{"id":1009,"date":"2023-06-22T01:39:09","date_gmt":"2023-06-22T01:39:09","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/introstatstest\/chapter\/more-conditional-probabilities-dig-deeper\/"},"modified":"2024-02-10T00:00:49","modified_gmt":"2024-02-10T00:00:49","slug":"more-conditional-probabilities-fresh-take","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/introstatstest\/chapter\/more-conditional-probabilities-fresh-take\/","title":{"raw":"Bayes' Theorem: Fresh Take","rendered":"Bayes&#8217; Theorem: Fresh Take"},"content":{"raw":"<section class=\"textbox learningGoals\">\r\n<ul>\r\n\t<li>Understand conditional probability and Bayes' theorem<\/li>\r\n<\/ul>\r\n<\/section>\r\n<section class=\"textbox recall\"><strong>Conditional probability<\/strong> is defined as\u00a0the likelihood of an event occurring, based on the occurrence of a given event.\r\n\r\n<p style=\"text-align: center;\">[latex] P(A | B) = \\dfrac{P(A \\text{ and } B)}{P(B)} [\/latex]<\/p>\r\n<p style=\"text-align: center;\">OR<\/p>\r\n<p style=\"text-align: center;\">[latex] P(A \\text{ and } B) = P(A | B) \\times P(B) [\/latex]<\/p>\r\n<\/section>\r\n<section class=\"textbox example\">Recall the example below: An urn has three red marbles and eight blue marbles in it. Draw two marbles, one at a time, this time without replacement, from the urn. Following is a tree diagram for this situation.<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\nFind\/calculate the following probabilities using the tree diagram:<strong>(a) <\/strong>[latex]P(R \\text{ on the 2nd} | B \\text{ on the 1st})[\/latex]<strong><br \/>\r\n<\/strong>\r\n<p><strong>(b) <\/strong>[latex]P(R \\text{ on the 2nd} | R \\text{ on the 1st})[\/latex]<\/p>\r\n<p><strong>(c)<\/strong> [latex]P(\\text{2nd one is }R)[\/latex]<\/p>\r\n<p><strong>(d) <\/strong>[latex]P(B \\text{ on the 1st} | R \\text{ on the 2nd})[\/latex]<\/p>\r\n<p><strong>(e) <\/strong>[latex]P(R \\text{ on the 1st} | R \\text{ on the 2nd})[\/latex]<\/p>\r\n<p>[reveal-answer q=\"172617\"]Show Answer[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"172617\"]<\/p>\r\n<p><strong>(a) <\/strong>[latex]P(R \\text{ on the 2nd} | B \\text{ on the 1st}) = \\dfrac{3}{10}[\/latex]<\/p>\r\n<p><strong>(b) <\/strong>[latex]P(R \\text{ on the 2nd} | R \\text{ on the 1st}) = \\dfrac{2}{10}[\/latex]<\/p>\r\n<p><strong>(c) <\/strong>[latex]P(\\text{2nd one is }R) = P(BR) + P(RR) = \\dfrac{24}{110} + \\dfrac{6}{110} = \\dfrac{30}{110}[\/latex]<\/p>\r\n<p><strong>(d)\u00a0<\/strong>[latex]P(B \\text{ on the 1st} | R \\text{ on the 2nd})[\/latex]<\/p>\r\n<p>Notice that for this question, the GIVEN information is on the second draw\/stage. Remember that the condition gave us a new sample space.<\/p>\r\n<p>This means that our new sample space are the events BR and RR only. The probability distribution table is as follows:<\/p>\r\n<table style=\"border-collapse: collapse; width: 56.7114%; height: 143px;\" border=\"0\">\r\n<tbody>\r\n<tr>\r\n<td style=\"width: 32.5444%;\"><strong>Events<\/strong><\/td>\r\n<td style=\"width: 67.4556%;\"><strong>Probability<\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 32.5444%;\">BR<\/td>\r\n<td style=\"width: 67.4556%;\">[latex]\\dfrac{24}{110}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 32.5444%;\">RR<\/td>\r\n<td style=\"width: 67.4556%;\">[latex]\\dfrac{6}{110}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 32.5444%;\"><strong>Total<\/strong><\/td>\r\n<td style=\"width: 67.4556%;\"><strong>[latex]\\dfrac{30}{110}[\/latex]<\/strong><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p>So, [latex]P(B \\text{ on the 1st} | R \\text{ on the 2nd}) = \\dfrac{P(BR)}{P(\\text{2nd one is }R)} = \\dfrac{\\dfrac{24}{110}}{\\dfrac{30}{110}}=\\dfrac{24}{30}[\/latex]<\/p>\r\n<p><strong>(e) <\/strong>[latex]P(R \\text{ on the 1st} | R \\text{ on the 2nd})[\/latex]<\/p>\r\n<p>This has similar ideas as part (d). Using the table on part (d), we have:<\/p>\r\n<p>[latex]P(R \\text{ on the 1st} | R \\text{ on the 2nd}) = \\dfrac{P(RR)}{P(\\text{2nd one is }R)} = \\dfrac{\\dfrac{6}{110}}{\\dfrac{30}{110}}=\\dfrac{6}{30}[\/latex]<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<p>Part (d) and (e) in the example above is also known as <strong>Bayes' Theorem Probability.<\/strong><\/p>\r\n<h4>Bayes' Theorem<\/h4>\r\n<p>Bayes' theorem or Bayes' rule, named after Thomas Bayes[footnote]https:\/\/en.wikipedia.org\/wiki\/Thomas_Bayes[\/footnote], describes the probability of an event, based on the occurrence of another event.<\/p>\r\n<section class=\"textbox keyTakeaway\">\r\n<h3>Bayes' Theorem<\/h3>\r\n<p>For events [latex]A[\/latex] and [latex]B[\/latex] and that [latex]P(B) \\ne 0[\/latex], the probability of event [latex]A[\/latex] occurring given that event [latex]B[\/latex] is true is:<\/p>\r\n<p style=\"text-align: center;\">[latex]P(A | B) = \\dfrac{P(A \\text{ and } B)}{P(B)} = \\dfrac{P(B|A) \\times P(A)}{P(B)}[\/latex]<\/p>\r\n<\/section>\r\n<section class=\"textbox example\">This tree diagram shows the tossing of an unfair coin followed by drawing one bead from a cup containing three red (R), four yellow (Y) and five blue (B) beads.<br \/>\r\n<br \/>\r\n<img class=\"aligncenter wp-image-6151 size-medium\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/27\/2023\/06\/22013908\/1ec4dade09e0f3edb3a5540aefa69e56f9b32d73.jpeg\" alt=\"This is a tree diagram with branches showing frequencies of drawing a colored bead after an unfair coin toss. The first branch shows two lines: 2\/3H and 1\/3T. The second branch has a set of three lines (3\/12R, 4\/12Y, and 5\/12B for H and 3\/12R, 4\/12Y, and 5\/12B for T).) for each line of the first branch. Multiply along each line to find 56\/110BB, 24\/110BR, 24\/110RB, and 6\/110RR.\" width=\"168\" height=\"300\" \/><br \/>\r\nFind the probability that a flipped coin shows a head, given that the bead drawn is blue.<br \/>\r\n[reveal-answer q=\"380524\"]Show Answer[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"380524\"]Mathematically, the question can be rewritten as: [latex]P(H | B)[\/latex]Based on the formula above:\r\n\r\n<p>[latex]P(H | B) = \\dfrac{P(H \\text{ and } B)}{P(B)} = \\dfrac{P(B|H) \\times P(H)}{P(B)}=\\dfrac{\\dfrac{5}{12} \\times \\dfrac{2}{3}}{P(B)} = \\dfrac{\\dfrac{10}{36}}{P(B)}[\/latex]<\/p>\r\n<p>How do we find [latex]P(B)[\/latex]?<\/p>\r\n<p>Notice that there are two different possibilities to get [latex]B[\/latex]. They are HB and TB.<br \/>\r\nSo:<\/p>\r\n<p>[latex]P(B) = P(HB) + P(TB) = (P(B|H) \\times P(H)) + (P(B|T) \\times P(T)) [\/latex]<\/p>\r\n<p>[latex]P(B) = (\\dfrac{5}{12} \\times \\dfrac{2}{3}) + (\\dfrac{5}{12} \\times \\dfrac{1}{3}) = \\dfrac{10}{36} + \\dfrac{5}{36} = \\dfrac{15}{36}[\/latex]<\/p>\r\n<p>So, [latex]P(H | B) = \\dfrac{\\dfrac{10}{36}}{P(B)} = \\dfrac{\\dfrac{10}{36}}{\\dfrac{15}{36}} = \\dfrac{10}{15}[\/latex]<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<p>A box contains 60 balls containing a variety of a colored balls, but all we know is 40% are red. In addition to colors, the balls are also made up of two sizes: small and large. Suppose we know that 50% of the red balls and 30% of the non-red balls are small. A person draws a ball at random from the box and notices it is a small ball. Let [latex] R = [\/latex] be the event the ball is red and [latex] S = [\/latex] the event the ball is small. What is the probability that it is red?