams are often used to represent sequences of independent events. Independent events are events where the outcome of one event does not affect the outcome of the other events.<\/p>\r\n<p>&nbsp;<\/p>\r\n<p>If two events [latex]A[\/latex] and [latex]B[\/latex] are\u00a0<strong>independent: <\/strong><\/p>\r\n<p>&nbsp;<\/p>\r\n<p style=\"text-align: center;\">[latex]P(A \\text{ and }B) = P(A) \\times P(B) [\/latex]<\/p>\r\n<p>&nbsp;<\/p>\r\n<p>So, in a tree diagram, probabilities are multiplied along the branches to find the probability at the end of a branch due to the <strong>Multiplication Rule of Probability<\/strong>.<\/p>\r\n<p>&nbsp;<\/p>\r\n<p>When these probabilities are multiplied, they are essentially being combined to represent the likelihood of a specific sequence of events occurring.<\/p>\r\n<\/section>\r\n<section>\r\n<section class=\"textbox example\">An urn has three red marbles and eight blue marbles in it. Draw two marbles, one at a time, this time <strong>without replacement<\/strong>, from the urn. <br \/>\r\n<br \/>\r\nNote: Without replacement means that you do not put the first ball back before you select the second marble. <br \/>\r\n<br \/>\r\nFollowing is a tree diagram for this situation.<br \/>\r\n<br \/>\r\n<strong><img class=\"aligncenter wp-image-5970 size-medium\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/27\/2023\/06\/22013901\/fig-ch03_07_02.jpg\" alt=\"This is a tree diagram with branches showing frequencies of each draw. The first branch shows two lines: 8\/11B and 3\/11R. The second branch has a set of two lines (7\/10B and 3\/10R for line B, 8\/10B and 2\/10R for line R) for each line of the first branch. Multiply along each line to find 56\/110BB, 24\/110BR, 24\/110RB, and 6\/110RR.\" width=\"300\" height=\"245\" \/><br \/>\r\n<\/strong>The branches are labeled with probabilities. Note: 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. <br \/>\r\n<br \/>\r\nThe numbers at the ends of the branches are calculated by multiplying the numbers on the two corresponding branches. <br \/>\r\n<br \/>\r\nFor example:\r\n\r\n<p style=\"text-align: center;\">[latex]P(BB)=\\frac{8}{11}*\\frac{7}{10}=\\frac{56}{110}[\/latex]<\/p>\r\n<p>Find\/calculate the following probabilities using the tree diagram:<\/p>\r\n<p><strong>(a) <\/strong>[latex]P(B)[\/latex]<\/p>\r\n<p><strong>(b) <\/strong>[latex]P(R \\text{ on the 1st} \\text{ and } B \\text{ on the 2nd})[\/latex]<\/p>\r\n<p><strong>(c) <\/strong>[latex]P(RB \\text{ or } BR)[\/latex]<\/p>\r\n<p><strong>(d) <\/strong>[latex]P(\\text{2nd one is }R)[\/latex]<\/p>\r\n<p><strong>(e) <\/strong>Show that the [latex]P(\\text{Sample Space})[\/latex] is [latex]1[\/latex] or [latex]100\\%[\/latex].<\/p>\r\n<p>[reveal-answer q=\"983476\"]Show Answer[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"983476\"]<\/p>\r\n<p><strong>(a) <\/strong>[latex]P(B)=\\frac{8}{11}[\/latex]<\/p>\r\n<p><strong>(b) <\/strong>[latex]P(R \\text{ on the 1st} \\text{ and } B \\text{ on the 2nd})=\\frac{3}{11}*\\frac{8}{10}=\\frac{24}{110}[\/latex]. This is also the same as writing it as [latex]P(RB)[\/latex].<\/p>\r\n<p><strong>(c) <\/strong>[latex]P(RB \\text{ or } BR)=P(RB)+P(BR)=\\frac{24}{110}+\\frac{24}{110}=\\frac{48}{110}[\/latex]<\/p>\r\n<p><strong>(d) <\/strong>[latex]P(\\text{2nd one is }R)=P(BR)+P(RR)=\\frac{24}{110}+\\frac{6}{110}=\\frac{30}{110}[\/latex]<\/p>\r\n<p><strong>(e) <\/strong>[latex]P(\\text{Sample Space})= P(BB)+P(BR)+P(RB)+P(RR)=\\frac{56}{110}+\\frac{24}{110}+\\frac{24}{110}+\\frac{6}{110}=\\frac{56+24+24+6}{110}=\\frac{110}{110}=1[\/latex]<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<\/section>\r\n<section><\/section>\r\n<section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]13823[\/ohm2_question]<\/section>","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li>Learn how to construct and interpret a tree diagram to represent sequential events or decisions<\/li>\n<li>Calculate conditional probabilities using tree diagrams, considering both dependent and independent events<\/li>\n<li>Apply tree diagrams to solve problems involving probability of multiple events, such as probability of compound events and conditional probability<\/li>\n<\/ul>\n<\/section>\n<p>Tree diagrams are particularly useful when investigating the probability of compound events. By creating a probability tree, we can visualize the sequence of events and determine the probability by multiplying the appropriate path along corresponding branches in the tree diagram. This method is beneficial when solving for two events that are either independent or dependent.\u00a0<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>multiplication rule of probability<\/h3>\n<p>Tree diagrams are often used to represent sequences of independent events. Independent events are events where the outcome of one event does not affect the outcome of the other events.