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Two applied examples use supplier defect rates and nurse pay-raise/training data to build trees, label events, and compute desired probabilities.",{"@graph":14,"@context":72},[15,34,55],{"@type":16,"itemListElement":17},"BreadcrumbList",[18,23,27,31],{"item":19,"name":20,"@type":21,"position":22},"https://docshare.wps.com","Home","ListItem",1,{"item":24,"name":25,"@type":21,"position":26},"https://docshare.wps.com/document/","Document",2,{"item":28,"name":29,"@type":21,"position":30},"https://docshare.wps.com/document/exam/","Exam",3,{"item":32,"name":10,"@type":21,"position":33},"https://docshare.wps.com/document/probability-and-probability-distributions-4-tree-diagrams-and-bayes-theorem-tree-diagrams/657361/",4,{"url":32,"name":10,"@type":35,"image":36,"author":41,"headline":10,"publisher":44,"fileFormat":47,"inLanguage":8,"description":12,"dateModified":48,"datePublished":49,"encodingFormat":47,"isAccessibleForFree":50,"interactionStatistic":51},"DigitalDocument",{"url":37,"@type":38,"width":39,"height":40},"https://docshare.wps.com/thumbnails/probability-and-probability-distributions-4-tree-diagrams-and-bayes-theorem-tree-diagrams/657361.png","ImageObject",300,407,{"name":42,"@type":43},"Mia  ","Person",{"url":19,"name":45,"@type":46},"DocShare","Organization","application/pdf","2026-10-30","2026-10-26",true,{"@type":52,"interactionType":53,"userInteractionCount":33},"InteractionCounter",{"@type":54},"ViewAction",{"@type":56,"mainEntity":57},"FAQPage",[58,64,68],{"name":59,"@type":60,"acceptedAnswer":61},"How do you compute the probability of a sequence using a tree diagram?","Question",{"text":62,"@type":63},"Multiply the probabilities along the branches that make up the sequence. The product gives the probability for that entire sequence.","Answer",{"name":65,"@type":60,"acceptedAnswer":66},"Why should the probabilities of all sequences in a complete tree add up to 1?",{"text":67,"@type":63},"A complete tree represents all possible sequences for the experiment, so exactly one sequence must occur. Therefore their probabilities sum to 1.",{"name":69,"@type":60,"acceptedAnswer":70},"How are conditional probabilities represented in the tree diagram examples?",{"text":71,"@type":63},"The tree branches include conditional probabilities such as P(D | S1) or P(B | A), with event labels used to compute the probability of outcomes under given conditions.","https://schema.org",{"og:url":32,"og:type":74,"og:title":10,"og:site_name":45,"og:description":12},"article",{"robots":76,"canonical":32},"index,follow",{"doc_id":78,"site_id":7},657361,1793073418,{"code":4,"msg":81,"data":82},"success",[83,87,91,94,99,104,109,114,119,122,126],{"id":22,"doc_module":4,"doc_module_name":25,"category_name":84,"show_sort_weight":85,"slug":86},"Story & Novel",90,"story-novel",{"id":26,"doc_module":4,"doc_module_name":25,"category_name":88,"show_sort_weight":89,"slug":90},"Literature",80,"literature",{"id":33,"doc_module":4,"doc_module_name":25,"category_name":29,"show_sort_weight":92,"slug":93},70,"exam",{"id":95,"doc_module":4,"doc_module_name":25,"category_name":96,"show_sort_weight":97,"slug":98},5,"Comic",60,"comic",{"id":100,"doc_module":4,"doc_module_name":25,"category_name":101,"show_sort_weight":102,"slug":103},6,"Technology",50,"technology",{"id":105,"doc_module":4,"doc_module_name":25,"category_name":106,"show_sort_weight":107,"slug":108},7,"Healthcare",40,"healthcare",{"id":110,"doc_module":4,"doc_module_name":25,"category_name":111,"show_sort_weight":112,"slug":113},8,"Research & Report",30,"research-report",{"id":115,"doc_module":4,"doc_module_name":25,"category_name":116,"show_sort_weight":117,"slug":118},9,"Religion & Spirituality",20,"religion-spirituality",{"id":117,"doc_module":4,"doc_module_name":25,"category_name":120,"show_sort_weight":117,"slug":121},"World Cup","world-cup",{"id":123,"doc_module":4,"doc_module_name":25,"category_name":124,"show_sort_weight":123,"slug":125},10,"Lifestyle","lifestyle",{"id":127,"doc_module":4,"doc_module_name":25,"category_name":128,"show_sort_weight":95,"slug":129},19,"General","general",{"code":4,"msg":81,"data":131},{"doc_id":78,"user_id":132,"nickname":42,"user_avatar":133,"doc_module":4,"category_id":33,"category_name":29,"doc_title":10,"doc_description":12,"doc_content":134,"file_id":135,"file_url":136,"file_type":137,"file_size":138,"view_count":33,"is_deleted":4,"is_public":22,"is_downloadable":22,"audit_status":22,"page_count":115,"language":139,"language_code":8,"site_id":7,"html_lang":8,"table_of_contents":140,"faqs":141,"seo_title":142,"seo_description":12,"update_tm":143,"read_time":144},687207024478,"https://ap-avatar.wpscdn.com/davatar_a8503ba1806abce46bf441b54a3ca4cd","A26 APPENDIX C Probability and Probability Distributions  \nC.4 Tree Diagrams and Bayes’ Theorem  \n■ Find probabilities using tree diagrams.  \n■ Find probabilities using Bayes’ Theorem.  \nTree Diagrams  \nA type of diagram that can be used as an aid in computing probabilities is a tree diagram. For example, consider an experiment of tossing a six-sided die. Each time the experiment is repeated, the probability of obtaining a 1 (event A) is P􀀂A􀀃 􀀃 1􀀄6. If you are only concerned with whether the number is 1 or not 1 (event A or event A􀀄), and the experiment is repeated three times, then eight different sequences of events are possible. The tree diagram in Figure C.11 shows the probabilities of these eight sequences of events.  \nTrial 1  \nTrial 2  \nTrial 3  \nP(A) = 16  \nP(A-A-A) = (16((16((16( = 2116  \nP(A-A-A′) = (16((16((56( = 2516  \nP(A-A′-A) = (16((56((16( = 2516  \nP(A-A′-A′) = (16((56((56( = 22516 P(A′-A-A) = (56((16((16( = 2516  \nP(A′-A-A′) = (56((16((56( = 22516  \nP(A′-A′-A) = (56((56((16( = 22516  \nP(A′-A′-A′) = (56((56((56( = 122156  \nFIGURE C .11  \nNote that the probability of each sequence of events is given by the product of the probabilities of the events in the sequence.  \nIf a tree diagram represents all possible sequences of events, as it does in Figure C.11, then the sum of the probabilities of the various sequences should be  \n1. For instance, the sum of the probabilities of the eight different sequences in Figure C.11 is  \n2116 􀀂 2516 􀀂 2516 􀀂 25216 􀀂 2516 􀀂 25216 􀀂 25216 􀀂 125216 􀀃 216216 􀀃 1.  \nTree diagrams often involve conditional probabilities, as demonstrated in the next two examples.  \nAPPENDIX C . 4 Tree Diagrams and Bayes’ Theorem A27  \nExample 1  \nA Tree Diagram Involving Conditional Probabilities  \nA manufacturer orders a biomedical part from three different suppliers. From past records, the manufacturer finds that the three suppliers have the following percentages of defective parts.  \nSupplier 1   Supplier 2   Supplier 3  \nDefective Parts 3% 5% 4%  \nNondefective Parts 97% 95% 96%  \nThe current inventory of parts contains 5000 units from Supplier 1, 3500 units from Supplier 2, and 2000 units from Supplier 3. If a part is chosen from the inventory at random, what is the probability that it is defective?  \nSOLUTION Because the current inventory contains 10,500 parts, the probability that a part chosen at random is from Supplier 1 is 5000􀀄10500 􀀃 10􀀄21, from Supplier 2 is 3500􀀄10500 􀀃 7􀀄21, and from Supplier 3 is 2000􀀄10500 􀀃 4􀀄21. In Figure C.12, let D represent the event that the chosen part is defective, and S1, S2 , and S3 represent the events that the chosen part came from Suppliers 1, 2, and 3, respectively.  \nP(S1) ⋅ P(D ⎪ S1) = (1021((0.03) ≈ 0.014 P(S1) ⋅ P(D ′ ⎪ S1) = (1021((0.97) ≈ 0.462 P(S2) ⋅ P(D ⎪ S2) = (271((0.05) ≈ 0.017  \nP(S2) ⋅ P(D ′ ⎪ S2) = (271((0.95) ≈ 0.317 P(S3) ⋅ P(D ⎪ S3) = (421((0.04) ≈ 0.008 P(S3) ⋅ P(D ′ ⎪ S3) = (421((0.96) ≈ 0.183  \nFrom the tree diagram, you can see that the probability of choosing a defective part is  \nP􀀂D􀀃 􀀅 0.014 􀀂 0.017 􀀂 0.008 􀀃 0.039.    \n✓CHECKPOINT 1  \nIn Example 1, suppose the current inventory of parts contains 4000 units from Supplier 1, 5000 units from Supplier 2, and 1500 units from Supplier 3. If apart is chosen from the inventory at random, what is the probability that it is defective? ■  \nA28 APPENDIX C Probability and Probability Distributions  \nExample 2  \nA Tree Diagram Involving Conditional Probabilities  \nA hospital has 300 nurses. During the past year, 48 of the nurses earned a pay raise. At the beginning of the year, the hospital offered a special training seminar, which was attended by 138 of the nurses. Twenty-seven of the nurses who earned a raise had attended the training seminar.  \na. What is the probability that a nurse who attended the seminar earned a pay raise?  \nb. If a nurse is selected at random, what is the probability that the person attended the seminar and earned a pay raise?  \nSOLUTION Create a tree diagram showing ","cbCainVR0Zuj4OxO","https://ap.wps.com/l/cbCainVR0Zuj4OxO","pdf",530345,"English","# Tree Diagrams\n## Example: Six-sided die sequences\n## Conditional probabilities (Example 1: suppliers)\n## Conditional probabilities (Example 2: nurses)\n# Bayes’ Theorem","[{\"question\":\"How do you compute the probability of a sequence using a tree diagram?\",\"answer\":\"Multiply the probabilities along the branches that make up the sequence. The product gives the probability for that entire sequence.\"},{\"question\":\"Why should the probabilities of all sequences in a complete tree add up to 1?\",\"answer\":\"A complete tree represents all possible sequences for the experiment, so exactly one sequence must occur. Therefore their probabilities sum to 1.\"},{\"question\":\"How are conditional probabilities represented in the tree diagram examples?\",\"answer\":\"The tree branches include conditional probabilities such as P(D | S1) or P(B | A), with event labels used to compute the probability of outcomes under given conditions.\"}]","Probability and Probability Distributions - 4 Tree Diagrams and Bayes’ Theorem - Tree Diagrams | PDF",1792977991,23]