<\/p>\r\n<p><br \/>\r\n[reveal-answer q=\"589862\"]Show Answer[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"589862\"]By Bayes' Theorem, [latex]P(R|S) = \\frac{0.40 \\times 0.50}{(0.40 \\times 0.50) + (0.60 \\times 0.30)} = \\frac{0.20}{0.20+0.18} \\approx 0.5263[\/latex][\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox youChoose\"><img class=\"wp-image-4578 aligncenter\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/10\/2022\/10\/13214944\/ChooseDatasetImage.png\" alt=\"\" width=\"680\" height=\"116\" \/>[choosedataset divId=\"tnh-choose-dataset\" title=\"Calculate Probability\" label=\"\" default=\"Choose a Dataset\"]<br \/>\r\n[datasetoption]<br \/>\r\n[displayname]Spam Mail[\/displayname]<br \/>\r\n[ohmid]13838[\/ohmid]<br \/>\r\n[\/datasetoption]<br \/>\r\n[datasetoption]<br \/>\r\n[displayname]Small or Large Car?[\/displayname]<br \/>\r\n[ohmid]13909[\/ohmid]<br \/>\r\n[\/datasetoption]<br \/>\r\n[datasetoption]<br \/>\r\n[displayname]Night and Day Shift[\/displayname]<br \/>\r\n[ohmid]13910[\/ohmid]<br \/>\r\n[\/datasetoption][\/choosedataset]<\/section>","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li>Understand conditional probability and Bayes&#8217; theorem<\/li>\n<\/ul>\n<\/section>\n<section class=\"textbox recall\"><strong>Conditional probability<\/strong> is defined as\u00a0the likelihood of an event occurring, based on the occurrence of a given event.<\/p>\n<p style=\"text-align: center;\">[latex]P(A | B) = \\dfrac{P(A \\text{ and } B)}{P(B)}[\/latex]<\/p>\n<p style=\"text-align: center;\">OR<\/p>\n<p style=\"text-align: center;\">[latex]P(A \\text{ and } B) = P(A | B) \\times P(B)[\/latex]<\/p>\n<\/section>\n<section class=\"textbox example\">Recall the example below: An urn has three red marbles and eight blue marbles in it. Draw two marbles, one at a time, this time without replacement, from the urn. Following is a tree diagram for this situation.<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 \/>\nFind\/calculate the following probabilities using the tree diagram:<strong>(a) <\/strong>[latex]P(R \\text{ on the 2nd} | B \\text{ on the 1st})[\/latex]<strong><br \/>\n<\/strong><\/p>\n<p><strong>(b) <\/strong>[latex]P(R \\text{ on the 2nd} | R \\text{ on the 1st})[\/latex]<\/p>\n<p><strong>(c)<\/strong> [latex]P(\\text{2nd one is }R)[\/latex]<\/p>\n<p><strong>(d) <\/strong>[latex]P(B \\text{ on the 1st} | R \\text{ on the 2nd})[\/latex]<\/p>\n<p><strong>(e) <\/strong>[latex]P(R \\text{ on the 1st} | R \\text{ on the 2nd})[\/latex]<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q172617\">Show Answer<\/button><\/p>\n<div id=\"q172617\" class=\"hidden-answer\" style=\"display: none\">\n<p><strong>(a) <\/strong>[latex]P(R \\text{ on the 2nd} | B \\text{ on the 1st}) = \\dfrac{3}{10}[\/latex]<\/p>\n<p><strong>(b) <\/strong>[latex]P(R \\text{ on the 2nd} | R \\text{ on the 1st}) = \\dfrac{2}{10}[\/latex]<\/p>\n<p><strong>(c) <\/strong>[latex]P(\\text{2nd one is }R) = P(BR) + P(RR) = \\dfrac{24}{110} + \\dfrac{6}{110} = \\dfrac{30}{110}[\/latex]<\/p>\n<p><strong>(d)\u00a0<\/strong>[latex]P(B \\text{ on the 1st} | R \\text{ on the 2nd})[\/latex]<\/p>\n<p>Notice that for this question, the GIVEN information is on the second draw\/stage. Remember that the condition gave us a new sample space.