{"id":3310,"date":"2023-10-02T05:59:53","date_gmt":"2023-10-02T05:59:53","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/introstatstest\/?post_type=chapter&#038;p=3310"},"modified":"2025-03-25T14:24:26","modified_gmt":"2025-03-25T14:24:26","slug":"probability-with-tree-diagrams-learn-it-3-update","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/introstatstest\/chapter\/probability-with-tree-diagrams-learn-it-3-update\/","title":{"raw":"Probability with Tree Diagrams: Learn It 3","rendered":"Probability with Tree Diagrams: Learn It 3"},"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>A very natural use of tree diagrams is to help visualize conditional probability for compound events.\u00a0<\/p>\r\n<section class=\"textbox keyTakeaway\">\r\n<h3>conditional probability<\/h3>\r\n<p>The\u00a0<strong>conditional probability\u00a0<\/strong>of [latex]B[\/latex] given [latex]A[\/latex] is denoted as [latex]P(B\\text{ given }A) = P(B|A).[\/latex]<\/p>\r\n<p>&nbsp;<\/p>\r\n<p>[latex]P(B\\text{ given }A) = P(B|A)[\/latex] is the probability that event [latex]B[\/latex] <span style=\"font-size: 1rem; text-align: initial; background-color: initial;\">will occur given that the event [latex]A[\/latex] <\/span><span style=\"font-size: 1rem; text-align: initial; background-color: initial;\">has already occurred.<\/span><\/p>\r\n<p>&nbsp;<\/p>\r\n<p>In a tree diagram, conditional probability focuses on a specific branch of this tree, considering that a particular event has already happened.<\/p>\r\n<p>&nbsp;<\/p>\r\n<p>In mathematical terms, if [latex]P(A)[\/latex] represents the probability of event [latex]A[\/latex], and [latex]P(B)[\/latex] represents the probability of event [latex]B[\/latex], the conditional probability of [latex]B[\/latex] given that event [latex]A[\/latex] has occurred is denoted as [latex]P(B|A)[\/latex], and it is calculated as:<\/p>\r\n<p style=\"text-align: center;\"><span class=\"math math-inline\"><span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist-s\">[latex]P(B|A) = \\dfrac{P(A\\text{ and }B)}{P(A)}[\/latex]<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\r\n<\/section>\r\n<p>Utilizing the tree diagram allows us to intuitively grasp conditional probability!<\/p>\r\n<p>By using the tree diagram, we can visualize the conditional probability [latex]P(B\\text{ given } A)[\/latex] by analyzing a particular route within the tree. So, when we consider events [latex]A[\/latex] and [latex]B[\/latex], we can interpret the conditional probability [latex]P(B\\text{ given } A)[\/latex] by first tracing the path associated with event [latex]A[\/latex] and then determining the probability associated with the subsequent path leading to event [latex]B[\/latex].<\/p>\r\n<section class=\"textbox example\">\r\n<p><span style=\"font-size: 16.8px;\">Suppose 10 marbles were placed in a bag with 4 being blue and 6 being red. Suppose two marbles were consecutively drawn without replacement from the bag.<\/span><\/p>\r\n<p>The tree diagram for the given scenario is given below.<\/p>\r\n<p><img class=\"alignnone wp-image-3313 size-medium\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/27\/2023\/10\/03050357\/Tree_Diagram_Conditional_Prob-300x189.png\" alt=\"This is a tree diagram with branches showing frequencies of each draw. The first branch shows two lines: 4\/10 blue and 6\/10 red. The second branch has a set of two lines, 3\/9 blue and 6\/9 red for the blue branch and 4\/9 blue and 5\/9 red for the red branch.\" width=\"300\" height=\"189\" \/><\/p>\r\n<p>Let's calculate the following conditional probability!<\/p>\r\n<p><strong>(a)<\/strong> If a red was drawn first, what is the probability that the second draw was blue? [reveal-answer q=\"378111\"]Show Answer[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"378111\"] First of all the probability can be represented as [latex]P(\\text{Blue drawn second}|\\text{Red drawn first})[\/latex]. This signifies the likelihood of drawing a blue marble as the second one given that a red marble was drawn first.<\/p>\r\n<p>In this scenario, after drawing a red marble first, the total number of marbles remaining in the bag decreases to [latex]9[\/latex] (since one red marble has been drawn). Among these 9 marbles, there are 4 blue ones. Therefore, the conditional probability [latex]P(\\text{Blue drawn second}|\\text{Red drawn first}) = \\dfrac{4}{9}[\/latex][\/hidden-answer]<\/p>\r\n<p><strong>(b) <\/strong>If a blue was drawn first, what is the probability that the second draw is red? [reveal-answer q=\"378112\"]Show Answer[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"378112\"] [latex]P(\\text{Red drawn second}|\\text{Blue drawn first}) = \\dfrac{6}{9}[\/latex][\/hidden-answer]<\/p>\r\n<p><strong>(c) <\/strong>If a blue was drawn first, what is the probability that the second draw is also blue? [reveal-answer q=\"378113\"]Show Answer[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"378113\"] [latex]P(\\text{Blue drawn second}|\\text{Blue drawn first}) = \\dfrac{3}{9}[\/latex][\/hidden-answer]<\/p>\r\n<\/section>\r\n<section>\r\n<section class=\"textbox tryIt\">\r\n<p>[ohm2_question hide_question_numbers=1]13825[\/ohm2_question]<\/p>\r\n<\/section>\r\n<\/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>A very natural use of tree diagrams is to help visualize conditional probability for compound events.\u00a0<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>conditional probability<\/h3>\n<p>The\u00a0<strong>conditional probability\u00a0<\/strong>of [latex]B[\/latex] given [latex]A[\/latex] is denoted as [latex]P(B\\text{ given }A) = P(B|A).[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>[latex]P(B\\text{ given }A) = P(B|A)[\/latex] is the probability that event [latex]B[\/latex] <span style=\"font-size: 1rem; text-align: initial; background-color: initial;\">will occur given that the event [latex]A[\/latex] <\/span><span style=\"font-size: 1rem; text-align: initial; background-color: initial;\">has already occurred.