ACTEX. SOA Exam STAM Study Manual. With StudyPlus + Spring 2018 Edition Volume I Samuel A. Broverman, Ph.D., ASA

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1 ACTEX SOA Exam STAM Study Manual With StudyPlus + StudyPlus + gives you digital access* to: Actuarial Exam & Career Strategy Guides Technical Skill elearning Tools Samples of Supplemental Textbooks And more! *See inside for keycode access and login instructions Spring 2018 Edition Volume I Samuel A. Broverman, Ph.D., ASA ACTEX Learning Learn Today. Lead Tomorro.

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3 ACTEX SOA Exam STAM Study Manual Spring 2018 Edition Samuel A. Broverman, Ph.D., ASA ACTEX Learning Ne Hartford, Connecticut

4 ACTEX Learning Learn Today. Lead Tomorro. Actuarial & Financial Risk Resource Materials Since 1972 Copyright 2018 SRBooks, Inc. ISBN: Printed in the United States of America. No portion of this ACTEX Study Manual may be reproduced or transmitted in any part or by any means ithout the permission of the publisher.

5 ACTEX is eager to provide you ith helpful study material to assist you in gaining the necessary knoledge to become a successful actuary. In turn e ould like your help in evaluating our manuals so e can help you meet that end. We invite you to provide us ith a critique of this manual by sending this form to us at your convenience. We appreciate your time and value your input. Publication: Your Opinion is Important to Us ACTEX STAM Study Manual, Spring 2018 Edition I found Actex by: (Check one) A Professor School/Internship Program Employer Friend Facebook/Titter In preparing for my exam I found this manual: (Check one) Very Good Good Satisfactory Unsatisfactory I found the folloing helpful: I found the folloing problems: (Please be specific as to area, i.e., section, specific item, and/or page number.) To improve this manual I ould: Name: Address: Phone: (Please provide this information in case clarification is needed.) Send to: Stephen Camilli ACTEX Learning P.O. Box 715 Ne Hartford, CT Or visit our ebsite at.actexmadriver.com to complete the survey on-line. Click on the Send Us Feedback link to access the online version. You can also your comments to Support@ActexMadRiver.com.

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9 ACTEX STAM EXAM MANUAL - TABLE OF CONTENTS INTRODUCTORY COMMENTS NOTES AND PROBLEM SETS SECTION 1 - Preliminary Revie - Probability 1 PROBLEM SET 1 9 SECTION 2 - Preliminary Revie - Random Variables I 19 PROBLEM SET 2 29 SECTION 3 - Preliminary Revie - Random Variables II 33 PROBLEM SET 3 41 SECTION 4 - Preliminary Revie - Random Variables III 49 PROBLEM SET 4 57 SECTION 5 - Parametric Distributions and Transformations 61 PROBLEM SET 5 69 SECTION 6 - Distribution Tail Behavior 79 PROBLEM SET 6 83 SECTION 7 - Mixture of To Distributions 85 PROBLEM SET 7 91 SECTION 8 - Mixture of 8 Distributions 99 PROBLEM SET SECTION 9 - Continuous Mixtures 113 PROBLEM SET SECTION 10 - Frequency Models 129 PROBLEM SET SECTION 11 - Policy Limits 157 PROBLEM SET SECTION 12 - Policy Deductible (1), The Cost Per Loss 163 PROBLEM SET SECTION 13 - Policy Deductible (2), The Cost Per Loss 183 PROBLEM SET SECTION 14 - Deductibles Applied to the Uniform, Exponential and Pareto Distributions 203 PROBLEM SET

10 SECTION 15 - Combined Limit and Deductible 215 PROBLEM SET SECTION 16 - Additional Policy Adjustments 243 PROBLEM SET SECTION 17 - Models for the Aggregate Loss, Compound Distributions (1) 255 PROBLEM SET SECTION 18 - Compound Distributions (2) 287 PROBLEM SET SECTION 19 - More Properties of the Aggregate Loss Random Variable 309 PROBLEM SET SECTION 20 - Stop Loss Insurance 333 PROBLEM SET SECTION 21 - Risk Measures 349 PROBLEM SET SECTION 22 - Data and Estimation Revie 357 SECTION 23 - Maximum Likelihood Estimation Based on Complete Data 361 PROBLEM SET SECTION 24 - Maximum Likelihood Estimation Based on Incomplete Data 375 PROBLEM SET SECTION 25 - Maximum Likelihood Estimation for the Exponential Distribution 391 PROBLEM SET SECTION 26 - MLE Applied to Pareto and Weibull Distributions 407 PROBLEM SET SECTION 27 - MLE Applied to STAM Exam Table Distributions 429 PROBLEM SET SECTION 28 - Revie of Mathematical Statistics 449 PROBLEM SET SECTION 29 - Properties of Maximum Likelihood Estimators 463 PROBLEM SET SECTION 30 - Hypothesis Tests For Fitted Models 483 PROBLEM SET SECTION 31 - Limited Fluctuation Credibility 521 PROBLEM SET

11 SECTION 32 - Bayesian Estimation, Discrete Prior 557 PROBLEM SET SECTION 33 - Bayesian Credibility, Discrete Prior 583 PROBLEM SET SECTION 34 - Bayesian Credibility, Continuous Prior 629 PROBLEM SET SECTION 35 - Bayesian Credibility Applied to Distributions in STAM Exam Table 669 PROBLEM SET SECTION 36 - Buhlmann Bayesian 707 PROBLEM SET SECTION 37 - Empirical Bayes Credibility 769 PROBLEM SET SECTION 38 - Major Medical and Dental Coverage 811 SECTION 39 - Property and Casualty Coverages 813 SECTION 40 - Loss Reserving 817 PROBLEM SET SECTION 41 - Ratemaking 841 PROBLEM SET SECTION 42 - Additional Casualty Insurance Topics 855 PROBLEM SET PRACTICE EXAMS PRACTICE EXAM PRACTICE EXAM PRACTICE EXAM PRACTICE EXAM PRACTICE EXAM 5 951

