{"id":3386,"date":"2023-10-08T22:11:59","date_gmt":"2023-10-08T22:11:59","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/introstatstest\/?post_type=chapter&#038;p=3386"},"modified":"2026-04-27T15:10:42","modified_gmt":"2026-04-27T15:10:42","slug":"bayes-theorem-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/introstatstest\/chapter\/bayes-theorem-learn-it-2\/","title":{"raw":"Bayes' Theorem: Learn It 2","rendered":"Bayes&#8217; Theorem: Learn It 2"},"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<p>Representing Bayes' Theorem as a tree diagram can provide a clear and intuitive visualization of how conditional probabilities work, especially in situations involving diagnostic tests.<\/p>\r\n<p>The tree diagram breaks down complex scenarios into simpler, sequential events. This breakdown helps in understanding the flow of events and the associated probabilities at each step. Tree diagrams explicitly show the conditional relationships between events. In the context of Bayes' Theorem, it illustrates how the probability of a specific event (having a disease) changes based on the outcome of another event (a positive test result).<\/p>\r\n<p>Representing Bayes' Theorem as a tree diagram simplifies the understanding of complex conditional probability relationships, making it an effective tool for both learning and problem-solving.\u00a0<\/p>\r\n<section class=\"textbox recall\">\r\n<p>Consider two events [latex] A [\/latex] and [latex] B [\/latex] whose tree diagram is shown below.<\/p>\r\n<p><img class=\"aligncenter wp-image-3389 size-medium\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/27\/2023\/10\/08222401\/Bayes_Theorem_Probability_Tree-300x209.png\" alt=\"This is a tree diagram for events A and B. The first branch shows two lines: A and A'. The second branch has a set of 2 lines for each first branch line. Below A is B and B'. Below A' is B and B'.\" width=\"300\" height=\"209\" \/><\/p>\r\n<p>In the tree diagram, we observe that event [latex] B [\/latex] is conditioned on the result of event [latex] A [\/latex]. Using this tree, we can find [latex] P(B|A) [\/latex] by first following the path of [latex] A [\/latex] and then examining the probability for the branch that leads to event [latex] B [\/latex].<\/p>\r\n<p><img class=\"aligncenter wp-image-4265 size-full\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/27\/2023\/10\/01192529\/Bayes_Theorem_Probability_Tree-300x209-1.png\" alt=\"This is a tree diagram for events A and B. The first branch shows two lines: A and A'. The second branch has a set of 2 lines for each first branch line. Below A is B and B'. Below A' is B and B'. Between A and B P(B|A) is written in red.\" width=\"300\" height=\"209\" \/><\/p>\r\n<\/section>\r\n<p>In the context of a tree diagram, Bayes' Theorem addresses conditional probabilities at the second stage. For the tree diagram above, we are interested in finding the probability of an event [latex]A[\/latex] occurring given that event [latex]B[\/latex] has occurred. That is, [latex]P(A|B)[\/latex].<\/p>\r\n<section class=\"textbox recall\">\r\n<p><strong>Bayes'<\/strong> <strong>Theorem<\/strong> states that given events [latex]A[\/latex] and [latex]B[\/latex], then<\/p>\r\n<p style=\"text-align: center;\">[latex]P(A|B) = \\dfrac{P(A)\\times P(B|A)}{P(B)}= \\dfrac{P(A)\\times P(B|A)}{P(A)\\times P(B|A) + P(A')\\times P(B|A')}[\/latex]<\/p>\r\n<\/section>\r\n<p>When computing Bayes' Theorem using the probability tree, we can see that the numerator is the the path containing events [latex] A [\/latex] and [latex] B [\/latex]. The denominator is then the sum of all the paths that contain the event [latex] B [\/latex].\u00a0<\/p>\r\n<section class=\"textbox keyTakeaway\">\r\n<h3>Bayes' Theorem for tree diagrams<\/h3>\r\n<p>When given two events [latex]A[\/latex] and [latex]B[\/latex] whose tree diagram is:<\/p>\r\n<p>&nbsp;<\/p>\r\n<p><img class=\"aligncenter wp-image-3389 size-medium\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/27\/2023\/10\/08222401\/Bayes_Theorem_Probability_Tree-300x209.png\" alt=\"This is a tree diagram for events A and B. The first branch shows two lines: A and A'. The second branch has a set of 2 lines for each first branch line. Below A is B and B'. Below A' is B and B'.