{"id":1257,"date":"2017-02-01T22:10:20","date_gmt":"2017-02-01T22:10:20","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/waymakermath4libarts\/?post_type=chapter&#038;p=1257"},"modified":"2022-01-07T00:29:10","modified_gmt":"2022-01-07T00:29:10","slug":"4-2-computing-the-probability-of-an-event","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/frontrange-mathforliberalartscorequisite1\/chapter\/4-2-computing-the-probability-of-an-event\/","title":{"raw":"4.2 Introduction to Computing the Probability of an Event","rendered":"4.2 Introduction to Computing the Probability of an Event"},"content":{"raw":"<h2>Learning to Calculate the Probability of Events<\/h2>\r\n<strong>Probability<\/strong> is the likelihood of a particular outcome or event happening. Statisticians and actuaries use probability to make predictions about events. \u00a0An actuary that works for a car insurance company would, for example, be interested in how likely a 17-year-old male would be to get in a car accident. They would use data from past events to make predictions about future events using the characteristics of probabilities, then use this information to calculate an insurance rate.\r\n\r\n<strong>Probabilities are between zero and one, inclusive<\/strong>\u00a0(that is, zero and one and all numbers between these values).\r\n<ul>\r\n \t<li><em>P<\/em>(<em>A<\/em>) = 0 means the event\u00a0<em>A<\/em>\u00a0can never happen.<\/li>\r\n \t<li><em>P<\/em>(<em>A<\/em>) = 1 means the event\u00a0<em>A<\/em>\u00a0always happens.<\/li>\r\n \t<li><em>P<\/em>(<em>A<\/em>) = 0.5 means the event\u00a0<em>A<\/em>\u00a0is equally likely to occur or not to occur.<\/li>\r\n<\/ul>\r\n<strong>Equally likely\u00a0<\/strong>means that each outcome of an experiment occurs with equal probability. For example, if you toss a\u00a0<strong>fair<\/strong>, six-sided die, each face (1, 2, 3, 4, 5, or 6) is as likely to occur as any other face. If you toss a fair coin, a Head (<em>H<\/em>) and a Tail (<em>T<\/em>) are equally likely to occur. If you randomly guess the answer to a true\/false question on an exam, you are equally likely to select a correct answer or an incorrect answer.\r\n\r\n<strong>To calculate the probability of an event\u00a0<em>A<\/em>\u00a0when all outcomes in the sample space are equally likely<\/strong>, count the number of outcomes for event\u00a0<em>A<\/em>\u00a0and divide by the total number of outcomes in the sample space. For example, if you toss a fair dime and a fair nickel, the sample space is {<em>HH<\/em>,\u00a0<em>TH<\/em>,\u00a0<em>HT<\/em>,\u00a0<em>TT<\/em>} where\u00a0<em>T<\/em>\u00a0= tails and\u00a0<em>H<\/em>\u00a0= heads. The sample space has four outcomes.\u00a0<em>A<\/em>\u00a0= getting one head. There are two outcomes that meet this condition {<em>HT<\/em>,\u00a0<em>TH<\/em>}, so P(A) = 2\/4 = 0.5.\r\n\r\nSuppose you roll one fair six-sided die, with the numbers {1, 2, 3, 4, 5, 6} on its faces. Let event\u00a0<em>E<\/em>\u00a0= rolling a number that is at least five. There are two outcomes {5, 6}.\u00a0<span id=\"MathJax-Element-2-Frame\" class=\"mjx-chtml MathJax_CHTML\" style=\"font-family: proxima-nova, sans-serif; padding: 1px 0px; margin: 0px; font-size: 21.8px; vertical-align: baseline; background: transparent; border: 0px; line-height: 0; text-indent: 0px; text-align: left; font-style: normal; font-weight: normal; letter-spacing: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px;\" role=\"presentation\"><span id=\"MJXc-Node-27\" class=\"mjx-math\"><span id=\"MJXc-Node-28\" class=\"mjx-mrow\"><span id=\"MJXc-Node-29\" class=\"mjx-mstyle\"><span id=\"MJXc-Node-30\" class=\"mjx-mrow\"><span id=\"MJXc-Node-31\" class=\"mjx-texatom\"><span id=\"MJXc-Node-32\" class=\"mjx-mrow\"><span id=\"MJXc-Node-33\" class=\"mjx-mi\"><span