{"id":336,"date":"2016-10-11T17:59:15","date_gmt":"2016-10-11T17:59:15","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/math4libarts\/?post_type=chapter&#038;p=336"},"modified":"2022-01-07T00:30:07","modified_gmt":"2022-01-07T00:30:07","slug":"4-3-probability-of-a-complementary-event","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/frontrange-mathforliberalartscorequisite1\/chapter\/4-3-probability-of-a-complementary-event\/","title":{"raw":"4.3 Probability of a Complementary Event","rendered":"4.3 Probability of a Complementary Event"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Calculate the probability of a complementary event<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Recall operations on fractions<\/h3>\r\nAdding and subtracting fractions with common denominators\r\n\r\n[latex]\\dfrac{a}{c}\\pm \\dfrac{b}{c}=\\dfrac{a\\pm b}{c}[\/latex]\r\n\r\nIn the two equations below, note that this relationship is described in both directions.\r\n\r\nThat is, it is also true that\r\n\r\n[latex]\\dfrac{a\\pm b}{c}=\\dfrac{a}{c}\\pm \\dfrac{b}{c}[\/latex]\r\n\r\nThe second equation furthermore includes the fact that\r\n\r\n[latex]\\dfrac{a}{a}=1[\/latex]\r\n\r\n<\/div>\r\n<h2>Complementary Events<\/h2>\r\nNow let us examine the probability that an event does <strong>not<\/strong> happen. As in the previous section, consider the situation of rolling a six-sided die and first compute the probability of rolling a six: the answer is <em>P<\/em>(six) =1\/6. Now consider the probability that we do <em>not<\/em> roll a six: there are 5 outcomes that are not a six, so the answer is <em>P<\/em>(not a six) = [latex]\\frac{5}{6}[\/latex]. Notice that\r\n<p style=\"text-align: center;\">[latex]P(\\text{six})+P(\\text{not a six})=\\frac{1}{6}+\\frac{5}{6}=\\frac{6}{6}=1[\/latex]<\/p>\r\nThis is not a coincidence.\u00a0 Consider a generic situation with <em>n<\/em> possible outcomes and an event <em>E<\/em> that corresponds to <em>m<\/em> of these outcomes. Then the remaining <em>n<\/em> - <em>m<\/em> outcomes correspond to <em>E<\/em> not happening, thus\r\n<p style=\"text-align: center;\">[latex]P(\\text{not}E)=\\frac{n-m}{n}=\\frac{n}{n}-\\frac{m}{n}=1-\\frac{m}{n}=1-P(E)[\/latex]<\/p>\r\n<p style=\"text-align: center;\"><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/28183109\/ace-167052_640.jpg\"><img class=\"aligncenter size-full wp-image-991\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/28183109\/ace-167052_640.jpg\" alt=\"scattered playing cards on a table. The Ace of Spades is on top.\" width=\"640\" height=\"426\" \/><\/a><\/p>\r\n\r\n<div class=\"textbox\">\r\n<h3>Complement of an Event<\/h3>\r\nThe <strong>complement<\/strong> of an event is the event \u201c<em>E<\/em> doesn\u2019t happen\u201d\r\n<ul>\r\n \t<li>The notation [latex]\\bar{E}[\/latex] is used for the complement of event <em>E<\/em>.\u00a0 Other commonly used notations for the complement of E are E' or E<sup>c<\/sup>.<\/li>\r\n \t<li>We can compute the probability of the complement using [latex]P\\left({\\bar{E}}\\right)=1-P(E)[\/latex]<\/li>\r\n \t<li>Notice also that [latex]P(E)=1-P\\left({\\bar{E}}\\right)[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\nIf you pull a random card from a deck of playing cards, what is the probability it is not a heart?\r\n[reveal-answer q=\"926985\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"926985\"]\r\n\r\nThere are 13 hearts in the deck, so [latex]P(\\text{heart})=\\frac{13}{52}=\\frac{1}{4}[\/latex].\r\n\r\nThe probability of <em>not<\/em> drawing a heart is the complement: [latex]P(\\text{not heart})=1-P(\\text{heart})=1-\\frac{1}{4}=\\frac{3}{4}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\nThis situation is explained in the following video.\r\n\r\nhttps:\/\/youtu.be\/RnljiW6ZM6A\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]7097[\/ohm_question]\r\n\r\n<\/div>\r\n<p style=\"text-align: center;\"><span style=\"color: #ff0000;\"><strong>This is the end of the section. Close this tab and proceed to the corresponding assignment.<\/strong><\/span><\/p>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Calculate the probability of a complementary event<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Recall operations on fractions<\/h3>\n<p>Adding and subtracting fractions with common denominators<\/p>\n<p>[latex]\\dfrac{a}{c}\\pm \\dfrac{b}{c}=\\dfrac{a\\pm b}{c}[\/latex]<\/p>\n<p>In the two equations below, note that this relationship is described in both directions.<\/p>\n<p>That is, it is also true that<\/p>\n<p>[latex]\\dfrac{a\\pm b}{c}=\\dfrac{a}{c}\\pm \\dfrac{b}{c}[\/latex]<\/p>\n<p>The second equation furthermore includes the fact that<\/p>\n<p>[latex]\\dfrac{a}{a}=1[\/latex]<\/p>\n<\/div>\n<h2>Complementary Events<\/h2>\n<p>Now let us examine the probability that an event does <strong>not<\/strong> happen. As in the previous section, consider the situation of rolling a six-sided die and first compute the probability of rolling a six: the answer is <em>P<\/em>(six) =1\/6. Now consider the probability that we do <em>not<\/em> roll a six: there are 5 outcomes that are not a six, so the answer is <em>P<\/em>(not a six) = [latex]\\frac{5}{6}[\/latex]. Notice that<\/p>\n<p style=\"text-align: center;\">[latex]P(\\text{six})+P(\\text{not a six})=\\frac{1}{6}+\\frac{5}{6}=\\frac{6}{6}=1[\/latex]<\/p>\n<p>This is not a coincidence.