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13 INTRODUCTORY COMMENTS This study guide is designed to help in the preparation for the Society of Actuaries STAM Exam. The first part of this manual consists of a summary of notes, illustrative examples and problem sets ith detailed solutions. The second part consists of 5 practice exams. The practice exams all have 35 questions. The level of difficulty of the practice exams has been designed to be similar to that of the past 3.5-hour exams. Some of the questions in the problem sets are taken from the relevant topics on SOA exams that have been released prior to 2009 but the practice exam questions are not from old SOA exams. I have attempted to be thorough in the coverage of the topics upon hich the exam is based, and consistent ith the notation and content of the official references. I have been, perhaps, more thorough than necessary on a couple of topics, such as maximum likelihood estimation and Bayesian credibility. Because of the time constraint on the exam, a crucial aspect of exam taking is the ability to ork quickly. I believe that orking through many problems and examples is a good ay to build up the speed at hich you ork. It can also be orthhile to ork through problems that have been done before, as this helps to reinforce familiarity, understanding and confidence. Working many problems ill also help in being able to more quickly identify topic and question types. I have attempted, herever possible, to emphasize shortcuts and efficient and systematic ays of setting up solutions. There are also occasional comments on interpretation of the language used in some exam questions. While the focus of the study guide is on exam preparation, from time to time there ill be comments on underlying theory in places that I feel those comments may provide useful insight into a topic. The notes and examples are divided into 42 sections of varying lengths, ith some suggested time frames for covering the material. There are almost 200 examples in the notes and over 900 exercises in the problem sets, all ith detailed solutions. The 5 practice exams have 35 questions each, also ith detailed solutions. Some of the examples and exercises are taken from previous SOA exams. Some of the in the problem sets that have come from previous SOA exams. Some of the problem set exercises are more in depth than actual exam questions, but the practice exam questions have been created in an attempt to replicate the level of depth and difficulty of actual exam questions. In total there are almost 1300 examples/problems/sample exam questions ith detailed solutions. ACTEX gratefully acknoledges the SOA for alloing the use of their exam problems in this study guide. I suggest that you ork through the study guide by studying a section of notes and then attempting the exercises in the problem set that follos that section. The order of the sections of notes is the order that I recommend in covering the material, although the material on pricing and reserving in Sections 39 to 42 is independent of the other material on the exam. The order of topics in this manual is not the same as the order presented on the exam syllabus. About 80% or more of the material on the STAM Exam as on the former Exam C. It has been my intention to make this study guide self-contained and comprehensive for the STAM Exam topics, hoever there are some exam topics for hich the study notes are essentially summaries of concepts. For that material, I have attempted to summarize concepts as ell, but it is best to refer to original reference material on all topics. At the time this study manual is being ritten, the STAM Exam Syllabus available on the SOA ebsite contains a statement indicating that the study note titled "STAM Supplement to Chapter 3 of Intro to Ratemaking and Loss Reserving for Property and Casualty Insurance. Fourth Edition" ill be available at a later date. When that study note is available, ACTEX ill release, if needed, a supplement to this study manual to cover the content of that study note. SOA STAM Exam - Short Term Actuarial Mathematics

14 While the ability to derive formulas used on the exam is usually not the focus of an exam question, it is useful in enhancing the understanding of the material and may be helpful in memorizing formulas. There may be an occasional reference in the revie notes to a derivation, but you are encouraged to revie the official reference material for more detail on formula derivations. In order for the revie notes in this study guide to be most effective, you should have some background at the junior or senior college level in probability and statistics. It ill be assumed that you are reasonably familiar ith differential and integral calculus. The prerequisite concepts to modeling and model estimation are revieed in this study guide. The study guide begins ith a detailed revie of probability distribution concepts such as distribution function, hazard rate, expectation and variance. Of the various calculators that are alloed for use on the exam, I am most familiar ith the BA II PLUS. It has several easily accessible memories. The TI-30X IIS has the advantage of a multiline display. Both have the functionality needed for the exam. There is a set of tables that has been provided ith the exam in past sittings. These tables consist of some detailed description of a number of probability distributions along ith tables for the standard normal and chi-squared distributions. The tables can be donloaded from the SOA ebsite.soa.org. If you have any questions, comments, criticisms or compliments regarding this study guide, please contact the publisher ACTEX, or you may contact me directly at the address belo. I apologize in advance for any errors, typographical or otherise, that you might find, and it ould be greatly appreciated if you ould bring them to my attention. ACTEX ill be maintaining a ebsite for errata that can be accessed from.actexmadriver.com. It is my sincere hope that you find this study guide helpful and useful in your preparation for the exam. I ish you the best of luck on the exam. Samuel A. Broverman January 2018 Department of Statistical Sciences.sambroverman.com University of Toronto sam@utstat.toronto.edu or 2brove@rogers.com SOA STAM Exam - Short Term Actuarial Mathematics

15 NOTES AND PROBLEM SETS

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17 SECTION 1 - PRELIMINARY REVIEW - PROBABILITY STAM-1 SECTION 1 - PRELIMINARY REVIEW - PROBABILITY Basic Probability, Conditional Probability and Independence A significant part of the STAM Exam involves probability and statistical methods applied to various aspects of loss modeling and model estimation. A good background in probability and statistics is necessary to fully understand models and the modeling that is done. In this section of the study guide, e ill revie fundamental probability rules. 1.1 Basic Probability Concepts Sample point and probability space A sample point is the simple outcome of a random experiment. The probability space (also called sample space) is the collection of all possible sample points related to a specified experiment. When the experiment is performed, one of the sample points ill be the outcome. An experiment could be observing the loss that occurs on an automobile insurance policy during the course of one year, or observing the number of claims arriving at an insurance office in one eek. The probability space is the "full set" of possible outcomes of the experiment. In the case of the automobile insurance policy, it ould be the range of possible loss amounts that could occur during the year, and in the case of the insurance office eekly number of claims, the probability space ould be the set of integers Ö!ß "ß ß ÞÞÞ. Event Any collection of sample points, or any subset of the probability space is referred to as an event. We say "event E has occurred" if the experimental outcome as one of the sample points in E. Union of events Eand F E F denotes the union of events Eand F, and consists of all sample points that are in either Eor F. A B Union of events E" ß E ß ÞÞÞß E8 8 E" E â E8 E3 denotes the union of the events E" ß E ß ÞÞÞß E8, and consists of all sample 3" points that are in at least one of the E 3 's. This definition can be extended to the union of infinitely many events.

