`EXCERPTS FROM ACTEX STUDY MANUAL FOR SOA EXAM P/CAS EXAM 1 SPRING 2008Table of Contents Introductory CommentsSection 2 - Conditional Probability and Independence© ACTEX 2008SOA Exam P/CAS Exam 1 - ProbabilityTABLE OF CONTENTSINTRODUCTORY COMMENTS SECTION 0 - REVIEW OF ALGEBRA AND CALCULUSSet Theory Graphing an Inequality in Two Dimensions Properties of Functions Limits and Continuity Differentiation Integration Geometric and Arithmetic Progressions Problem Set 0 and Solutions 1 9 10 13 14 17 23 25SECTION 1 - BASIC PROBABILITY CONCEPTSProbability Spaces and Events Probability 35 39 49Problem Set 1 and Solutions SECTION 2 - CONDITIONAL PROBABILITY AND INDEPENDENCEDefinition of Conditional Probability Bayes' Rule, Bayes' Theorem and the Law of Total Probability Independent EventsProblem Set 2 and Solutions SECTION 3 - COMBINATORIAL PRINCIPLESPermutations and Combinations57 59 63 67Problem Set 3 and Solutions SECTION 4 - RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONSDiscrete Random Variable Continuous Random Variable Mixed Distribution Cumulative Distribution Function Independent Random Variables83 89Problem Set 4 and Solutions SECTION 5 - EXPECTATION AND OTHER DISTRIBUTION PARAMETERSExpected Value Moments of a Random Variable Variance and Standard Deviation Moment Generating Function Percentiles, Median and Mode93 94 96 96 101 109Problem Set 5 and Solutions © ACTEX 2008115 116 117 119 121 131SOA Exam P/CAS Exam 1 - ProbabilitySECTION 6 - FREQUENTLY USED DISCRETE DISTRIBUTIONSDiscrete Uniform Distribution Binomial Distribution Poisson Distribution Geometric Distribution Negative Binomial Distribution Hypergeometric Distribution Multinomial Distribution Summary of Discrete Distributions 143 144 147 150 151 153 154 155 157Problem Set 6 and Solutions SECTION 7 - FREQUENTLY USED CONTINUOUS DISTRIBUTIONSContinuous Uniform Distribution Normal Distribution Approximating a Distribution Using a Normal Distribution Exponential Distribution Gamma Distribution Pareto Distribution Beta Distribution Lognormal and Weibull Distributions Chi-Square Distribution Summary of Continuous DistributionsProblem Set 7 and Solutions SECTION 8 - JOINT, MARGINAL, AND CONDITIONAL DISTRIBUTIONSDefinition of Joint Distribution Expectation of a Function of Jointly Distributed Random Variables Marginal Distributions Independence of Random Variables Conditional Distributions Covariance and Correlation Between Random Variables Moment Generating Function for a Joint Distribution Bivariate Normal Distribution169 169 172 176 179 180 181 182 183 186 187Problem Set 8 and Solutions SECTION 9 - TRANSFORMATIONS OF RANDOM VARIABLESDistribution of a Transformation of \ Distribution of a Transformation of Joint Distribution of \ and ] Distribution of a Sum of Random Variables Distribution of the Maximum or Minimum of Independent Ö\&quot; ß \# ß ÞÞÞß \8 × Order Statistics Mixtures of Distributions197 200 201 204 205 209 220 220 215Problem Set 9 and Solutions237 238 239 243 244 247 249© ACTEX 2008SOA Exam P/CAS Exam 1 - ProbabilitySECTION 10 - RISK MANAGEMENT CONCEPTSLoss Distributions and Insurance The Normal Approximation to Aggregate Claims Insurance Policy Deductible Insurance Policy Limit Proportional Insurance 267 272 275 277 278 281Problem Set 10 and SolutionsTABLE FOR THE NORMAL DISTRIBUTIONPRACTICE EXAM 1 PRACTICE EXAM 2 PRACTICE EXAM 3 PRACTICE EXAM 4 PRACTICE EXAM 5 PRACTICE EXAM 6 PRACTICE EXAM 7303 323 345 373 393 411 431© ACTEX 2008SOA Exam P/CAS Exam 1 - ProbabilityINTRODUCTORY COMMENTSThis study guide is designed to help in the preparation for the Society of Actuaries Exam PCasualty Actuarial Society Exam 1. The study manual is divided into two main parts. The first part consists of a summary of notes and illustrative examples related to the material described in the exam catalog as well as a series of problem sets and detailed solutions related to each topic. The second part of the study manual consists of seven practice exams, with detailed solutions, which are designed to cover the range of material that will be found