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Solutions use event definitions, conditional probability, Bayes-style manipulations, inclusion-exclusion style reasoning, independence and complement rules, and algebraic solving to compute targeted probabilities and final answer choices.",{"@graph":14,"@context":73},[15,34,56],{"@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/soceity-of-actuaries-exam-p-probability-exam-p-sample-solutions/398917/",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/soceity-of-actuaries-exam-p-probability-exam-p-sample-solutions/398917.png","ImageObject",300,407,{"name":42,"@type":43},"Jiven","Person",{"url":19,"name":45,"@type":46},"DocShare","Organization","application/pdf","2026-11-10","2026-09-27",true,{"@type":52,"interactionType":53,"userInteractionCount":55},"InteractionCounter",{"@type":54},"ViewAction",17,{"@type":57,"mainEntity":58},"FAQPage",[59,65,69],{"name":60,"@type":61,"acceptedAnswer":62},"What does the document explain before the solutions start?","Question",{"text":63,"@type":64},"It introduces “Society of Actuaries Exam P Probability” sample solutions and notes that some questions were taken or reformatted from past examinations, with specified additions across 2014–2015.","Answer",{"name":66,"@type":61,"acceptedAnswer":67},"How are solutions typically structured in the sample set?",{"text":68,"@type":64},"Each problem defines relevant events/variables, applies probability rules (such as conditional probability, independence, and complements), then performs algebra to reach the final numeric probability and answer choice.",{"name":70,"@type":61,"acceptedAnswer":71},"Are the probabilities in the solutions computed using only the given definitions?",{"text":72,"@type":64},"Yes. The solutions rely on explicitly defined events (e.g., named by letters) and then use established probability identities and the numbers provided in each question to compute results.","https://schema.org",{"og:url":32,"og:type":75,"og:title":10,"og:site_name":45,"og:description":12},"article",{"robots":77,"canonical":32},"index,follow",{"doc_id":79,"site_id":7},398917,1790578551,{"code":4,"msg":82,"data":83},"success",[84,88,92,95,100,105,110,115,120,123,127],{"id":22,"doc_module":4,"doc_module_name":25,"category_name":85,"show_sort_weight":86,"slug":87},"Story & Novel",90,"story-novel",{"id":26,"doc_module":4,"doc_module_name":25,"category_name":89,"show_sort_weight":90,"slug":91},"Literature",80,"literature",{"id":33,"doc_module":4,"doc_module_name":25,"category_name":29,"show_sort_weight":93,"slug":94},70,"exam",{"id":96,"doc_module":4,"doc_module_name":25,"category_name":97,"show_sort_weight":98,"slug":99},5,"Comic",60,"comic",{"id":101,"doc_module":4,"doc_module_name":25,"category_name":102,"show_sort_weight":103,"slug":104},6,"Technology",50,"technology",{"id":106,"doc_module":4,"doc_module_name":25,"category_name":107,"show_sort_weight":108,"slug":109},7,"Healthcare",40,"healthcare",{"id":111,"doc_module":4,"doc_module_name":25,"category_name":112,"show_sort_weight":113,"slug":114},8,"Research & Report",30,"research-report",{"id":116,"doc_module":4,"doc_module_name":25,"category_name":117,"show_sort_weight":118,"slug":119},9,"Religion & Spirituality",20,"religion-spirituality",{"id":118,"doc_module":4,"doc_module_name":25,"category_name":121,"show_sort_weight":118,"slug":122},"World Cup","world-cup",{"id":124,"doc_module":4,"doc_module_name":25,"category_name":125,"show_sort_weight":124,"slug":126},10,"Lifestyle","lifestyle",{"id":128,"doc_module":4,"doc_module_name":25,"category_name":129,"show_sort_weight":96,"slug":130},19,"General","general",{"code":4,"msg":82,"data":132},{"doc_id":79,"user_id":133,"nickname":42,"user_avatar":134,"doc_module":4,"category_id":33,"category_name":29,"doc_title":10,"doc_description":12,"doc_content":135,"file_id":136,"file_url":137,"file_type":138,"file_size":139,"view_count":55,"is_deleted":4,"is_public":22,"is_downloadable":22,"audit_status":22,"page_count":140,"language":141,"language_code":8,"site_id":7,"html_lang":8,"table_of_contents":142,"faqs":143,"seo_title":144,"seo_description":12,"update_tm":145,"read_time":146,"preview_page_count":4},1099513958607,"https://ap-avatar.wpscdn.com/avatar/100002390cf8733938c?x-image-process=image/resize,m_fixed,w_180,h_180&k=1778829742770036399","SOCIETY OF ACTUARIES  \nEXAM P PROBABILITY  \nEXAM P SAMPLE SOLUTIONS  \nCopyright 2015 by the Society of Actuaries  \nSome of the questions in this study note are taken from past examinations.  \nSome of the questions have been reformatted from previous versions of this note. Questions 154 and 155 were added in October 2014.  \nQuestions 156 – 206 were added January 2015.  \nQuestions 207 – 237 were added April 2015.  \nQuestions 238-240 were added May 2015.  \nQuestions 241-242 were added November 2015.  \n1. Solution: D  \nLet G = viewer watched gymnastics, B = viewer watched baseball, S = viewer watched soccer. Then we want to find  \nPr 􀀋 G 􀂉 B 􀂉 S 􀀌 c  􀀠 1􀀐 Pr 􀀋 G 􀂉 B 􀂉 S 􀀌  \n􀀠 1 􀀐 Pr 􀀋 G 􀀌􀀎 Pr 􀀋 B 􀀌􀀎 Pr 􀀋 S 􀀌􀀐 Pr 􀀋 G 􀂈 B 􀀌􀀐 Pr 􀀋 G 􀂈 S 􀀌􀀐 Pr 􀀋 B 􀂈 S 􀀌􀀎 Pr 􀀋 G 􀂈 B 􀂈 S 􀀌   \n􀀠 1􀀐 􀀋0.28 􀀎 0.29 􀀎 0.19 􀀐 0.14 􀀐 0.10 􀀐 0.12 􀀎 0.08 􀀌 􀀠 1􀀐 0.48 􀀠 0.52  \n2. Solution: A  \nLet R = referral to a specialist and L = lab work. Then  \nP[R 􀂈 L] 􀀠 P[R] 􀀎 P[L] 􀀐 P[R 􀂉 L] 􀀠 P[R] 􀀎 P[L] 􀀐1􀀎 P[(R 􀂉 L)c ]  \n􀀠 p[R] 􀀎 P[L] 􀀐1􀀎 P[Rc 􀂈 L􀁣 ] 􀀠 0.30 􀀎 0 .40 􀀐1􀀎 0 .35 􀀠 0.05.  \n3. Solution: D  \nFirst note  \nP 􀀾 A 􀂉 B 􀁀 􀀠 P 􀀾 A􀁀􀀎 P 􀀾 B 􀁀􀀐 P 􀀾 A 􀂈 B 􀁀 P A 􀂉 Bc  􀀠 P 􀀾 A􀁀􀀎 P Bc  􀀐 P A 􀂈 Bc  Then add these two equations to get  \nP 􀀾 A 􀂉 B 􀁀􀀎 P A 􀂉 Bc  􀀠 2P 􀀾 A􀁀􀀎 􀀋 P 􀀾 B 􀁀􀀎 P Bc  􀀌􀀐 􀀋P 􀀾 A 􀂈 B 􀁀􀀎 P A 􀂈 Bc  􀀌  \n0.7 􀀎 0.9 􀀠 2P 􀀾 A􀁀􀀎1􀀐 P 􀀋 A 􀂈 B 􀀌􀂉 􀀋 A 􀂈 Bc 􀀌   \n1.6 􀀠 2P 􀀾 A􀁀􀀎1􀀐 P 􀀾 A􀁀 P 􀀾 A􀁀 􀀠 0.6  \n4. Solution: A  \nFor i = 1,2, let Ri = event that a red ball is drawn from urn i and let Bi = event that a blue ball is drawn from urn i. Then, if x is the number of blue balls in urn 2,  \n0.44 􀀠 Pr[􀀋R1 􀂈 R2 􀀌 􀀋 B1 􀂈 B2 􀀌] 􀀠 Pr[R1 􀂈 R2 ] 􀀎 Pr 􀀾 B1 􀂈 B2 􀁀  \n􀀠 Pr 􀀾 R1 􀁀 Pr 􀀾 R2 􀁀􀀎 Pr 􀀾 B1 􀁀 Pr 􀀾 B2 􀁀  \n4 􀂧 16 􀂷 6 􀂧 x 􀂷  \n􀀠 ~~ ~~ 􀂨 ~~ ~~ 􀂸 􀀎 ~~ ~~ 􀂨 ~~ ~~ 􀂸  \n10 􀂩 x 􀀎16 􀂹 10 􀂩 x 􀀎16 􀂹  \nTherefore,  \n 32   3x   3x 􀀎 32  \n2.2 􀀠 􀀎 􀀠  \nx 􀀎16 x 􀀎16 x 􀀎16  \n2.2x 􀀎 35.2 􀀠 3x 􀀎 32  \n0.8x 􀀠 3.2 x 􀀠 4  \n5. Solution: D  \nLet N(C) denote the number of policyholders in classification C. Then N(Young and Female and Single)  \n= N(Young and Female)–N(Young and Female and Married)  \n= N(Young)–N(Young and Male)– [N(Young and Married)–N(Young and Married and Male)]  \n= 3000 – 1320 –(1400 – 600) = 880.  \n6. Solution: B  \nLet  \nH = event that a death is due to heart disease  \nF = event that at least one parent suffered from heart disease Then based on the medical records,  \nP H 􀂈 Fc  􀀠 21091702 􀀠 190387 P Fc  􀀠 93793712 􀀠 692357  \nP H 􀂈 Fc  108 625 108 and P H | Fc  􀀠  P Fc   􀀠 937    937 􀀠 625 􀀠 0.173  \n7. Solution: D  \nLet A = event that a policyholder has an auto policy and H = event that a policyholder has a homeowners policy. Then,  \nPr 􀀋 A 􀂈 H 􀀌 􀀠 0.15  \nPr 􀀋 A 􀂈 Hc 􀀌 􀀠 Pr 􀀋 A􀀌􀀐 Pr 􀀋 A 􀂈 H 􀀌 􀀠 0.65 􀀐 0.15 􀀠 0.50  \nPr 􀀋 Ac 􀂈 H􀀌 􀀠 Pr 􀀋 H 􀀌􀀐 Pr 􀀋 A 􀂈 H 􀀌 􀀠 0.50 􀀐 0.15 􀀠 0.35  \nand the portion of policyholders that will renew at least one policy is given by 0.4 Pr 􀀋 A 􀂈 Hc 􀀌 􀀎 0.6 Pr 􀀋 Ac 􀂈 H􀀌 􀀎 0.8 Pr 􀀋 A 􀂈 H 􀀌  \n􀀠 􀀋0.4 􀀌􀀋0.5 􀀌􀀎 􀀋0.6 􀀌􀀋0.35 􀀌􀀎 􀀋0.8 􀀌􀀋0.15 􀀌 􀀠 0.53  \n8. Solution: D  \n􀀋 􀀠 53% 􀀌  \nLet C = event that patient visits a chiropractor and T = event that patient visits a physical therapist. We are given that  \nPr 􀀾 C 􀁀 􀀠 Pr 􀀾 T 􀁀􀀎 0.14 Pr 􀀋 C 􀂈T 􀀌 􀀠 0.22 Pr 􀀋 Cc 􀂈Tc 􀀌 􀀠 0.12  \nTherefore,  \n0.88 􀀠 1􀀐 Pr Cc 􀂈Tc  􀀠 Pr 􀀾 C 􀂉T 􀁀 􀀠 Pr 􀀾 C 􀁀􀀎 Pr 􀀾 T 􀁀􀀐 Pr 􀀾 C 􀂈T 􀁀  \n􀀠 Pr 􀀾 T 􀁀􀀎 0.14 􀀎 Pr 􀀾 T 􀁀􀀐 0.22  \n􀀠 2 Pr 􀀾 T 􀁀􀀐 0.08 or  \nP 􀀾 T 􀁀 􀀠 􀀋0.88 􀀎 0.08􀀌2 􀀠 0.48  \n9. Solution: B  \nLet M= event that customer insures more than one car and S = event that customer insurers a sports car. Then applying DeMorgan’s Law, compute the desired probability as:  \nPr 􀀋 Mc 􀂈 Sc 􀀌 􀀠 Pr 􀀋 M 􀂉 S 􀀌 c  􀀠 1􀀐 Pr 􀀋 M 􀂉 S 􀀌 􀀠 1 􀀐 Pr 􀀋 M 􀀌􀀎 Pr 􀀋 S 􀀌􀀐 Pr 􀀋 M 􀂈 S 􀀌   \n􀀠 1􀀐 Pr 􀀋 M 􀀌􀀐 Pr 􀀋 S 􀀌􀀎 Pr 􀀋 S  M 􀀌 Pr 􀀋 M 􀀌 􀀠 1􀀐 0.70 􀀐 0.20 􀀎 􀀋0.15 􀀌􀀋0.70 􀀌 􀀠 0.205  \n10. This question duplicates Question 9 and has been deleted  \n11. Solution: B  \nLet C = Event that a policyholder buys collision coverage and D = Event that a policyholder buys disability coverage. Then we are given that P[C] 􀀠 2P[D] and P[C 􀂈 D] 􀀠 0.15 .  \nBy the independenc","cbCaipLy4pIHTNYO","https://ap.wps.com/l/cbCaipLy4pIHTNYO","pdf",753838,68,"English","# Solution Set Overview\n## Problem history and additions\n## Worked solutions (e.g., Questions 1–16)","[{\"question\":\"What does the document explain before the solutions start?\",\"answer\":\"It introduces “Society of Actuaries Exam P Probability” sample solutions and notes that some questions were taken or reformatted from past examinations, with specified additions across 2014–2015.\"},{\"question\":\"How are solutions typically structured in the sample set?\",\"answer\":\"Each problem defines relevant events/variables, applies probability rules (such as conditional probability, independence, and complements), then performs algebra to reach the final numeric probability and answer choice.\"},{\"question\":\"Are the probabilities in the solutions computed using only the given definitions?\",\"answer\":\"Yes. The solutions rely on explicitly defined events (e.g., named by letters) and then use established probability identities and the numbers provided in each question to compute results.\"}]","SOCEITY OF ACTUARIES - EXAM P PROBABILITY - EXAM P SAMPLE SOLUTIONS | PDF",1790490062,171]