{"id":3223,"date":"2023-09-20T19:49:29","date_gmt":"2023-09-20T19:49:29","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/introstatstest\/?post_type=chapter&#038;p=3223"},"modified":"2025-05-11T23:31:05","modified_gmt":"2025-05-11T23:31:05","slug":"probability-of-events-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/introstatstest\/chapter\/probability-of-events-learn-it-3\/","title":{"raw":"Probability of Compound Events: Learn It 3","rendered":"Probability of Compound Events: Learn It 3"},"content":{"raw":"<section class=\"textbox learningGoals\">\r\n<ul>\r\n\t<li>Calculate and interpret probabilities of simple and compound events.<\/li>\r\n\t<li>Understand the concept of mutually exclusive events.<\/li>\r\n<\/ul>\r\n<\/section>\r\n<section class=\"textbox youChoose\">[choosedataset divId=\"tnh-choose-dataset\" title=\"Choose Your Own Dataset\" label=\"For this problem, you'll determine probabilities from your choice of tables.\" default=\"Choose a Dataset\"]<br \/>\r\n[datasetoption]<br \/>\r\n[displayname]Korean Dog Owners[\/displayname]<br \/>\r\n[ohmid]2763[\/ohmid]<br \/>\r\n[\/datasetoption]<br \/>\r\n[datasetoption]<br \/>\r\n[displayname]Why Didn't You Vote?[\/displayname]<br \/>\r\n[ohmid]2764[\/ohmid]<br \/>\r\n[\/datasetoption]<br \/>\r\n[datasetoption]<br \/>\r\n[displayname]Endangered Species[\/displayname]<br \/>\r\n[ohmid]2765[\/ohmid]<br \/>\r\n[\/datasetoption][\/choosedataset]<\/section>\r\n<h2>Mutually Exclusive Events<\/h2>\r\n<p>When both events [latex]A[\/latex] and [latex]B[\/latex] do not occur or happen at the same time, we can say that event [latex]A[\/latex] and event [latex]B[\/latex] are <strong>mutually exclusive\u00a0<\/strong>or\u00a0<strong>disjoint.<\/strong><\/p>\r\n<p>To find the probability of event [latex]A[\/latex] OR event [latex]B[\/latex], add the probability of event [latex]A[\/latex] and the probability of event [latex]B[\/latex], as long as both events are mutually exclusive.<\/p>\r\n<section class=\"textbox keyTakeaway\">\r\n<h3>mutually exclusive<\/h3>\r\n<p style=\"text-align: left;\">If event [latex]A[\/latex] and event [latex]B[\/latex] are mutually exclusive, then<\/p>\r\n<p style=\"text-align: center;\">[latex]P(A \\text{ or } B) = P(A \\cup B) = P(A)+P(B)[\/latex]<\/p>\r\n<\/section>\r\n<section class=\"textbox example\">Klaus is trying to choose where to go on vacation. His two choices are New Zealand and Alaska. Klaus can only afford one vacation. \u00a0[latex]P(\\text{New Zealand}) = 0.60[\/latex] and [latex]P(\\text{Alaska}) = 0.35.[\/latex]\r\n\r\n<p><strong>(a)<\/strong> What is the probability that Klaus will go on vacation to New Zealand OR Alaska?<br \/>\r\n[reveal-answer q=\"89535\"]Show answer[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"89535\"]Because New Zealand and Alaska are mutually exclusive:[latex]P(\\text{New Zealand } \\text{or} \\text{ Alaska}) = P(\\text{New Zealand})+P(\\text{Alaska}) = 0.60+0.35 = 0.95[\/latex].[\/hidden-answer]<\/p>\r\n<p><strong>(b)<\/strong> What is the probability that Klaus will not go to New Zealand or Alaska for vacation?[reveal-answer q=\"278178\"]Show answer[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"278178\"]If [latex]P(\\text{New Zealand } \\text{or} \\text{ Alaska}) = 0.95[\/latex], then [latex]P(\\text{not }(\\text{New Zealand } \\text{or} \\text{ Alaska})) = 1-0.95 =0.05[\/latex].[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section>\r\n<section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]12295[\/ohm2_question]<\/section>\r\n<\/section>\r\n<section>\r\n<h2>Independent Events<\/h2>\r\n<p>Sometimes, there are pairs of events for which one event has no effect on the probability of another event occurring. When this is the case, we say the events are <strong>independent<\/strong>.<\/p>\r\n<section class=\"textbox keyTakeaway\">\r\n<h3>independent events<\/h3>\r\n<p style=\"text-align: left;\">Two events are independent if the outcome of the first event does not influence the probability of the second event. We can test for independence using the rule:<\/p>\r\n<p style=\"text-align: center;\">[latex]P(A \\text{ and } B)=P(A) \\times P(B)[\/latex]<\/p>\r\n<\/section>\r\n<section>\r\n<section class=\"textbox example\">A researcher conducts a survey of 120 randomly selected college students to try to answer the questions: If someone has a laptop, are they likely to own a desktop computer? If someone has a desktop computer, are they likely to own a laptop?<br \/>\r\nThe results of the survey are displayed in the following contingency table.