18 STAM-2 SECTION 1 - PRELIMINARY REVIEW - PROBABLITY Intersection of events E" ß E ß ÞÞÞß E8 8 E" E â E8 E3 denotes the intersection of the events E" ß E ß ÞÞÞß E8, and consists of all 3" sample points that are simultaneously in all of the E 3 's. A B Mutually exclusive events E" ß E ß ÞÞÞß E8 To events are mutually exclusive if they have no sample points in common, or equivalently, if they have empty intersection. Events E" ß E ß ÞÞÞß E8 are mutually exclusive if E3 E4 g for all 3 Á 4, here g denotes the empty set ith no sample points. Mutually exclusive events cannot occur simultaneously. Exhaustive events F" ß F ß ÞÞÞß F8 If F" F â F8 W, the entire probability space, then the events F" ß F ß ÞÞÞß F8 are referred to as exhaustive events. Complement of event E The complement of event E consists of all sample points in the probability space that are not in E. The - complement is denoted Eß µeße or E and is equal to ÖBÀBÂE. When the underlying random experiment is performed, to say that the complement of E has occurred is the same as saying that E has not occurred. Subevent (or subset) E of event F If event F contains all the sample points in event E, then E is a subevent of F, denoted E F. The occurrence of event E implies that event F has occurred. Partition of event E 8 Events G" ß G ß ÞÞÞß G8 form a partition of event E if E G3 and the G3's are mutually exclusive. 3" DeMorgan's Las (i) ÐE FÑ E F, to say that E Fhas not occurred is to say that Ehas not occurred and F has not occurred ; this rule generalizes to any number of events; 8 8 E3 ÐE E â E Ñ E E â E E 3" 3" " 8 " 8 3 (ii) ÐE FÑ E F, to say that E Fhas not occurred is to say that either Ehas not occurred or F has not occurred (or both have not occurred) ; this rule generalizes to any 8 8 number of events, E3 ÐE" E â E8Ñ E" E â E8 E3 3" 3"

19 SECTION 1 - PRELIMINARY REVIEW - PROBABILITY STAM-3 Indicator function for event E The function M ÐBÑ is the indicator function for event E, here Bdenotes a sample point. E " if B E! if BÂE ME ÐBÑ is 1 if event E has occurred. Some important rules concerning probability are given belo. (i) TÒWÓ " if W is the entire probability space (hen the underlying experiment is performed, some outcome must occur ith probability 1). (ii) TÒgÓ! (the probability of no face turning up hen e toss a die is 0). (iii) If events E ß E ß ÞÞÞß E " 8 are mutually exclusive (also called disjoint) then 8 8 TÒ E3 Ó TÒE" E â E8 Ó TÒE" Ó TÒE Ó â TÒE8Ó TÒE3Ó 3" 3" (1.1) This extends to infinitely many mutually exclusive events. (iv) For any event E,! Ÿ TÒEÓ Ÿ " (v) If E F then TÒEÓŸTÒFÓ (vi) For any events E, F and G, T ÒE FÓ T ÒEÓ T ÒFÓ T ÒE FÓ (1.2) (vii) For any event E, TÒE Ó " TÒEÓ (1.3) (viii) For any events Eand F, TÒEÓ TÒE FÓ TÒE F Ó (1.4) 8 (ix) For exhaustive events F" ß F ß ÞÞÞß F8, T Ò F3Ó " (1.5) 3" If F" ß F ß ÞÞÞß F8 are exhaustive and mutually exclusive, they form a partition of the entire probability space, and for any event E, TÒEÓ TÒE F Ó TÒE F Ó â TÒE F Ó TÒE F Ó " 8 3 3" 8 (1.6) (x) The ords "percentage" and "proportion" are used as alternatives to "probability". As an example, if e are told that the percentage or proportion of a group of people that are of a certain type is 20%, this is generally interpreted to mean that a randomly chosen person from the group has a 20% probability of being of that type. This is the "long-run frequency" interpretation of probability. As another example, suppose that e are tossing a fair die. In the long-run frequency interpretation of " probability, to say that the probability of tossing a 1 is ' is the same as saying that if e repeatedly " toss the die, the proportion of tosses that are 1's ill approach '.

20 STAM Conditional Probability and Independence of Events SECTION 1 - PRELIMINARY REVIEW - PROBABLITY Conditional probability arises throughout the STAM Exam material. It is important to be familiar and comfortable ith the definitions and rules of conditional probability. Conditional probability of event E given event F If TÐFÑ!, then TÐElFÑ TÐE FÑ is the conditional probability that event Eoccurs given that TÐFÑ event F has occurred. By reriting the equation e get T ÐE FÑ T ÐElFÑ T ÐFÑ. Partition of a Probability Space Events F" ß F ß ÞÞÞß F8 are said to form a partition of a probability space W if (i) F F â F W and (ii) F F g for any pair ith 3 Á 4. " A partition is a disjoint collection of events hich combines to be the full probability space. A simple example of a partition is any event F and its complement F. If E is any event in probability space W and ÖF" ß F ß ÞÞÞß F 8 is a partition of probability space W, then TÐEÑ TÐE F" Ñ TÐE F Ñ â TÐE F8 Ñ. A special case of this rule is TÐEÑ TÐE FÑ TÐE F Ñ for any to events Eand F. B 2 B 1 A B A B A B 1 A B 2 A B 3 A B 4 A B B B 3 4 Bayes rule and Bayes Theorem For any events Eand F ith TÐEÑ!, TÐFlEÑ TÐElFÑ TÐFÑ TÐEÑ (1.7) If F ß F ß ÞÞÞß F form a partition of the entire sample space W, then " 8 TÐElF4Ñ TÐF4Ñ T ÐF4lEÑ 8 for each 4 "ß ß ÞÞÞß 8 (1.8) TÐElF Ñ TÐF Ñ 3" 3 3 The values of TÐF4Ñare called prior probabilities, and the value of TÐF4lEÑ is called a posterior probability. Variations on this rule are very important in Bayesian credibility.