<\/p>\n<p>This means that our new sample space are the events BR and RR only. The probability distribution table is as follows:<\/p>\n<table style=\"border-collapse: collapse; width: 56.7114%; height: 143px;\">\n<tbody>\n<tr>\n<td style=\"width: 32.5444%;\"><strong>Events<\/strong><\/td>\n<td style=\"width: 67.4556%;\"><strong>Probability<\/strong><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 32.5444%;\">BR<\/td>\n<td style=\"width: 67.4556%;\">[latex]\\dfrac{24}{110}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 32.5444%;\">RR<\/td>\n<td style=\"width: 67.4556%;\">[latex]\\dfrac{6}{110}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 32.5444%;\"><strong>Total<\/strong><\/td>\n<td style=\"width: 67.4556%;\"><strong>[latex]\\dfrac{30}{110}[\/latex]<\/strong><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>So, [latex]P(B \\text{ on the 1st} | R \\text{ on the 2nd}) = \\dfrac{P(BR)}{P(\\text{2nd one is }R)} = \\dfrac{\\dfrac{24}{110}}{\\dfrac{30}{110}}=\\dfrac{24}{30}[\/latex]<\/p>\n<p><strong>(e) <\/strong>[latex]P(R \\text{ on the 1st} | R \\text{ on the 2nd})[\/latex]<\/p>\n<p>This has similar ideas as part (d). Using the table on part (d), we have:<\/p>\n<p>[latex]P(R \\text{ on the 1st} | R \\text{ on the 2nd}) = \\dfrac{P(RR)}{P(\\text{2nd one is }R)} = \\dfrac{\\dfrac{6}{110}}{\\dfrac{30}{110}}=\\dfrac{6}{30}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<p>Part (d) and (e) in the example above is also known as <strong>Bayes&#8217; Theorem Probability.<\/strong><\/p>\n<h4>Bayes&#8217; Theorem<\/h4>\n<p>Bayes&#8217; theorem or Bayes&#8217; rule, named after Thomas Bayes<a class=\"footnote\" title=\"https:\/\/en.wikipedia.org\/wiki\/Thomas_Bayes\" id=\"return-footnote-1009-1\" href=\"#footnote-1009-1\" aria-label=\"Footnote 1\"><sup class=\"footnote\">[1]<\/sup><\/a>, describes the probability of an event, based on the occurrence of another event.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>Bayes&#8217; Theorem<\/h3>\n<p>For events [latex]A[\/latex] and [latex]B[\/latex] and that [latex]P(B) \\ne 0[\/latex], the probability of event [latex]A[\/latex] occurring given that event [latex]B[\/latex] is true is:<\/p>\n<p style=\"text-align: center;\">[latex]P(A | B) = \\dfrac{P(A \\text{ and } B)}{P(B)} = \\dfrac{P(B|A) \\times P(A)}{P(B)}[\/latex]<\/p>\n<\/section>\n<section class=\"textbox example\">This tree diagram shows the tossing of an unfair coin followed by drawing one bead from a cup containing three red (R), four yellow (Y) and five blue (B) beads.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-6151 size-medium\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/27\/2023\/06\/22013908\/1ec4dade09e0f3edb3a5540aefa69e56f9b32d73.jpeg\" alt=\"This is a tree diagram with branches showing frequencies of drawing a colored bead after an unfair coin toss. The first branch shows two lines: 2\/3H and 1\/3T. The second branch has a set of three lines (3\/12R, 4\/12Y, and 5\/12B for H and 3\/12R, 4\/12Y, and 5\/12B for T).) for each line of the first branch. Multiply along each line to find 56\/110BB, 24\/110BR, 24\/110RB, and 6\/110RR.\" width=\"168\" height=\"300\" \/><br \/>\nFind the probability that a flipped coin shows a head, given that the bead drawn is blue.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q380524\">Show Answer<\/button><\/p>\n<div id=\"q380524\" class=\"hidden-answer\" style=\"display: none\">Mathematically, the question can be rewritten as: [latex]P(H | B)[\/latex]Based on the formula above:<\/p>\n<p>[latex]P(H | B) = \\dfrac{P(H \\text{ and } B)}{P(B)} = \\dfrac{P(B|H) \\times P(H)}{P(B)}=\\dfrac{\\dfrac{5}{12} \\times \\dfrac{2}{3}}{P(B)} = \\dfrac{\\dfrac{10}{36}}{P(B)}[\/latex]<\/p>\n<p>How do we find [latex]P(B)[\/latex]?