\u00a0 Consider a generic situation with <em>n<\/em> possible outcomes and an event <em>E<\/em> that corresponds to <em>m<\/em> of these outcomes. Then the remaining <em>n<\/em> &#8211; <em>m<\/em> outcomes correspond to <em>E<\/em> not happening, thus<\/p>\n<p style=\"text-align: center;\">[latex]P(\\text{not}E)=\\frac{n-m}{n}=\\frac{n}{n}-\\frac{m}{n}=1-\\frac{m}{n}=1-P(E)[\/latex]<\/p>\n<p style=\"text-align: center;\"><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/28183109\/ace-167052_640.jpg\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-991\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/28183109\/ace-167052_640.jpg\" alt=\"scattered playing cards on a table. The Ace of Spades is on top.\" width=\"640\" height=\"426\" \/><\/a><\/p>\n<div class=\"textbox\">\n<h3>Complement of an Event<\/h3>\n<p>The <strong>complement<\/strong> of an event is the event \u201c<em>E<\/em> doesn\u2019t happen\u201d<\/p>\n<ul>\n<li>The notation [latex]\\bar{E}[\/latex] is used for the complement of event <em>E<\/em>.\u00a0 Other commonly used notations for the complement of E are E&#8217; or E<sup>c<\/sup>.<\/li>\n<li>We can compute the probability of the complement using [latex]P\\left({\\bar{E}}\\right)=1-P(E)[\/latex]<\/li>\n<li>Notice also that [latex]P(E)=1-P\\left({\\bar{E}}\\right)[\/latex]<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>If you pull a random card from a deck of playing cards, what is the probability it is not a heart?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q926985\">Show Solution<\/span><\/p>\n<div id=\"q926985\" class=\"hidden-answer\" style=\"display: none\">\n<p>There are 13 hearts in the deck, so [latex]P(\\text{heart})=\\frac{13}{52}=\\frac{1}{4}[\/latex].<\/p>\n<p>The probability of <em>not<\/em> drawing a heart is the complement: [latex]P(\\text{not heart})=1-P(\\text{heart})=1-\\frac{1}{4}=\\frac{3}{4}[\/latex]<\/p>\n<\/div>\n<\/div>\n<p>This situation is explained in the following video.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Probability - Complements\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/RnljiW6ZM6A?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm7097\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=7097&theme=oea&iframe_resize_id=ohm7097&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p style=\"text-align: center;\"><span style=\"color: #ff0000;\"><strong>This is the end of the section. Close this tab and proceed to the corresponding assignment.<\/strong><\/span><\/p>\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-336\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Revisoin and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Working With Events. <strong>Authored by<\/strong>: David Lippman. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/www.opentextbookstore.com\/mathinsociety\/\">http:\/\/www.opentextbookstore.com\/mathinsociety\/<\/a>. <strong>Project<\/strong>: Math in Society. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-sa\/4.0\/\">CC BY-SA: Attribution-ShareAlike<\/a><\/em><\/li><li>ace-playing-cards-deck-spades. <strong>Authored by<\/strong>: PDPics. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/pixabay.com\/en\/ace-playing-cards-deck-spades-167052\/\">https:\/\/pixabay.com\/en\/ace-playing-cards-deck-spades-167052\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/about\/cc0\">CC0: No Rights Reserved<\/a><\/em><\/li><li>Probability - Complements. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/RnljiW6ZM6A\">https:\/\/youtu.be\/RnljiW6ZM6A<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Probability of two events: P(A or B). <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/klbPZeH1np4\">https:\/\/youtu.be\/klbPZeH1np4<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Joint probabilities of independent events: P(A and B). <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/6F17WLp-EL8\">https:\/\/youtu.be\/6F17WLp-EL8<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Probabilities from a table: AND and OR. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/HWrGoM1yRaU\">https:\/\/youtu.be\/HWrGoM1yRaU<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Basic conditional probability. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/b6tstekMlb8\">https:\/\/youtu.be\/b6tstekMlb8<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Conditional probability with cards. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/ngyGsgV4_0U\">https:\/\/youtu.be\/ngyGsgV4_0U<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Conditional probability from a table. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/LH0cuHS9Ez0\">https:\/\/youtu.be\/LH0cuHS9Ez0<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Question ID 7111, 7113, 7114, 7115. <strong>Authored by<\/strong>: unknown. <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>: IMathAS Community License CC-BY + GPL<\/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":20,"menu_order":4,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Working With Events\",\"author\":\"David Lippman\",\"organization\":\"\",\"url\":\"http:\/\/www.opentextbookstore.com\/mathinsociety\/\",\"project\":\"Math in Society\",\"license\":\"cc-by-sa\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"ace-playing-cards-deck-spades\",\"author\":\"PDPics\",\"organization\":\"\",\"url\":\"https:\/\/pixabay.com\/en\/ace-playing-cards-deck-spades-167052\/\",\"project\":\"\",\"license\":\"cc0\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Probability - 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