21 SECTION 1 - PRELIMINARY REVIEW - PROBABILITY STAM-5 Independent events Eand F If events Eand F satisfy the relationship TÐE FÑTÐEÑ TÐFÑ, then the events are said to be independent or stochastically independent or statistically independent. The independence of (non-empty) events E and F is equivalent to T ÐElFÑ T ÐEÑ or T ÐFlEÑ T ÐFÑ. Mutually independent events E ß E ß ÞÞÞß E " 8 The events are mutually independent if (i) for any E3 and E4, TÐE3 E4Ñ TÐE3Ñ TÐE4Ñ, and (ii) for any E3, E4 and E5, TÐE3 E4 E5ÑTÐEÑ TÐEÑ TÐE 3 4 5Ñ, and so on for any subcollection of the events, including all events: TÐE E â E Ñ TÐE Ñ TÐE Ñ â TÐE Ñ TÐE Ñ " 8 " 8 3 3" 8 (1.9) Here are some rules concerning conditional probability and independence. These can be verified in a fairly straightforard ay from the definitions given above. (i) TÐE FÑ TÐEÑ TÐFÑ TÐE FÑ for any events Eand F (1.10) (ii) TÐE FÑ TÐFlEÑ TÐEÑ TÐElFÑ TÐFÑ for any events Eand F (1.11) (iii) If F" ß F ß ÞÞÞß F8 form a partition of the sample space W, then for any event E 8 8 TÐEÑ TÐE F Ñ TÐElF Ñ TÐF Ñ " 3" (1.12) As a special case, for any events E and F, e have TÐEÑ TÐE FÑ TÐE F Ñ TÐElFÑ TÐFÑ TÐElF Ñ TÐF Ñ (1.13) (iv) If TÐE" E â E8 " Ñ!, then TÐE E â E Ñ TÐE Ñ TÐE le Ñ TÐE le E Ñ â TÐE le E â E " 8 " " $ " 8 " 8 " (v) T ÐE Ñ " T ÐEÑ and T ÐE lfñ " T ÐElFÑ (1.14) TÐE FÑ TÐEÑ (vi) if E F then T ÐElFÑ TÐFÑ TÐFÑ, and T ÐFlEÑ " (vii) if E and F are independent events then E and F are independent events, Eand F are independent events, and E and F are independent events (viii) since for any event, it follos that is independent of TÐgÑ TÐg EÑ! TÐgÑ TÐEÑ E g any event E Ñ

22 STAM-6 Example 1-1: Suppose a fair six-sided die is tossed. We define the folloing events: SECTION 1 - PRELIMINARY REVIEW - PROBABLITY E "the number tossed is Ÿ $ " Ö"ß ß $, F "the number tossed is even" Öß %ß ' G "the number tossed is a " or a " Ö"ß H "the number tossed doesn't start ith the letters 'f' or 't'" Ö"ß ' The conditional probability of E given F is T ÐÖ"ßß$ Öß%ß' Ñ T ÐÖ Ñ "Î' TÐElFÑ T ÐÖß%ß' Ñ T ÐÖß%ß' Ñ "Î " $. Events E and F are not independent, since " T ÐE FÑ Á T ÐEÑ T ÐFÑ " " " ' %, or alternatively, events E and F are not independent since TÐElFÑ Á TÐEÑ. " TÐElGÑ " Á TÐEÑ, so that Eand G are not independent. " TÐFlGÑ TÐFÑ, so that F and G are independent " " " (alternatively, T ÐF GÑ T ÐÖ Ñ ' $ T ÐFÑ T ÐGÑ ). It is not difficult to check that both E and F are independent of H. IMPORTANT NOTE: The folloing manipulation of event probabilities arises from time to time: T ( E) T ( ElF ) T ÐFÑ T ( ElF ) T ÐF Ñ. If e kno the conditional probabilities for event E given some other event F and its complement F, and if e kno the (unconditional) probability of event F, then e can find the probability of event E. One of the important aspects of applying this relationship is the determination of the appropriate events E and F. Example 1-2: Urn I contains 2 hite and 2 black balls and Urn II contains 3 hite and 2 black balls. An Urn is chosen at random, and a ball is randomly selected from that Urn. Find the probability that the ball chosen is hite. Solution: Let F be the event that Urn I is chosen and F is the event that Urn II is chosen. The implicit assumption is that both Urns are equally likely to be chosen (this is the meaning of "an Urn is chosen at random"). " " Therefore, TÐFÑ and TÐF Ñ. Let Ebe the event that the ball chosen in hite. If e kno that " Urn I as chosen, then there is probability of choosing a hite ball (2 hite out of 4 balls, it is assumed that each ball has the same chance of being chosen); this can be described as TÐElFÑ ". $ In a similar ay, if Urn II is chosen, then TÐElF Ñ & (3 hite out of 5 balls). We can no apply the " " " relationship described prior to this example. T ÐE FÑ T ÐElFÑ T ÐFÑ Ð ÑÐ Ñ %, and $ " $ T ÐE F Ñ T ÐElF Ñ T ÐF Ñ Ð& ÑÐ Ñ "!. Finally, " $ "" TÐEÑ TÐE FÑ TÐE F Ñ % "!!. The order of calculations can be summarized in the folloing table F F E 1. T ÐE FÑ T ÐElFÑ T ÐFÑ 2. TÐE F Ñ TÐElF Ñ TÐF Ñ 3. TÐEÑ TÐE FÑ TÐE F Ñ