\" width=\"300\" height=\"209\" \/><\/p>\r\n<p>&nbsp;<\/p>\r\n<p>Then,<\/p>\r\n<p>&nbsp;<\/p>\r\n<p style=\"text-align: center;\">[latex]P(A|B) = \\dfrac{P(\\text{the path of events A and B})}{\\text{ sum of } P(\\text{paths containing event B})}[\/latex].<\/p>\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<p>Use the tree diagram to help compute the following probabilities.<\/p>\r\n<p><img class=\"aligncenter wp-image-3393 size-medium\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/27\/2023\/10\/08232231\/Bayes_Theorem_Probability_Tree_Example-300x187.png\" alt=\"This is a tree diagram for events A and B. The first branch shows two lines: A and A'. A is labeled 0.77 and A' is labeled 0.23. The second branch has a set of 2 lines for each first branch line. Below A is B and B', labeled 0.43 and 0.57 respectively. Below A' is B and B', labeled 0.61 and 0.39 respectively..\" width=\"300\" height=\"187\" \/><\/p>\r\n<p class=\"font-claude-response-body break-words whitespace-normal leading-[1.7]\"><strong>(a)<\/strong> [reveal-answer q=\"752068\"]Show Answer[\/reveal-answer] [hidden-answer a=\"752068\"][latex]P(A \\text{ AND } B) = 0.77 \\times 0.43 = 0.3311[\/latex][\/hidden-answer]<\/p>\r\n<p class=\"font-claude-response-body break-words whitespace-normal leading-[1.7]\"><strong>(b)<\/strong> [reveal-answer q=\"778545\"]Show Answer[\/reveal-answer] [hidden-answer a=\"778545\"][latex]P(B \\text{ GIVEN } A) = 0.43[\/latex][\/hidden-answer]<\/p>\r\n<p class=\"font-claude-response-body break-words whitespace-normal leading-[1.7]\"><strong>(c)<\/strong> [reveal-answer q=\"965551\"]Show Answer[\/reveal-answer] [hidden-answer a=\"965551\"][latex]P(B \\text{ GIVEN } A') = 0.61[\/latex][\/hidden-answer]<\/p>\r\n<p class=\"font-claude-response-body break-words whitespace-normal leading-[1.7]\"><strong>(d)<\/strong> [reveal-answer q=\"196619\"]Show Answer[\/reveal-answer] [hidden-answer a=\"196619\"][latex]P(A \\text{ GIVEN } B) = \\frac{0.77 \\times 0.43}{0.77 \\times 0.43 + 0.23 \\times 0.61} = \\frac{0.3311}{0.3311+0.1403} \\approx 0.7026[\/latex][\/hidden-answer]<\/p>\r\n<p class=\"font-claude-response-body break-words whitespace-normal leading-[1.7]\"><strong>(e)<\/strong> [reveal-answer q=\"346396\"]Show Answer[\/reveal-answer] [hidden-answer a=\"346396\"][latex]P(A' \\text{ GIVEN } B) = \\frac{0.23 \\times 0.61}{0.77 \\times 0.43 + 0.23 \\times 0.61} = \\frac{0.1403}{0.3311+0.1403} \\approx 0.2974[\/latex][\/hidden-answer]<\/p>\r\n<\/section>\r\n<p>In general, tree diagrams can can be helpful when we try and solve Bayes' Theorem problem by giving us a way to visualize the necessary paths needed to compute a conditional probability.<\/p>\r\n<section class=\"textbox example\">\r\n<p>Recall the example: Suppose a certain disease has an incidence rate of 0.1% (that is, it afflicts 0.1% of the population). A test has been devised to detect this disease. The test does not produce false negatives (that is, anyone who has the disease will test positive for it), but the false positive rate is 5% (that is, about 5% of people who take the test will test positive, even though they do not have the disease).<\/p>\r\n<p>Suppose a randomly selected person takes the test and tests positive.\u00a0 What is the probability that this person actually has the disease?<\/p>\r\n<p>The tree diagram associated to this scenario is as follows:<\/p>\r\n<p><img class=\"aligncenter wp-image-4273 size-full\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/27\/2023\/10\/01203832\/Tree-blank.png\" alt=\"Appropriate alternative text can be found in the description above.\" width=\"400\" height=\"316\" \/><\/p>\r\n<p>Using the formula of Bayes' Theorem:<\/p>\r\n<p>[latex]P(\\text{disease} | \\text{positive}) = \\frac{P(\\text{disease}) \\times P(\\text{positive} | \\text{disease})}{P(\\text{positive})} =\\frac{P(\\text{disease}) \\times P(\\text{positive} | \\text{disease})}{P(\\text{disease}) \\times P(\\text{positive} | \\text{disease}) + P(\\text{no disease}) \\times P(\\text{positive} | \\text{no disease})} = \\frac{(0.001)(1)}{(0.001)(1) + (0.999)(0.05)} \\approx 0.0196<br \/>\r\n[\/latex]<\/p>\r\n<p>Is it easier to find the [latex]P(\\text{disease} | \\text{positive}) [\/latex] using the tree diagram rather than the prompt?