class=\"mjx-char MJXc-TeX-math-I\">P<\/span><\/span><\/span><\/span><span id=\"MJXc-Node-34\" class=\"mjx-texatom\"><span id=\"MJXc-Node-35\" class=\"mjx-mrow\"><span id=\"MJXc-Node-36\" class=\"mjx-mo\"><span class=\"mjx-char MJXc-TeX-main-R\">(<\/span><\/span><span id=\"MJXc-Node-37\" class=\"mjx-texatom\"><span id=\"MJXc-Node-38\" class=\"mjx-mrow\"><span id=\"MJXc-Node-39\" class=\"mjx-mi\"><span class=\"mjx-char MJXc-TeX-math-I\">E<\/span><\/span><\/span><\/span><span id=\"MJXc-Node-40\" class=\"mjx-mo\"><span class=\"mjx-char MJXc-TeX-main-R\">)<\/span><\/span><\/span><\/span><span id=\"MJXc-Node-41\" class=\"mjx-mo MJXc-space3\"><span class=\"mjx-char MJXc-TeX-main-R\">=<\/span><\/span><span id=\"MJXc-Node-42\" class=\"mjx-mfrac MJXc-space3\"><span class=\"mjx-box MJXc-stacked\"><span class=\"mjx-numerator\"><span id=\"MJXc-Node-43\" class=\"mjx-texatom\"><span id=\"MJXc-Node-44\" class=\"mjx-mrow\"><span id=\"MJXc-Node-45\" class=\"mjx-mn\"><span class=\"mjx-char MJXc-TeX-main-R\">2\/<\/span><\/span><\/span><\/span><\/span><span class=\"mjx-denominator\"><span id=\"MJXc-Node-46\" class=\"mjx-texatom\"><span id=\"MJXc-Node-47\" class=\"mjx-mrow\"><span id=\"MJXc-Node-48\" class=\"mjx-mn\"><span class=\"mjx-char MJXc-TeX-main-R\">6.\u00a0 This simplifies to P(E) = 1\/3<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>\r\n\r\nThis important characteristic of probability experiments is known as the\u00a0<strong>law of large numbers<\/strong>\u00a0which states that as the number of repetitions of an experiment is increased, the empirical probability of the experiment tends to become closer and closer to the theoretical probability. Even though the outcomes do not happen according to any set pattern or order, overall, the long-term empirical probability will approach the theoretical probability.\r\n<div class=\"textbox\">\r\n<h3>Basic Probability<\/h3>\r\nGiven that all outcomes are equally likely, we can compute the probability of an event <em>E<\/em> using this formula:\r\n<p style=\"text-align: center;\">[latex]P(E)=\\frac{\\text{Number of favorable outcomes corresponding to the event E}}{\\text{Number of equally-likely outcomes (sample space)}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>examples<\/h3>\r\nIf we roll a 6-sided die, calculate\r\n<ol>\r\n \t<li>P(rolling a 1)<\/li>\r\n \t<li>P(rolling a number bigger than 4)<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"417705\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"417705\"]\r\n\r\nRecall that the sample space is {1,2,3,4,5,6}\r\n<ol>\r\n \t<li>There is one outcome corresponding to \u201crolling a 1,\u201d so the probability is\u00a0 [latex]\\frac{2}{6}[\/latex]\u00a0[latex]\\frac{1}{6}[\/latex]<\/li>\r\n \t<li>There are two outcomes bigger than a 4, so the probability is [latex]\\frac{2}{6}=\\frac{1}{3}[\/latex]<\/li>\r\n<\/ol>\r\nProbabilities are essentially fractions, and can be simplified to lower terms like fractions.\r\n\r\n[\/hidden-answer]\r\n\r\nThis video describes this example and the previous one in detail.\r\n\r\nhttps:\/\/youtu.be\/37P01dt0zsE\r\n\r\nLet's say you have a bag with 20 cherries, 14 sweet and 6 sour. If you pick a cherry at random, what is the probability that it will be sweet?\r\n[reveal-answer q=\"9680\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"9680\"]\r\n\r\nThere are 20 possible cherries that could be picked, so the number of possible outcomes is 20. Of these 20 possible outcomes, 14 are favorable (sweet), so the probability that the cherry will be sweet is [latex]\\frac{14}{20}=\\frac{7}{10}[\/latex].