23 SECTION 1 - PRELIMINARY REVIEW - PROBABILITY STAM-7 Example 1-3: Urn I contains 2 hite and 2 black balls and Urn II contains 3 hite and 2 black balls. One ball is chosen at random from Urn I and transferred to Urn II, and then a ball is chosen at random from Urn II. The ball chosen from Urn II is observed to be hite. Find the probability that the ball transferred from Urn I to Urn II as hite. Solution: Let F denote the event that the ball transferred from Urn I to Urn II as hite and let E denote the event that the ball chosen from Urn II is hite. We are asked to find TÐFlEÑ. From the simple nature of the situation (and the usual assumption of uniformity in such a situation, meaning all balls are equally likely to be chosen from Urn I in the first step), e have " " TÐFÑ (2 of the 4 balls in Urn I are hite), and by implication, it follos that TÒF Ó. If the ball transferred is hite, then Urn II has 4 hite and 2 black balls, and the probability of choosing a hite ball out of Urn II is ; this is $ TÐElFÑ $. If the ball transferred is black, then Urn II has 3 hite and 3 black balls, and the probability of choosing a hite ball out of Urn II is " ; this is " TÐElF Ñ. All of the information needed has been identified. We do calculations in the folloing order: " " 1. T ÒE FÓ T ÒElFÓ T ÒFÓ Ð$ ÑÐ Ñ $ " " " 2. T ÒE F Ó T ÒElF Ó T ÒF Ó Ð ÑÐ Ñ % " " ( 3. TÒEÓ TÒE FÓ TÒE F Ó $ % " T ÒF EÓ "Î$ % 4. TÒFlEÓ T ÒEÓ (Î" ( Example 1-4: Three dice have the folloing probabilities of throing a "six": :ß;ß<ß respectively. One of the dice is chosen at random and thron (each is equally likely to be chosen). A "six" appeared. What is the probability that the die chosen as the first one? Solution: The event " a 6 is thron" is denoted by "6" But TÒÐ die "Ñ Ð"'" ÑÓ TÒ "'" l die "Ó TÒ die "Ó : $ TÒ die "l"'" Ó TÒ "" ' Ó TÒ "" ' Ó TÒ "" ' Ó TÒ'ÓTÒÐ'Ñ Ð " " " " die "ÑÓ TÒÐ'Ñ Ð " " die ÑÓ TÒÐ'Ñ Ð " " die$ñó TÒ'l " " die"ó TÒ die "Ó TÒ'l " " dieó TÒ die Ó TÒ'l " " die$ó TÒ die$ó " " " " " : ; < : $ : $ : : $ ; $ < $ $ p TÒ die "l"'" Ó T Ò "" ' Ó " Ð: ; <Ñ : ; < " $

24 STAM-8 SECTION 1 - PRELIMINARY REVIEW - PROBABLITY

25 SECTION 1 PROBLEM SET SECTION 1 PROBLEM SET Preliminary Revie - Probability STAM-9 1. A survey of 1000 people determines that 80% like alking and 60% like biking, and all like at least one of the to activities. Ho many people in the survey like biking but not alking? A) 0 B).1 C).2 D).3 E).4 2. A life insurer classifies insurance applicants according to the folloing attributes: Q - the applicant is male L - the applicant is a homeoner Out of a large number of applicants the insurer has identified the folloing information: 40% of applicants are male, 40% of applicants are homeoners and 20% of applicants are female homeoners. Find the percentage of applicants ho are male and do not on a home. A).1 B).2 C).3 D).4 E).5 3. Let Eß Fß G and H be events such that F E ß G H g, and T ÒEÓ ", T ÒFÓ $, T ÒGlEÓ ", T ÒGlFÓ $, T ÒHlEÓ ", T ÒHlFÓ " % % % % ) Calculate TÒG HÓ. & " ( $ A) $ B) % C) $ D) % E) " 4. You are given that T ÒEÓ Þ& and T ÒE FÓ Þ(. Actuary 1 assumes that E and Fare independent and calculates TÒFÓ based on that assumption. Actuary 2 assumes that E and Fmutually exclusive and calculates TÒFÓ based on that assumption. Find the absolute difference beteen the to calculations. A) 0 B).05 C).10 D).15 E) A test for a disease correctly diagnoses a diseased person as having the disease ith probability.85. The test incorrectly diagnoses someone ithout the disease as having the disease ith a probability of.10. If 1% of the people in a population have the disease, hat is the chance that a person from this population ho tests positive for the disease actually has the disease? A) Þ!!)& B) Þ!(*" C) Þ"!(& D) Þ"&!! E) Þ*!!! 6. To bols each contain 5 black and 5 hite balls. A ball is chosen at random from bol 1 and put into bol 2. A ball is then chosen at random from bol 2 and put into bol 1. Find the probability that bol 1 still has 5 black and 5 hite balls. A) B) C) D) E) $ ' " ' $ & "" "$

26 STAM-10 SECTION 1 PROBLEM SET 7. People passing by a city intersection are asked for the month in hich they ere born. It is assumed that the population is uniformly divided by birth month, so that any randomly passing person has an equally likely chance of being born in any particular month. Find the minimum number of people needed so that the probability that no to people have the same birth month is less than.5. A) 2 B) 3 C) 4 D) 5 E) 6 8. In a T-maze, a laboratory rat is given the choice of going to the left and getting food or going to the right and receiving a mild electric shock. Assume that before any conditioning (in trial number 1) rats are equally likely to go the left or to the right. After having received food on a particular trial, the probability of going to the left and right become.6 and.4, respectively on the folloing trial. Hoever, after receiving a shock on a particular trial, the probabilities of going to the left and right on the next trial are.8 and.2, respectively. What is the probability that the animal ill turn left on trial number 2? A).1 B).3 C).5 D).7 E).9 9. In the game sho "Let's Make a Deal", a contestant is presented ith 3 doors. There is a prize behind one of the doors, and the host of the sho knos hich one. When the contestant makes a choice of door, at least one of the other doors ill not have a prize, and the host ill open a door (one not chosen by the contestant) ith no prize. The contestant is given the option to change his choice after the host shos the door ithout a prize. If the contestant sitches doors, hat is the probability that he gets the door ith the prize? " " " A)! B) ' C) $ D) E) $ 10. A supplier of a testing device for a type of component claims that the device is highly reliable, ith T ÒElFÓ T ÒE lf Ó Þ*&, here Edevice indicates component is faulty, and Fcomponent is faulty. You plan to use the testing device on a large batch of components of hich 5% are faulty. Find the probability that the component is faulty given that the testing device indicates that the component is faulty. A) 0 B).05 C).15 D).25 E).50