\r\n\r\n<div align=\"center\">\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td>&nbsp;<\/td>\r\n<td>Owns a laptop<\/td>\r\n<td>Does not own a laptop<\/td>\r\n<td><strong>Total<\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Owns a desktop<\/td>\r\n<td>20<\/td>\r\n<td>20<\/td>\r\n<td><strong>40<\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Does not own a desktop<\/td>\r\n<td>60<\/td>\r\n<td>20<\/td>\r\n<td><strong>80<\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Total<\/strong><\/td>\r\n<td><strong>80<\/strong><\/td>\r\n<td><strong>40<\/strong><\/td>\r\n<td><strong>120<\/strong><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<p>Is owning a desktop independent of whether the student owns a laptop? Or is there a relationship between these two events?<\/p>\r\n<p>[reveal-answer q=\"715949\"]Show answer[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"715949\"]<\/p>\r\n<p>To answer this question, we want to test if [latex]P(\\text{Owns a laptop})\\times P(\\text{Owns a desktop})=P(\\text{Owns a laptop AND a desktop})[\/latex], then we can conclude that owning a desktop is <strong>independent\u00a0<\/strong>of owning a laptop. If the probabilities are different, then we say the variables are <strong>dependent<\/strong>.<\/p>\r\n<p style=\"text-align: center;\">[latex] P(\\text{Owns a desktop}) \\times P(\\text{Owns a laptop}) = P(\\text{Owns a desktop} \\text{ and } \\text{Owns a laptop})[\/latex].<\/p>\r\n<p style=\"text-align: center;\">[latex]\\frac{40}{120} \\times \\frac{80}{120} ? \\frac{20}{120}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\frac{3200}{14400} ? \\frac{20}{120}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\frac{2}{9} \\ne \\frac{1}{6}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">We have found that owning a desktop and owning a laptop are <strong>dependent<\/strong>.<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox tryIt\">\r\n<p>[ohm2_question hide_question_numbers=1]17610[\/ohm2_question]<\/p>\r\n<\/section>\r\n<\/section>\r\n<\/section>\r\n<section><\/section>","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li>Calculate and interpret probabilities of simple and compound events.<\/li>\n<li>Understand the concept of mutually exclusive events.<\/li>\n<\/ul>\n<\/section>\n<section class=\"textbox youChoose\">\n<div id=\"tnh-choose-dataset\" class=\"chooseDataset\">\n<h3>Choose Your Own Dataset<\/h3>\n<h4>For this problem, you'll determine probabilities from your choice of tables.<\/h4>\n<form><select name=\"dataset\"><option value=\"\">Choose a Dataset<\/option><option value=\"2763\">Korean Dog Owners<\/option><option value=\"2764\">Why Didn&#8217;t You Vote?<\/option><option value=\"2765\">Endangered Species<\/option><\/select><\/form>\n<div class=\"ohmContainer\"><\/div>\n<\/p><\/div>\n<\/section>\n<h2>Mutually Exclusive Events<\/h2>\n<p>When both events [latex]A[\/latex] and [latex]B[\/latex] do not occur or happen at the same time, we can say that event [latex]A[\/latex] and event [latex]B[\/latex] are <strong>mutually exclusive\u00a0<\/strong>or\u00a0<strong>disjoint.<\/strong><\/p>\n<p>To find the probability of event [latex]A[\/latex] OR event [latex]B[\/latex], add the probability of event [latex]A[\/latex] and the probability of event [latex]B[\/latex], as long as both events are mutually exclusive.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>mutually exclusive<\/h3>\n<p style=\"text-align: left;\">If event [latex]A[\/latex] and event [latex]B[\/latex] are mutually exclusive, then<\/p>\n<p style=\"text-align: center;\">[latex]P(A \\text{ or } B) = P(A \\cup B) = P(A)+P(B)[\/latex]<\/p>\n<\/section>\n<section class=\"textbox example\">Klaus is trying to choose where to go on vacation. His two choices are New Zealand and Alaska. Klaus can only afford one vacation. \u00a0[latex]P(\\text{New Zealand}) = 0.60[\/latex] and [latex]P(\\text{Alaska}) = 0.35.[\/latex]<\/p>\n<p><strong>(a)<\/strong> What is the probability that Klaus will go on vacation to New Zealand OR Alaska?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q89535\">Show answer<\/button><\/p>\n<div id=\"q89535\" class=\"hidden-answer\" style=\"display: none\">Because New Zealand and Alaska are mutually exclusive:[latex]P(\\text{New Zealand } \\text{or} \\text{ Alaska}) = P(\\text{New Zealand})+P(\\text{Alaska}) = 0.60+0.35 = 0.95[\/latex].<\/div>\n<\/div>\n<p><strong>(b)<\/strong> What is the probability that Klaus will not go to New Zealand or Alaska for vacation?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q278178\">Show answer<\/button><\/p>\n<div id=\"q278178\" class=\"hidden-answer\" style=\"display: none\">If [latex]P(\\text{New Zealand } \\text{or} \\text{ Alaska}) = 0.95[\/latex], then [latex]P(\\text{not }(\\text{New Zealand } \\text{or} \\text{ Alaska})) = 1-0.95 =0.05[\/latex].