\r\nThere is one potential complication to this example, however. It must be assumed that the probability of picking any of the cherries is the same as the probability of picking any other. This wouldn't be true if (let us imagine) the sweet cherries are smaller than the sour ones. (The sour cherries would come to hand more readily when you sampled from the bag.) Let us keep in mind, therefore, that when we assess probabilities in terms of the ratio of favorable to all potential cases, we rely heavily on the assumption of equal probability for all outcomes.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Cards<\/h3>\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5548\/2016\/10\/04191241\/Deck-of-Cards.png\"><img class=\"alignnone size-medium wp-image-5985\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5548\/2016\/10\/04191241\/Deck-of-Cards-223x300.png\" alt=\"\" width=\"223\" height=\"300\" \/><\/a>\r\n\r\nA standard deck of 52 playing cards consists of four <strong>suits<\/strong> (hearts, spades, diamonds and clubs). Spades and clubs are black while hearts and diamonds are red. Each suit contains 13 cards, each of a different <strong>rank<\/strong>: an Ace (which in many games functions as both a low card and a high card), cards numbered 2 through 10, a Jack, a Queen and a King.\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\nCompute the probability of randomly drawing one card from a deck and getting an Ace.\r\n[reveal-answer q=\"652517\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"652517\"]\r\n\r\nThere are 52 cards in the deck and 4 Aces so\u00a0[latex]P(Ace)=\\frac{4}{52}=\\frac{1}{13}\\approx 0.0769[\/latex]\r\n\r\nNotice that the smallest possible probability is 0 \u2013 if there are no outcomes that correspond with the event. The largest possible probability is 1 \u2013 if all possible outcomes correspond with the event.\r\n\r\n[\/hidden-answer]\r\n\r\nThis video demonstrates both this example and the previous cherry example\u00a0on the page.\u00a0 Note that sometimes a probability is interpreted as a percent.\u00a0 In our class we will give the probability as a simplified fraction or a number rounded to 3 decimal digits.\u00a0 (i.e. 0.729)\r\n\r\nhttps:\/\/youtu.be\/EBqj_R3dzd4\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]7130[\/ohm_question]\r\n\r\n<\/div>\r\n&nbsp;\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":"<h2>Learning to Calculate the Probability of Events<\/h2>\n<p><strong>Probability<\/strong> is the likelihood of a particular outcome or event happening. Statisticians and actuaries use probability to make predictions about events. \u00a0An actuary that works for a car insurance company would, for example, be interested in how likely a 17-year-old male would be to get in a car accident. They would use data from past events to make predictions about future events using the characteristics of probabilities, then use this information to calculate an insurance rate.<\/p>\n<p><strong>Probabilities are between zero and one, inclusive<\/strong>\u00a0(that is, zero and one and all numbers between these values).<\/p>\n<ul>\n<li><em>P<\/em>(<em>A<\/em>) = 0 means the event\u00a0<em>A<\/em>\u00a0can never happen.<\/li>\n<li><em>P<\/em>(<em>A<\/em>) = 1 means the event\u00a0<em>A<\/em>\u00a0always happens.<\/li>\n<li><em>P<\/em>(<em>A<\/em>) = 0.5 means the event\u00a0<em>A<\/em>\u00a0is equally likely to occur or not to occur.<\/li>\n<\/ul>\n<p><strong>Equally likely\u00a0<\/strong>means that each outcome of an experiment occurs with equal probability. For example, if you toss a\u00a0<strong>fair<\/strong>, six-sided die, each face (1, 2, 3, 4, 5, or 6) is as likely to occur as any other face. If you toss a fair coin, a Head (<em>H<\/em>) and a Tail (<em>T<\/em>) are equally likely to occur. If you randomly guess the answer to a true\/false question on an exam, you are equally likely to select a correct answer or an incorrect answer.