<\/p>\r\n<\/section>\r\n<section class=\"textbox tryIt\">\r\n<p>[ohm2_question hide_question_numbers=1]13832[\/ohm2_question]<\/p>\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li>Understand conditional probability and Bayes&#8217; theorem<\/li>\n<\/ul>\n<\/section>\n<p>Representing Bayes&#8217; Theorem as a tree diagram can provide a clear and intuitive visualization of how conditional probabilities work, especially in situations involving diagnostic tests.<\/p>\n<p>The tree diagram breaks down complex scenarios into simpler, sequential events. This breakdown helps in understanding the flow of events and the associated probabilities at each step. Tree diagrams explicitly show the conditional relationships between events. In the context of Bayes&#8217; Theorem, it illustrates how the probability of a specific event (having a disease) changes based on the outcome of another event (a positive test result).<\/p>\n<p>Representing Bayes&#8217; Theorem as a tree diagram simplifies the understanding of complex conditional probability relationships, making it an effective tool for both learning and problem-solving.\u00a0<\/p>\n<section class=\"textbox recall\">\n<p>Consider two events [latex]A[\/latex] and [latex]B[\/latex] whose tree diagram is shown below.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-3389 size-medium\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/27\/2023\/10\/08222401\/Bayes_Theorem_Probability_Tree-300x209.png\" alt=\"This is a tree diagram for events A and B. The first branch shows two lines: A and A'. The second branch has a set of 2 lines for each first branch line. Below A is B and B'. Below A' is B and B'.\" width=\"300\" height=\"209\" srcset=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/27\/2023\/10\/08222401\/Bayes_Theorem_Probability_Tree-300x209.png 300w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/27\/2023\/10\/08222401\/Bayes_Theorem_Probability_Tree-65x45.png 65w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/27\/2023\/10\/08222401\/Bayes_Theorem_Probability_Tree-225x157.png 225w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/27\/2023\/10\/08222401\/Bayes_Theorem_Probability_Tree-350x244.png 350w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/27\/2023\/10\/08222401\/Bayes_Theorem_Probability_Tree.png 720w\" sizes=\"(max-width: 300px) 100vw, 300px\" \/><\/p>\n<p>In the tree diagram, we observe that event [latex]B[\/latex] is conditioned on the result of event [latex]A[\/latex]. Using this tree, we can find [latex]P(B|A)[\/latex] by first following the path of [latex]A[\/latex] and then examining the probability for the branch that leads to event [latex]B[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-4265 size-full\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/27\/2023\/10\/01192529\/Bayes_Theorem_Probability_Tree-300x209-1.png\" alt=\"This is a tree diagram for events A and B. The first branch shows two lines: A and A'. The second branch has a set of 2 lines for each first branch line. Below A is B and B'. Below A' is B and B'. Between A and B P(B|A) is written in red.\" width=\"300\" height=\"209\" srcset=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/27\/2023\/10\/01192529\/Bayes_Theorem_Probability_Tree-300x209-1.png 300w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/27\/2023\/10\/01192529\/Bayes_Theorem_Probability_Tree-300x209-1-65x45.png 65w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/27\/2023\/10\/01192529\/Bayes_Theorem_Probability_Tree-300x209-1-225x157.png 225w\" sizes=\"(max-width: 300px) 100vw, 300px\" \/><\/p>\n<\/section>\n<p>In the context of a tree diagram, Bayes&#8217; Theorem addresses conditional probabilities at the second stage. For the tree diagram above, we are interested in finding the probability of an event [latex]A[\/latex] occurring given that event [latex]B[\/latex] has occurred. That is, [latex]P(A|B)[\/latex].<\/p>\n<section class=\"textbox recall\">\n<p><strong>Bayes&#8217;<\/strong> <strong>Theorem<\/strong> states that given events [latex]A[\/latex] and [latex]B[\/latex], then<\/p>\n<p style=\"text-align: center;\">[latex]P(A|B) = \\dfrac{P(A)\\times P(B|A)}{P(B)}= \\dfrac{P(A)\\times P(B|A)}{P(A)\\times P(B|A) + P(A')\\times P(B|A')}[\/latex]<\/p>\n<\/section>\n<p>When computing Bayes&#8217; Theorem using the probability tree, we can see that the numerator is the the path containing events [latex]A[\/latex] and [latex]B[\/latex]. The denominator is then the sum of all the paths that contain the event [latex]B[\/latex].