<\/span><\/p>\n<p>&nbsp;<\/p>\n<p>In a tree diagram, conditional probability focuses on a specific branch of this tree, considering that a particular event has already happened.<\/p>\n<p>&nbsp;<\/p>\n<p>In mathematical terms, if [latex]P(A)[\/latex] represents the probability of event [latex]A[\/latex], and [latex]P(B)[\/latex] represents the probability of event [latex]B[\/latex], the conditional probability of [latex]B[\/latex] given that event [latex]A[\/latex] has occurred is denoted as [latex]P(B|A)[\/latex], and it is calculated as:<\/p>\n<p style=\"text-align: center;\"><span class=\"math math-inline\"><span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist-s\">[latex]P(B|A) = \\dfrac{P(A\\text{ and }B)}{P(A)}[\/latex]<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n<\/section>\n<p>Utilizing the tree diagram allows us to intuitively grasp conditional probability!<\/p>\n<p>By using the tree diagram, we can visualize the conditional probability [latex]P(B\\text{ given } A)[\/latex] by analyzing a particular route within the tree. So, when we consider events [latex]A[\/latex] and [latex]B[\/latex], we can interpret the conditional probability [latex]P(B\\text{ given } A)[\/latex] by first tracing the path associated with event [latex]A[\/latex] and then determining the probability associated with the subsequent path leading to event [latex]B[\/latex].<\/p>\n<section class=\"textbox example\">\n<p><span style=\"font-size: 16.8px;\">Suppose 10 marbles were placed in a bag with 4 being blue and 6 being red. Suppose two marbles were consecutively drawn without replacement from the bag.<\/span><\/p>\n<p>The tree diagram for the given scenario is given below.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-3313 size-medium\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/27\/2023\/10\/03050357\/Tree_Diagram_Conditional_Prob-300x189.png\" alt=\"This is a tree diagram with branches showing frequencies of each draw. The first branch shows two lines: 4\/10 blue and 6\/10 red. The second branch has a set of two lines, 3\/9 blue and 6\/9 red for the blue branch and 4\/9 blue and 5\/9 red for the red branch.\" width=\"300\" height=\"189\" srcset=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/27\/2023\/10\/03050357\/Tree_Diagram_Conditional_Prob-300x189.png 300w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/27\/2023\/10\/03050357\/Tree_Diagram_Conditional_Prob-65x41.png 65w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/27\/2023\/10\/03050357\/Tree_Diagram_Conditional_Prob-225x142.png 225w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/27\/2023\/10\/03050357\/Tree_Diagram_Conditional_Prob-350x220.png 350w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/27\/2023\/10\/03050357\/Tree_Diagram_Conditional_Prob.png 612w\" sizes=\"(max-width: 300px) 100vw, 300px\" \/><\/p>\n<p>Let&#8217;s calculate the following conditional probability!<\/p>\n<p><strong>(a)<\/strong> If a red was drawn first, what is the probability that the second draw was blue? <\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q378111\">Show Answer<\/button><\/p>\n<div id=\"q378111\" class=\"hidden-answer\" style=\"display: none\"> First of all the probability can be represented as [latex]P(\\text{Blue drawn second}|\\text{Red drawn first})[\/latex]. This signifies the likelihood of drawing a blue marble as the second one given that a red marble was drawn first.<\/p>\n<p>In this scenario, after drawing a red marble first, the total number of marbles remaining in the bag decreases to [latex]9[\/latex] (since one red marble has been drawn). Among these 9 marbles, there are 4 blue ones. Therefore, the conditional probability [latex]P(\\text{Blue drawn second}|\\text{Red drawn first}) = \\dfrac{4}{9}[\/latex]<\/p><\/div>\n<\/div>\n<p><strong>(b) <\/strong>If a blue was drawn first, what is the probability that the second draw is red? <\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q378112\">Show Answer<\/button><\/p>\n<div id=\"q378112\" class=\"hidden-answer\" style=\"display: none\"> [latex]P(\\text{Red drawn second}|\\text{Blue drawn first}) = \\dfrac{6}{9}[\/latex]<\/div>\n<\/div>\n<p><strong>(c) <\/strong>If a blue was drawn first, what is the probability that the second draw is also blue? <\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q378113\">Show Answer<\/button><\/p>\n<div id=\"q378113\" class=\"hidden-answer\" style=\"display: none\"> [latex]P(\\text{Blue drawn second}|\\text{Blue drawn first}) = \\dfrac{3}{9}[\/latex]<\/div>\n<\/div>\n<\/section>\n<section>\n<section class=\"textbox tryIt\">\n<iframe loading=\"lazy\" id=\"ohm13825\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=13825&theme=lumen&iframe_resize_id=ohm13825&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><br \/>\n<\/section>\n<\/section>\n","protected":false},"author":51,"menu_order":13,"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\/3310"}],"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":25,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/chapters\/3310\/revisions"}],"predecessor-version":[{"id":6358,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/chapters\/3310\/revisions\/6358"}],"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\/3310\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/wp\/v2\/media?parent=3310"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/chapter-type?post=3310"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/wp\/v2\/contributor?post=3310"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/wp\/v2\/license?post=3310"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}