<\/p>\n<p><strong>To calculate the probability of an event\u00a0<em>A<\/em>\u00a0when all outcomes in the sample space are equally likely<\/strong>, count the number of outcomes for event\u00a0<em>A<\/em>\u00a0and divide by the total number of outcomes in the sample space. For example, if you toss a fair dime and a fair nickel, the sample space is {<em>HH<\/em>,\u00a0<em>TH<\/em>,\u00a0<em>HT<\/em>,\u00a0<em>TT<\/em>} where\u00a0<em>T<\/em>\u00a0= tails and\u00a0<em>H<\/em>\u00a0= heads. The sample space has four outcomes.\u00a0<em>A<\/em>\u00a0= getting one head. There are two outcomes that meet this condition {<em>HT<\/em>,\u00a0<em>TH<\/em>}, so P(A) = 2\/4 = 0.5.<\/p>\n<p>Suppose you roll one fair six-sided die, with the numbers {1, 2, 3, 4, 5, 6} on its faces. Let event\u00a0<em>E<\/em>\u00a0= rolling a number that is at least five. There are two outcomes {5, 6}.\u00a0<span id=\"MathJax-Element-2-Frame\" class=\"mjx-chtml MathJax_CHTML\" style=\"font-family: proxima-nova, sans-serif; padding: 1px 0px; margin: 0px; font-size: 21.8px; vertical-align: baseline; background: transparent; border: 0px; line-height: 0; text-indent: 0px; text-align: left; font-style: normal; font-weight: normal; letter-spacing: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px;\" role=\"presentation\"><span id=\"MJXc-Node-27\" class=\"mjx-math\"><span id=\"MJXc-Node-28\" class=\"mjx-mrow\"><span id=\"MJXc-Node-29\" class=\"mjx-mstyle\"><span id=\"MJXc-Node-30\" class=\"mjx-mrow\"><span id=\"MJXc-Node-31\" class=\"mjx-texatom\"><span id=\"MJXc-Node-32\" class=\"mjx-mrow\"><span id=\"MJXc-Node-33\" class=\"mjx-mi\"><span class=\"mjx-char MJXc-TeX-math-I\">P<\/span><\/span><\/span><\/span><span id=\"MJXc-Node-34\" class=\"mjx-texatom\"><span id=\"MJXc-Node-35\" class=\"mjx-mrow\"><span id=\"MJXc-Node-36\" class=\"mjx-mo\"><span class=\"mjx-char MJXc-TeX-main-R\">(<\/span><\/span><span id=\"MJXc-Node-37\" class=\"mjx-texatom\"><span id=\"MJXc-Node-38\" class=\"mjx-mrow\"><span id=\"MJXc-Node-39\" class=\"mjx-mi\"><span class=\"mjx-char MJXc-TeX-math-I\">E<\/span><\/span><\/span><\/span><span id=\"MJXc-Node-40\" class=\"mjx-mo\"><span class=\"mjx-char MJXc-TeX-main-R\">)<\/span><\/span><\/span><\/span><span id=\"MJXc-Node-41\" class=\"mjx-mo MJXc-space3\"><span class=\"mjx-char MJXc-TeX-main-R\">=<\/span><\/span><span id=\"MJXc-Node-42\" class=\"mjx-mfrac MJXc-space3\"><span class=\"mjx-box MJXc-stacked\"><span class=\"mjx-numerator\"><span id=\"MJXc-Node-43\" class=\"mjx-texatom\"><span id=\"MJXc-Node-44\" class=\"mjx-mrow\"><span id=\"MJXc-Node-45\" class=\"mjx-mn\"><span class=\"mjx-char MJXc-TeX-main-R\">2\/<\/span><\/span><\/span><\/span><\/span><span class=\"mjx-denominator\"><span id=\"MJXc-Node-46\" class=\"mjx-texatom\"><span id=\"MJXc-Node-47\" class=\"mjx-mrow\"><span id=\"MJXc-Node-48\" class=\"mjx-mn\"><span class=\"mjx-char MJXc-TeX-main-R\">6.\u00a0 This simplifies to P(E) = 1\/3<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n<p>This important characteristic of probability experiments is known as the\u00a0<strong>law of large numbers<\/strong>\u00a0which states that as the number of repetitions of an experiment is increased, the empirical probability of the experiment tends to become closer and closer to the theoretical probability. Even though the outcomes do not happen according to any set pattern or order, overall, the long-term empirical probability will approach the theoretical probability.