<\/p>\n<p>&nbsp;<\/p>\n<p>If two events [latex]A[\/latex] and [latex]B[\/latex] are\u00a0<strong>independent: <\/strong><\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: center;\">[latex]P(A \\text{ and }B) = P(A) \\times P(B)[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>So, in a tree diagram, probabilities are multiplied along the branches to find the probability at the end of a branch due to the <strong>Multiplication Rule of Probability<\/strong>.<\/p>\n<p>&nbsp;<\/p>\n<p>When these probabilities are multiplied, they are essentially being combined to represent the likelihood of a specific sequence of events occurring.<\/p>\n<\/section>\n<section>\n<section class=\"textbox example\">An urn has three red marbles and eight blue marbles in it. Draw two marbles, one at a time, this time <strong>without replacement<\/strong>, from the urn. <\/p>\n<p>Note: Without replacement means that you do not put the first ball back before you select the second marble. <\/p>\n<p>Following is a tree diagram for this situation.<\/p>\n<p><strong><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-5970 size-medium\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/27\/2023\/06\/22013901\/fig-ch03_07_02.jpg\" alt=\"This is a tree diagram with branches showing frequencies of each draw. The first branch shows two lines: 8\/11B and 3\/11R. The second branch has a set of two lines (7\/10B and 3\/10R for line B, 8\/10B and 2\/10R for line R) for each line of the first branch. Multiply along each line to find 56\/110BB, 24\/110BR, 24\/110RB, and 6\/110RR.\" width=\"300\" height=\"245\" \/><br \/>\n<\/strong>The branches are labeled with probabilities. Note: 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<p>The numbers at the ends of the branches are calculated by multiplying the numbers on the two corresponding branches. <\/p>\n<p>For example:<\/p>\n<p style=\"text-align: center;\">[latex]P(BB)=\\frac{8}{11}*\\frac{7}{10}=\\frac{56}{110}[\/latex]<\/p>\n<p>Find\/calculate the following probabilities using the tree diagram:<\/p>\n<p><strong>(a) <\/strong>[latex]P(B)[\/latex]<\/p>\n<p><strong>(b) <\/strong>[latex]P(R \\text{ on the 1st} \\text{ and } B \\text{ on the 2nd})[\/latex]<\/p>\n<p><strong>(c) <\/strong>[latex]P(RB \\text{ or } BR)[\/latex]<\/p>\n<p><strong>(d) <\/strong>[latex]P(\\text{2nd one is }R)[\/latex]<\/p>\n<p><strong>(e) <\/strong>Show that the [latex]P(\\text{Sample Space})[\/latex] is [latex]1[\/latex] or [latex]100\\%[\/latex].<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q983476\">Show Answer<\/button><\/p>\n<div id=\"q983476\" class=\"hidden-answer\" style=\"display: none\">\n<p><strong>(a) <\/strong>[latex]P(B)=\\frac{8}{11}[\/latex]<\/p>\n<p><strong>(b) <\/strong>[latex]P(R \\text{ on the 1st} \\text{ and } B \\text{ on the 2nd})=\\frac{3}{11}*\\frac{8}{10}=\\frac{24}{110}[\/latex]. This is also the same as writing it as [latex]P(RB)[\/latex].<\/p>\n<p><strong>(c) <\/strong>[latex]P(RB \\text{ or } BR)=P(RB)+P(BR)=\\frac{24}{110}+\\frac{24}{110}=\\frac{48}{110}[\/latex]<\/p>\n<p><strong>(d) <\/strong>[latex]P(\\text{2nd one is }R)=P(BR)+P(RR)=\\frac{24}{110}+\\frac{6}{110}=\\frac{30}{110}[\/latex]<\/p>\n<p><strong>(e) <\/strong>[latex]P(\\text{Sample Space})= P(BB)+P(BR)+P(RB)+P(RR)=\\frac{56}{110}+\\frac{24}{110}+\\frac{24}{110}+\\frac{6}{110}=\\frac{56+24+24+6}{110}=\\frac{110}{110}=1[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<\/section>\n<section><\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm13823\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=13823&theme=lumen&iframe_resize_id=ohm13823&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n","protected":false},"author":51,"menu_order":12,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":2910,"module-header":"learn_it","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/chapters\/3172"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/wp\/v2\/users\/51"}],"version-history":[{"count":23,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/chapters\/3172\/revisions"}],"predecessor-version":[{"id":6357,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/chapters\/3172\/revisions\/6357"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/parts\/2910"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/chapters\/3172\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/wp\/v2\/media?parent=3172"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/chapter-type?post=3172"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/wp\/v2\/contributor?post=3172"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/wp\/v2\/license?post=3172"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}