\u00a0<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>Bayes&#8217; Theorem for tree diagrams<\/h3>\n<p>When given two events [latex]A[\/latex] and [latex]B[\/latex] whose tree diagram is:<\/p>\n<p>&nbsp;<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-3389 size-medium\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/27\/2023\/10\/08222401\/Bayes_Theorem_Probability_Tree-300x209.png\" alt=\"This is a tree diagram for events A and B. The first branch shows two lines: A and A'. The second branch has a set of 2 lines for each first branch line. Below A is B and B'. Below A' is B and B'.\" width=\"300\" height=\"209\" srcset=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/27\/2023\/10\/08222401\/Bayes_Theorem_Probability_Tree-300x209.png 300w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/27\/2023\/10\/08222401\/Bayes_Theorem_Probability_Tree-65x45.png 65w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/27\/2023\/10\/08222401\/Bayes_Theorem_Probability_Tree-225x157.png 225w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/27\/2023\/10\/08222401\/Bayes_Theorem_Probability_Tree-350x244.png 350w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/27\/2023\/10\/08222401\/Bayes_Theorem_Probability_Tree.png 720w\" sizes=\"(max-width: 300px) 100vw, 300px\" \/><\/p>\n<p>&nbsp;<\/p>\n<p>Then,<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: center;\">[latex]P(A|B) = \\dfrac{P(\\text{the path of events A and B})}{\\text{ sum of } P(\\text{paths containing event B})}[\/latex].<\/p>\n<\/section>\n<section class=\"textbox example\">\n<p>Use the tree diagram to help compute the following probabilities.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-3393 size-medium\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/27\/2023\/10\/08232231\/Bayes_Theorem_Probability_Tree_Example-300x187.png\" alt=\"This is a tree diagram for events A and B. The first branch shows two lines: A and A'. A is labeled 0.77 and A' is labeled 0.23. The second branch has a set of 2 lines for each first branch line. Below A is B and B', labeled 0.43 and 0.57 respectively. Below A' is B and B', labeled 0.61 and 0.39 respectively..\" width=\"300\" height=\"187\" srcset=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/27\/2023\/10\/08232231\/Bayes_Theorem_Probability_Tree_Example-300x187.png 300w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/27\/2023\/10\/08232231\/Bayes_Theorem_Probability_Tree_Example-768x479.png 768w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/27\/2023\/10\/08232231\/Bayes_Theorem_Probability_Tree_Example-65x41.png 65w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/27\/2023\/10\/08232231\/Bayes_Theorem_Probability_Tree_Example-225x140.png 225w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/27\/2023\/10\/08232231\/Bayes_Theorem_Probability_Tree_Example-350x218.png 350w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/27\/2023\/10\/08232231\/Bayes_Theorem_Probability_Tree_Example.png 779w\" sizes=\"(max-width: 300px) 100vw, 300px\" \/><\/p>\n<p class=\"font-claude-response-body break-words whitespace-normal leading-[1.7]\"><strong>(a)<\/strong> <\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q752068\">Show Answer<\/button> <\/p>\n<div id=\"q752068\" class=\"hidden-answer\" style=\"display: none\">[latex]P(A \\text{ AND } B) = 0.77 \\times 0.43 = 0.3311[\/latex]<\/div>\n<\/div>\n<p class=\"font-claude-response-body break-words whitespace-normal leading-[1.7]\"><strong>(b)<\/strong> <\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q778545\">Show Answer<\/button> <\/p>\n<div id=\"q778545\" class=\"hidden-answer\" style=\"display: none\">[latex]P(B \\text{ GIVEN } A) = 0.43[\/latex]<\/div>\n<\/div>\n<p class=\"font-claude-response-body break-words whitespace-normal leading-[1.7]\"><strong>(c)<\/strong> <\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q965551\">Show Answer<\/button> <\/p>\n<div id=\"q965551\" class=\"hidden-answer\" style=\"display: none\">[latex]P(B \\text{ GIVEN } A') = 0.61[\/latex]<\/div>\n<\/div>\n<p class=\"font-claude-response-body