<\/p>\n<p>Notice that there are two different possibilities to get [latex]B[\/latex]. They are HB and TB.<br \/>\nSo:<\/p>\n<p>[latex]P(B) = P(HB) + P(TB) = (P(B|H) \\times P(H)) + (P(B|T) \\times P(T))[\/latex]<br \/>\n[latex]P(B) = (\\dfrac{5}{12} \\times \\dfrac{2}{3}) + (\\dfrac{5}{12} \\times \\dfrac{1}{3}) = \\dfrac{10}{36} + \\dfrac{5}{36} = \\dfrac{15}{36}[\/latex]<\/p>\n<p>So, [latex]P(H | B) = \\dfrac{\\dfrac{10}{36}}{P(B)} = \\dfrac{\\dfrac{10}{36}}{\\dfrac{15}{36}} = \\dfrac{10}{15}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p>A box contains 60 balls containing a variety of a colored balls, but all we know is 40% are red. In addition to colors, the balls are also made up of two sizes: small and large. Suppose we know that 50% of the red balls and 30% of the non-red balls are small. A person draws a ball at random from the box and notices it is a small ball. Let [latex]R =[\/latex] be the event the ball is red and [latex]S =[\/latex] the event the ball is small. What is the probability that it is red?<\/p>\n<div class=\"wp-nocaption \"><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q589862\">Show Answer<\/button><\/p>\n<div id=\"q589862\" class=\"hidden-answer\" style=\"display: none\">By Bayes&#8217; Theorem, [latex]P(R|S) = \\frac{0.40 \\times 0.50}{(0.40 \\times 0.50) + (0.60 \\times 0.30)} = \\frac{0.20}{0.20+0.18} \\approx 0.5263[\/latex]<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox youChoose\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-4578 aligncenter\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/10\/2022\/10\/13214944\/ChooseDatasetImage.png\" alt=\"\" width=\"680\" height=\"116\" \/><\/p>\n<div id=\"tnh-choose-dataset\" class=\"chooseDataset\">\n<h3>Calculate Probability<\/h3>\n<form><select name=\"dataset\"><option value=\"\">Choose a Dataset<\/option><option value=\"13838\">Spam Mail<\/option><option value=\"13909\">Small or Large Car?<\/option><option value=\"13910\">Night and Day Shift<\/option><\/select><\/form>\n<div class=\"ohmContainer\"><\/div>\n<\/p><\/div>\n<\/section>\n<hr class=\"before-footnotes clear\" \/><div class=\"footnotes\"><ol><li id=\"footnote-1009-1\">https:\/\/en.wikipedia.org\/wiki\/Thomas_Bayes <a href=\"#return-footnote-1009-1\" class=\"return-footnote\" aria-label=\"Return to footnote 1\">&crarr;<\/a><\/li><\/ol><\/div>","protected":false},"author":8,"menu_order":19,"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":"fresh_take","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":[{"divId":"tnh-choose-dataset","title":"Calculate Probability","label":"","default":"Choose a Dataset","try_it_collection":[{"displayName":"Spam Mail","value":"13838"},{"displayName":"Small or Large Car?","value":"13909"},{"displayName":"Night and Day Shift","value":"13910"}]}],"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/chapters\/1009"}],"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\/8"}],"version-history":[{"count":17,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/chapters\/1009\/revisions"}],"predecessor-version":[{"id":5528,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/chapters\/1009\/revisions\/5528"}],"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\/1009\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/wp\/v2\/media?parent=1009"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/chapter-type?post=1009"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/wp\/v2\/contributor?post=1009"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/wp\/v2\/license?post=1009"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}