27 SECTION 1 PROBLEM SET 11. An insurer classifies flood hazard based on geographical areas, ith hazard categorized as lo, medium and high. The probability of a flood occurring in a year in each of the three areas is Area Hazard lo medium high Prob. of Flood The insurer's portfolio of policies consists of a large number of policies ith 80% lo hazard policies, 18% medium hazard policies and 2% high hazard policies. Suppose that a policy had a flood claim during a year. Find the probability that it is a high hazard policy. A).50 B).53 C).56 D).59 E).62 STAM One of the questions asked by an insurer on an application to purchase a life insurance policy is hether or not the applicant is a smoker. The insurer knos that the proportion of smokers in the general population is.30, and assumes that this represents the proportion of applicants ho are smokers. The insurer has also obtained information regarding the honesty of applicants: 40% of applicants that are smokers say that they are non-smokers on their applications, none of the applicants ho are non-smokers lie on their applications. What proportion of applicants ho say they are non-smokers are actually non-smokers? ' " $& A)! B) %" C) %" D) %" E) " 13. When sent a questionnaire, 50% of the recipients respond immediately. Of those ho do not respond immediately, 40% respond hen sent a follo-up letter. If the questionnaire is sent to 4 persons and a follo-up letter is sent to any of the 4 ho do not respond immediately, hat is the probability that at least 3 never respond? A) % $ ÐÞ$Ñ %ÐÞ$Ñ ÐÞ(Ñ B) $ %ÐÞ$Ñ ÐÞ(Ñ C) % $ ÐÞ"Ñ %ÐÞ"Ñ ÐÞ*Ñ D) Þ%ÐÞ$ÑÐÞ(Ñ ÐÞ(Ñ E) % $ ÐÞ*Ñ %ÐÞ*Ñ ÐÞ"Ñ $ % 14. A fair coin is tossed. If a head occurs, 1 fair die is rolled; if a tail occurs, 2 fair dice are rolled. If ] is the total on the die or dice, then TÒ] 'Ó " & "" " "" A) * B) $' C) ( D) ' E) $' 15. In Canada's national 6-49 lottery, a ticket has 6 numbers each from 1 to 49, ith no repeats. Find the probability of matching exactly 4 of the 6 inning numbers if the inning numbers are all randomly chosen. A) B) C) D) E).00103

28 STAM-12 SECTION 1 PROBLEM SET SOLUTIONS SECTION 1 PROBLEM SET 1. Let E"like alking" and F"like biking". We use the interpretation that "percentage" and "proportion" are taken to mean "probability". We are given TÐEÑ Þ)ß TÐFÑ Þ' and TÐE FÑ ". From the diagram belo e can see that since E FE ÐF EÑ e have TÐE FÑ TÐEÑ TÐE FÑ p TÐE FÑ Þ is the proportion of people ho like biking but (and) not alking. In a similar ay e get TÐE F Ñ Þ% A B A B An algebraic approach is the folloing. Using the rule TÐE FÑ TÐEÑ TÐFÑ TÐE FÑ, e get " Þ) Þ' T ÐE FÑ p T ÐE FÑ Þ%. Then, using the rule TÐFÑ TÐF EÑ TÐF EÑ, e get TÐF EÑ Þ' Þ% Þ. Anser: C 2. TÒQÓ Þ%ß TÒQ Ó Þ'ß TÒLÓ Þ%ß TÒL Ó Þ'ß TÒQ LÓ Þß We ish to find TÒQ L Ó. From probability rules, e have Þ' T ÒL Ó T ÒQ L Ó T ÒQ L Ó, and Þ' T ÒQ Ó T ÒQ LÓ T ÒQ L Ó Þ T ÒQ L Ó. Thus, T ÒQ L Ó Þ% and then T ÒQ L Ó Þ. The folloing diagram identifies the component probabilities. M H.4 M H M H M H The calculations above can also be summarized in the folloing table. The events across the top of the table categorize individuals as male ( Q) or female ( Q ), and the events don the left side of the table categorize individuals as homeoners ( L) or non-homeoners ( L ). TÐQÑÞ%, given TÐQÑ" Þ%Þ' TÐLÑ Þ% TÐQ LÑ É TÐQ LÑ Þ, given given TÐLÑ TÐQ LÑÞ% ÞÞ Ì TÐLÑ" Þ%Þ' TÐQ LÑTÐQÑ TÐQ LÑÞ% ÞÞ Anser: B