break-words whitespace-normal leading-[1.7]\"><strong>(d)<\/strong> <\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q196619\">Show Answer<\/button> <\/p>\n<div id=\"q196619\" class=\"hidden-answer\" style=\"display: none\">[latex]P(A \\text{ GIVEN } B) = \\frac{0.77 \\times 0.43}{0.77 \\times 0.43 + 0.23 \\times 0.61} = \\frac{0.3311}{0.3311+0.1403} \\approx 0.7026[\/latex]<\/div>\n<\/div>\n<p class=\"font-claude-response-body break-words whitespace-normal leading-[1.7]\"><strong>(e)<\/strong> <\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q346396\">Show Answer<\/button> <\/p>\n<div id=\"q346396\" class=\"hidden-answer\" style=\"display: none\">[latex]P(A' \\text{ GIVEN } B) = \\frac{0.23 \\times 0.61}{0.77 \\times 0.43 + 0.23 \\times 0.61} = \\frac{0.1403}{0.3311+0.1403} \\approx 0.2974[\/latex]<\/div>\n<\/div>\n<\/section>\n<p>In general, tree diagrams can can be helpful when we try and solve Bayes&#8217; Theorem problem by giving us a way to visualize the necessary paths needed to compute a conditional probability.<\/p>\n<section class=\"textbox example\">\n<p>Recall the example: Suppose a certain disease has an incidence rate of 0.1% (that is, it afflicts 0.1% of the population). A test has been devised to detect this disease. The test does not produce false negatives (that is, anyone who has the disease will test positive for it), but the false positive rate is 5% (that is, about 5% of people who take the test will test positive, even though they do not have the disease).<\/p>\n<p>Suppose a randomly selected person takes the test and tests positive.\u00a0 What is the probability that this person actually has the disease?<\/p>\n<p>The tree diagram associated to this scenario is as follows:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-4273 size-full\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/27\/2023\/10\/01203832\/Tree-blank.png\" alt=\"Appropriate alternative text can be found in the description above.\" width=\"400\" height=\"316\" srcset=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/27\/2023\/10\/01203832\/Tree-blank.png 400w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/27\/2023\/10\/01203832\/Tree-blank-300x237.png 300w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/27\/2023\/10\/01203832\/Tree-blank-65x51.png 65w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/27\/2023\/10\/01203832\/Tree-blank-225x178.png 225w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/27\/2023\/10\/01203832\/Tree-blank-350x277.png 350w\" sizes=\"(max-width: 400px) 100vw, 400px\" \/><\/p>\n<p>Using the formula of Bayes&#8217; Theorem:<\/p>\n<p>[latex]P(\\text{disease} | \\text{positive}) = \\frac{P(\\text{disease}) \\times P(\\text{positive} | \\text{disease})}{P(\\text{positive})} =\\frac{P(\\text{disease}) \\times P(\\text{positive} | \\text{disease})}{P(\\text{disease}) \\times P(\\text{positive} | \\text{disease}) + P(\\text{no disease}) \\times P(\\text{positive} | \\text{no disease})} = \\frac{(0.001)(1)}{(0.001)(1) + (0.999)(0.05)} \\approx 0.0196<br \/>[\/latex]<\/p>\n<p>Is it easier to find the [latex]P(\\text{disease} | \\text{positive})[\/latex] using the tree diagram rather than the prompt?<\/p>\n<\/section>\n<section class=\"textbox tryIt\">\n<iframe loading=\"lazy\" id=\"ohm13832\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=13832&theme=lumen&iframe_resize_id=ohm13832&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><br \/>\n<\/section>\n","protected":false},"author":51,"menu_order":17,"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\/3386"}],"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":31,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/chapters\/3386\/revisions"}],"predecessor-version":[{"id":7161,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/chapters\/3386\/revisions\/7161"}],"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\/3386\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/wp\/v2\/media?parent=3386"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/chapter-type?post=3386"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/wp\/v2\/contributor?post=3386"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/wp\/v2\/license?post=3386"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}