<\/p>\n<div class=\"textbox\">\n<h3>Basic Probability<\/h3>\n<p>Given that all outcomes are equally likely, we can compute the probability of an event <em>E<\/em> using this formula:<\/p>\n<p style=\"text-align: center;\">[latex]P(E)=\\frac{\\text{Number of favorable outcomes corresponding to the event E}}{\\text{Number of equally-likely outcomes (sample space)}}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>examples<\/h3>\n<p>If we roll a 6-sided die, calculate<\/p>\n<ol>\n<li>P(rolling a 1)<\/li>\n<li>P(rolling a number bigger than 4)<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q417705\">Show Solution<\/span><\/p>\n<div id=\"q417705\" class=\"hidden-answer\" style=\"display: none\">\n<p>Recall that the sample space is {1,2,3,4,5,6}<\/p>\n<ol>\n<li>There is one outcome corresponding to \u201crolling a 1,\u201d so the probability is\u00a0 [latex]\\frac{2}{6}[\/latex]\u00a0[latex]\\frac{1}{6}[\/latex]<\/li>\n<li>There are two outcomes bigger than a 4, so the probability is [latex]\\frac{2}{6}=\\frac{1}{3}[\/latex]<\/li>\n<\/ol>\n<p>Probabilities are essentially fractions, and can be simplified to lower terms like fractions.<\/p>\n<\/div>\n<\/div>\n<p>This video describes this example and the previous one in detail.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Basics of Probability - events and outcomes\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/37P01dt0zsE?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>Let&#8217;s say you have a bag with 20 cherries, 14 sweet and 6 sour. If you pick a cherry at random, what is the probability that it will be sweet?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q9680\">Show Solution<\/span><\/p>\n<div id=\"q9680\" class=\"hidden-answer\" style=\"display: none\">\n<p>There are 20 possible cherries that could be picked, so the number of possible outcomes is 20. Of these 20 possible outcomes, 14 are favorable (sweet), so the probability that the cherry will be sweet is [latex]\\frac{14}{20}=\\frac{7}{10}[\/latex].<br \/>\nThere is one potential complication to this example, however. It must be assumed that the probability of picking any of the cherries is the same as the probability of picking any other. This wouldn&#8217;t be true if (let us imagine) the sweet cherries are smaller than the sour ones. (The sour cherries would come to hand more readily when you sampled from the bag.) Let us keep in mind, therefore, that when we assess probabilities in terms of the ratio of favorable to all potential cases, we rely heavily on the assumption of equal probability for all outcomes.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>Cards<\/h3>\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5548\/2016\/10\/04191241\/Deck-of-Cards.png\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-5985\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5548\/2016\/10\/04191241\/Deck-of-Cards-223x300.png\" alt=\"\" width=\"223\" height=\"300\" \/><\/a><\/p>\n<p>A standard deck of 52 playing cards consists of four <strong>suits<\/strong> (hearts, spades, diamonds and clubs). Spades and clubs are black while hearts and diamonds are red. Each suit contains 13 cards, each of a different <strong>rank<\/strong>: an Ace (which in many games functions as both a low card and a high card), cards numbered 2 through 10, a Jack, a Queen and a King.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>Compute the probability of randomly drawing one card from a deck and getting an Ace.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q652517\">Show Solution<\/span><\/p>\n<div id=\"q652517\" class=\"hidden-answer\" style=\"display: none\">\n<p>There are 52 cards in the deck and 4 Aces so\u00a0[latex]P(Ace)=\\frac{4}{52}=\\frac{1}{13}\\approx 0.0769[\/latex]<\/p>\n<p>Notice that the smallest possible probability is 0 \u2013 if there are no outcomes that correspond with the event. The largest possible probability is 1 \u2013 if all possible outcomes correspond with the event.<\/p>\n<\/div>\n<\/div>\n<p>This video demonstrates both this example and the previous cherry example\u00a0on the page.