29 SECTION 1 PROBLEM SET 3. Since G and Hhave empty intersection, TÒG HÓ TÒGÓ TÒHÓÞ STAM-13 Also, since E and Fare "exhaustive" events (since they are complementary events, their union is the entire sample space, ith a combined probability of TÒE FÓ TÒEÓ TÒFÓ " ). TÒE GÓ We use the rule TÒGÓ TÒG EÓ TÒG EÓ, and the rule TÒGlEÓ TÒEÓ to get " " $ $ "" T ÒGÓ T ÒGlEÓ T ÒEÓ T ÒGlE Ó T ÒE Ó % % % "' and " " " $ & T ÒHÓ T ÒHlEÓ T ÒEÓ T ÒHlE Ó T ÒE Ó % % ) % $ Þ ( Then, TÒG HÓ TÒGÓ TÒHÓ $ Þ Anser: C. 4. Actuary 1: Since E and Fare independent, so are E and F. T ÒE F Ó " T ÒE FÓ Þ$. But Þ$ T ÒE F Ó T ÒE Ó T ÒF Ó ÐÞ&ÑT ÒF Ó p T ÒF Ó Þ' p T ÒFÓ Þ%. Actuary 2: Þ( T ÒE FÓ T ÒEÓ T ÒFÓ Þ& T ÒFÓ p T ÒFÓ Þ. Absolute difference is lþ% Þl Þ. Anser: E 5. We define the folloing events: H - a person has the disease, XT - a person tests positive for the disease. We are given TÒXTlHÓ Þ)& and T ÒX T lh Ó Þ"! and T ÒHÓ Þ!". We ish to find T ÒHlX T Ó. Using the formulation for conditional probability e have TÒHlXTÓ TÒH XTÓ TÒXTÓ. But T ÒH X T Ó T ÒX T lhó T ÒHÓ ÐÞ)&ÑÐÞ!"Ñ Þ!!)&, and T ÒH X T Ó T ÒX T lh Ó T ÒH Ó ÐÞ"!ÑÐÞ**Ñ Þ!**. Then, Þ!!)& T ÒX T Ó T ÒH X T Ó T ÒH X T Ó Þ"!(& p T ÒHlX T Ó Þ"!(& Þ!(*". The folloing table summarizes the calculations. TÒHÓ Þ!", given Ê TÒH Ó " TÒHÓ Þ** Ì Ì TÒH XTÓ TÒH XTÓ T ÒX T lhó T ÒHÓ Þ!!)& T ÒX T lh Ó T ÒH Ó Þ!** Ì T ÒX T Ó T ÒH X T Ó T ÒH X T Ó Þ"!(& Ì TÒH XTÓ Þ!!)& T ÒHlX T Ó T ÒX T Ó Þ"!(& Þ!(*". Anser: B

30 STAM-14 SECTION 1 PROBLEM SET 6. Let G be the event that bol 1 has 5 black balls after the exchange. Let F" be the event that the ball chosen from bol 1 is black, and let F be the event that the ball chosen from bol 2 is black. Event G is the disjoint union of F" F and F" F (black-black or hite-hite picks), so that TÒGÓ TÒF" F Ó TÒF" F Ó. ' " & The black-black combination has probability Ð "" ÑÐ Ñ, since there is a "! chance of picking black ' from bol 1, and then (ith 6 black in bol 2, hich no has 11 balls) "" is the probability of picking black from bol 2. This is ' " T ÒF" F Ó T ÒF lf" Ó T ÒF" Ó Ð "" ÑÐ Ñ. ' " In a similar ay, the hite-hite combination has probability Ð "" ÑÐ Ñ. ' " ' " ' Then TÒGÓ Ð "" ÑÐ Ñ Ð "" ÑÐ Ñ "". Anser: C 7. E event that second person has different birth month from the first. "" T ÐE Ñ " Þ*"'(Þ E $ event that third person has different birth month from first and second. Then, the probability that all three have different birthdays is "! "" T ÒE$ E Ó T ÒE$ le Ó T ÐE Ñ Ð " ÑÐ " Ñ Þ('$*. E % event that fourth person has different birth month from first three. Then, the probability that all four have different birthdays is T ÒE% E$ E Ó T ÒE% le$ E Ó T ÒE$ E Ó * "! "" T ÒE% le$ E Ó T ÒE$ le Ó T ÐE Ñ Ð " ÑÐ " ÑÐ " Ñ Þ&(*. E & event that fifth person has different birth month from first four. Then, the probability that all five have different birthdays is TÒE& E% E$ E Ó TÒE& le% E$ E Ó TÒE% E$ E Ó T ÒE& le% E$ E Ó T ÒE% le$ E Ó T ÒE$ le Ó T ÐE Ñ ) * "! "" Ð " ÑÐ " ÑÐ " ÑÐ " Ñ Þ$)"*. Anser: D 8. P" turn left on trial 1, V" turn right on trial 1, P turn left on trial 2. We are given that TÒP"Ó TÒV"Ó Þ&. TÒPÓ TÒP P"Ó TÒP V"Ó since P"ßV" form a partition. TÒPlP"Ó Þ' (if the rat turns left on trial 1 then it gets food and has a.6 chance of turning left on trial 2). Then T ÒP P"Ó T ÒPlP"Ó T ÒP"Ó ÐÞ'ÑÐÞ&Ñ Þ$. In a similar ay, T ÒP V"Ó T ÒPlV"Ó T ÒV"Ó ÐÞ)ÑÐÞ&Ñ Þ%. Then, T ÒPÓ Þ$ Þ% Þ(. Anser: D