<\/div>\n<\/div>\n<\/section>\n<section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm12295\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=12295&theme=lumen&iframe_resize_id=ohm12295&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<\/section>\n<section>\n<h2>Independent Events<\/h2>\n<p>Sometimes, there are pairs of events for which one event has no effect on the probability of another event occurring. When this is the case, we say the events are <strong>independent<\/strong>.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>independent events<\/h3>\n<p style=\"text-align: left;\">Two events are independent if the outcome of the first event does not influence the probability of the second event. We can test for independence using the rule:<\/p>\n<p style=\"text-align: center;\">[latex]P(A \\text{ and } B)=P(A) \\times P(B)[\/latex]<\/p>\n<\/section>\n<section>\n<section class=\"textbox example\">A researcher conducts a survey of 120 randomly selected college students to try to answer the questions: If someone has a laptop, are they likely to own a desktop computer? If someone has a desktop computer, are they likely to own a laptop?<br \/>\nThe results of the survey are displayed in the following contingency table.<\/p>\n<div style=\"margin: auto;\">\n<table>\n<tbody>\n<tr>\n<td>&nbsp;<\/td>\n<td>Owns a laptop<\/td>\n<td>Does not own a laptop<\/td>\n<td><strong>Total<\/strong><\/td>\n<\/tr>\n<tr>\n<td>Owns a desktop<\/td>\n<td>20<\/td>\n<td>20<\/td>\n<td><strong>40<\/strong><\/td>\n<\/tr>\n<tr>\n<td>Does not own a desktop<\/td>\n<td>60<\/td>\n<td>20<\/td>\n<td><strong>80<\/strong><\/td>\n<\/tr>\n<tr>\n<td><strong>Total<\/strong><\/td>\n<td><strong>80<\/strong><\/td>\n<td><strong>40<\/strong><\/td>\n<td><strong>120<\/strong><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<p>Is owning a desktop independent of whether the student owns a laptop? Or is there a relationship between these two events?<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q715949\">Show answer<\/button><\/p>\n<div id=\"q715949\" class=\"hidden-answer\" style=\"display: none\">\n<p>To answer this question, we want to test if [latex]P(\\text{Owns a laptop})\\times P(\\text{Owns a desktop})=P(\\text{Owns a laptop AND a desktop})[\/latex], then we can conclude that owning a desktop is <strong>independent\u00a0<\/strong>of owning a laptop. If the probabilities are different, then we say the variables are <strong>dependent<\/strong>.<\/p>\n<p style=\"text-align: center;\">[latex]P(\\text{Owns a desktop}) \\times P(\\text{Owns a laptop}) = P(\\text{Owns a desktop} \\text{ and } \\text{Owns a laptop})[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{40}{120} \\times \\frac{80}{120} ? \\frac{20}{120}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{3200}{14400} ? \\frac{20}{120}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{2}{9} \\ne \\frac{1}{6}[\/latex]<\/p>\n<p style=\"text-align: left;\">We have found that owning a desktop and owning a laptop are <strong>dependent<\/strong>.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\">\n<iframe loading=\"lazy\" id=\"ohm17610\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=17610&theme=lumen&iframe_resize_id=ohm17610&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><br \/>\n<\/section>\n<\/section>\n<\/section>\n<section><\/section>\n","protected":false},"author":12,"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":974,"module-header":"learn_it","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":[{"divId":"tnh-choose-dataset","title":"Choose Your Own Dataset","label":"For this problem, you'll determine probabilities from your choice of tables.","default":"Choose a Dataset","try_it_collection":[{"displayName":"Korean Dog Owners","value":"2763"},{"displayName":"Why Didn't You Vote?","value":"2764"},{"displayName":"Endangered Species","value":"2765"}]}],"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/chapters\/3223"}],"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\/12"}],"version-history":[{"count":15,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/chapters\/3223\/revisions"}],"predecessor-version":[{"id":6668,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/chapters\/3223\/revisions\/6668"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/parts\/974"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/chapters\/3223\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/wp\/v2\/media?parent=3223"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/chapter-type?post=3223"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/wp\/v2\/contributor?post=3223"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/wp\/v2\/license?post=3223"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}