\u00a0 Note that sometimes a probability is interpreted as a percent.\u00a0 In our class we will give the probability as a simplified fraction or a number rounded to 3 decimal digits.\u00a0 (i.e. 0.729)<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Basic Probabilities\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/EBqj_R3dzd4?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm7130\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=7130&theme=oea&iframe_resize_id=ohm7130&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\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-1257\">\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>Introduction and outcomes. <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\">Public domain content<\/div><ul class=\"citation-list\"><li> A car crash on Jagtvej, a road in Copenhagen, Denmark. <strong>Authored by<\/strong>: By Thue (Own work) . <strong>Provided by<\/strong>: Wikimedia Commons. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/commons.wikimedia.org\/wiki\/File%3ACar_crash_1.jpg,%20https:\/\/upload.wikimedia.org\/wikipedia\/commons\/e\/e1\/Car_crash_1.jpg\">https:\/\/commons.wikimedia.org\/wiki\/File%3ACar_crash_1.jpg,%20https:\/\/upload.wikimedia.org\/wikipedia\/commons\/e\/e1\/Car_crash_1.jpg<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/about\/pdm\">Public Domain: No Known Copyright<\/a><\/em><\/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":3,"template":"","meta":{"_candela_citation":"[{\"type\":\"pd\",\"description\":\" A car crash on Jagtvej, a road in Copenhagen, Denmark\",\"author\":\"By Thue (Own work) \",\"organization\":\"Wikimedia Commons\",\"url\":\"https:\/\/commons.wikimedia.org\/wiki\/File%3ACar_crash_1.jpg, https:\/\/upload.wikimedia.org\/wikipedia\/commons\/e\/e1\/Car_crash_1.jpg\",\"project\":\"\",\"license\":\"pd\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Introduction and outcomes\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"41231e5e-611e-4e7f-a4b6-4d66ee106908","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1257","chapter","type-chapter","status-publish","hentry"],"part":329,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/frontrange-mathforliberalartscorequisite1\/wp-json\/pressbooks\/v2\/chapters\/1257","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/frontrange-mathforliberalartscorequisite1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/frontrange-mathforliberalartscorequisite1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/frontrange-mathforliberalartscorequisite1\/wp-json\/wp\/v2\/users\/21"}],"version-history":[{"count":36,"href":"https:\/\/courses.lumenlearning.com\/frontrange-mathforliberalartscorequisite1\/wp-json\/pressbooks\/v2\/chapters\/1257\/revisions"}],"predecessor-version":[{"id":7036,"href":"https:\/\/courses.lumenlearning.com\/frontrange-mathforliberalartscorequisite1\/wp-json\/pressbooks\/v2\/chapters\/1257\/revisions\/7036"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/frontrange-mathforliberalartscorequisite1\/wp-json\/pressbooks\/v2\/parts\/329"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/frontrange-mathforliberalartscorequisite1\/wp-json\/pressbooks\/v2\/chapters\/1257\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/frontrange-mathforliberalartscorequisite1\/wp-json\/wp\/v2\/media?parent=1257"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/frontrange-mathforliberalartscorequisite1\/wp-json\/pressbooks\/v2\/chapter-type?post=1257"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/frontrange-mathforliberalartscorequisite1\/wp-json\/wp\/v2\/contributor?post=1257"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/frontrange-mathforliberalartscorequisite1\/wp-json\/wp\/v2\/license?post=1257"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}