31 SECTION 1 PROBLEM SET 9. We define the events Eprize door is chosen after contestant sitches doors, Fprize door is initial one chosen by contestant. Then TÒFÓ " $, since each door is equally likely to hold the prize initially. To find TÒEÓe use the La of Total Probability. " T ÒEÓ T ÒElFÓ T ÒFÓ T ÒElF Ó T ÒF Ó Ð!ÑÐ $ Ñ Ð"ÑÐ $ Ñ $ STAM-15 If the prize door is initially chosen, then after sitching, the door chosen is not the prize door, so that TÒElFÓ!. If the prize door is not initially chosen, then since the host shos the other non prize door, after sitching the contestant definitely has the prize door, so that TÒElF Ó ". Anser: E 10. We are given TÒFÓ Þ!&. We can calculate entries in the folloing table in the order indicated. E E F T ÒElFÓ Þ*& (given) T ÒE lf Ó Þ*& (given) T ÒFÓ Þ!& 1. T ÒE FÓ T ÒElFÓ T ÒFÓ Þ!%(& (given) F 3. T ÒE F Ó 2. T ÒE F Ó T ÒF Ó T ÒF Ó T ÒE F Ó T ÒE lf Ó T ÒF Ó " T ÒFÓ Þ*& Þ*!& Þ!%(& Þ*& Þ*!& Þ*& 4. TÒEÓ TÒE FÓ TÒE F Ó Þ!*& TÒF EÓ Þ!%(& 5. TÒFlEÓ T ÒEÓ Þ!*& Þ& Anser: E 11. This is a classical Bayesian probability situation. Let G denote the event that a flood claim occurred. We ish to find TÐLlGÑ. We can summarize the information in the folloing table, ith the order of calculations indicated. PßTÐPÑÞ) QßTÐQÑÞ") LßTÐLÑÞ! (given) (given) (given) G T ÐGlPÑ Þ!!" T ÐGlQÑ Þ! T ÐGlLÑ Þ& (given) (given) (given) 1. TÐG PÑ 2. TÐG QÑ 3. TÐG LÑ TÐGlPÑ TÐPÑ TÐGlQÑ TÐQÑ TÐGlLÑ TÐLÑ Þ!!!) Þ!!$' Þ!!& 4. T ÐGÑ T ÐG PÑ T ÐG QÑ T ÐG LÑ Þ!!*% TÐL GÑ Þ!!& 5. T ÐLlGÑ T ÐGÑ Þ!!*% Þ&$ Anser: B

32 STAM We identify the folloing events: SECTION 1 PROBLEM SET W - the applicant is a smoker, RW - the applicant is a non-smoker W HW - the applicant declares to be a smoker on the application HR - the applicant declares to be non-smoker on the application HW. The information e are given is TÒWÓ Þ$ß TÒRWÓ Þ(ß TÒHRlWÓ Þ%ß TÒHWlRWÓ!. We ish to find TÒRWlHRÓ TÒRW HRÓ TÒHRÓ. TÒHR WÓ TÒHR WÓ We calculate Þ% T ÒHRlWÓ TÒWÓ Þ$ p T ÒHR WÓ Þ", TÒHW RWÓ TÒHW RWÓ and! T ÒHWlRWÓ TÒRWÓ Þ( p T ÒHW RWÓ!. Using the rule TÒEÓTÒE FÓ TÒE FÓ, and noting that HWHR and WRW e have T ÒHW WÓ T ÒWÓ T ÒHR WÓ Þ$ Þ" Þ"), and T ÒHR RWÓ T ÒRWÓ T ÒHW RWÓ Þ(! Þ(, and T ÒHRÓ T ÒHR RWÓ T ÒHR WÓ Þ( Þ" Þ). TÒRW HRÓ Þ( $& Then, TÒRWlHRÓ T ÒHRÓ Þ) %". These calculations can be summarized in the order indicated in the folloing table. TÐWÑßÞ$ Ê 1. TÐRWÑ " TÐWÑ Þ( given Ì 6. HW É 5. T ÐHW WÑ 2. T ÐHWlRWÑ!, given TÐHWÑ TÐWÑ TÐHR WÑ TÐHW RWÑ T ÐHW WÑ Þ$ Þ" Þ") T ÐHWlRWÑ T ÐRWÑ T ÐHW RWÑ Ð!ÑÐÞ(Ñ! Þ")!Þ") Ì Ë 7. HR 4. T ÐHRlWÑ Þ% 3. T ÐHR RWÑ TÐHRÑ given TÐRWÑ TÐHW RWÑ " TÐHWÑ TÐHR WÑ Þ(!Þ( " Þ") TÐHRlWÑ TÐWÑ Þ) ÐÞ%ÑÐÞ$ÑÞ" Then, TÒRW HRÓ Þ( $& 8. TÒRWlHRÓ T ÒHRÓ Þ) %" Anser: D

33 SECTION 1 PROBLEM SET STAM The probability that an individual ill not respond to either the questionnaire or the follo-up letter is ÐÞ&ÑÐÞ'Ñ Þ$. The probability that all 4 ill not respond to either the questionnaire or the follo-up letter is ÐÞ$Ñ %. TÒ$ don't respond Ó TÒ" response on 1st round, no additional responses on 2nd roundó TÒno responses on 1st round, 1 response on 2nd roundó % $ % $ $ %ÒÐÞ&Ñ ÐÞ'Ñ Ó %ÒÐÞ&Ñ ÐÞ'Ñ ÐÞ%ÑÓ %ÐÞ$Ñ ÐÞ(Ñ. Then, % $ T Òat least 3 don't respond Ó ÐÞ$Ñ %ÐÞ$Ñ ÐÞ(Ñ. Anser: A " 14. If 1 fair die is rolled, the probability of rolling a 6 is ', and if 2 fair dice are rolled, the & probability of rolling a 6 is $' (of the 36 possible rolls from a pair of dice, the rolls 1-5, 2-4, 3-3, 4-2 and 5-1 result in a total of 6), Since the coin is fair, the probability of rolling a " & "" head or tail is.5. Thus, the probability that ] ' is ÐÞ&ÑÐ ' Ñ ÐÞ&ÑÐ $' Ñ ( Þ Anser: C 15. Suppose you have bought a lottery ticket. There are ' % "& ays of picking 4 numbers from the 6 numbers on your ticket. Suppose e look at one of those subsets of 4 numbers from your ticket. In order for the inning ticket number to match exactly those 4 of your 6 numbers, the other 2 inning ticket numbers must come from the 43 numbers beteen 1 and 49 that are not numbers on your ticket. There are %$ %$ % " *!$ ays of doing that, and since there are 15 subsets of 4 numbers on your ticket, there are "& *!$ "$ß &%& ays in hich the inning ticket numbers match exactly 3 of your ticket numbers. Since there are a total of 13,983,816 ays of picking 6 out of 49 numbers, your chance of matching exactly % of the inning numbers is 13,545 "$ß*)$ß)"' Þ!!!*')'. Anser: B

34 STAM-18 SECTION 1 PROBLEM SET