on the exam. The questions on these practice exams are mainly taken from old Society exams on the probability topic. Between the section of notes and the section with practice exams I have included the normal distribution table provided with the exam.In a couple of places in this manual there are paragraphs that are indented from the usual margin. These are topics that are mainly for additional background and are less likely to arise inan exam question. I have attempted to be thorough in the coverage of the topics upon which the exam is based. I have been, perhaps, more thorough than necessary on a couple of topics, particularly order statistics in Section 9 of the notes and some risk management topics in Section 10 of the notes. Section 0 of the notes provides a brief review of a few important topics in calculus and algebra. This manual will be most effective, however, for those who have had courses in college calculus at least to the sophomore level and courses in probability to the sophomore or junior level. If you are taking the Exam P for the first time, be aware that a most crucial aspect of the exam is the limited time given to take the exam (3 hours). It is important to be able to work very quickly and accurately. Continual drill on important concepts and formulas by working through many problems will be helpful. It is also very important to be disciplined enough while taking the exam so that an inordinate amount of time is not spent on any one question. If the formulas and reasoning that will be needed to solve a particular question are not clear within 2 or 3 minutes of starting the question, it should be abandoned (and returned to later if time permits). Using the exams in the second part of this study manual and simulating exam conditions will also help give you a feeling for the actual exam experience. If you have any comments, criticisms or compliments regarding this study guide, please contact the publisher, ACTEX, or you may contact me directly at the address below. I apologize in advance for any errors, typographical or otherwise, that you might find, and it would be greatly appreciated if you bring them to my attention. Any errors that are found will be posted in an errata file at the ACTEX website, www.actexmadriver.com . It is my sincere hope that you find this study guide helpful and useful in your preparation for the exam. I wish you the best of luck on the exam. Samuel A. Broverman Department of Statistics University of Toronto www.sambroverman.com October, 2007 E-mail: [email protected] or [email protected]© ACTEX 2008SOA Exam P/CAS Exam 1 - ProbabilitySECTION 2 - CONDITIONAL PROBABILITY AND INDEPENDENCEConditional probability of event F given event E: If T ÒEÓ  !, then T ÒFlEÓ oeT ÒFEÓ T ÒEÓis defined to be the conditional probability that event F occurs given that event E has occurred. Rewriting the equation results in T ÒF  EÓ oe T ÒFlEÓ  T ÒEÓ , which is referred to as the multiplication rule. When we condition on event E, we are assuming that event F has occurred so that E becomes the new probability space, and all conditional events must take place within event E (the new probability space). Dividing by T ÒEÓ scales all probabilities so that E is the entire probability space, and T ÒElEÓ oe &quot;. To say that event F has occurred given that event E has occurred means that both F and E (F  E) have occurred within the probability space E. This is the explanation behind the definition of the conditional probability T ÒFlEÓ.Example 2-1: Suppose that a fair six-sided die is tossed. The probability space is W oe Ö&quot;ß #ß \$ß %ß &amp;ß '×. We define the following events: E oe &quot;the number tossed is Y \$&quot; oe Ö&quot;ß #ß \$× , F oe &quot;the number tossed is even&quot; oe Ö#ß %ß '× , G oe &quot;the number tossed is a &quot; or a #&quot; oe Ö&quot;ß #× , H oe &quot;the number tossed doesn't start with the letters 'f' or 't'&quot; oe Ö&quot;ß '× . The conditional probability of E given F is T ÒElFÓ oeT ÒÖ&quot;ß#ß\$×Ö#ß%ß'×Ó T ÒÖ#ß%ß'×Ó &quot; oe T ÒÖ#ß%ß'×Ó oe &quot;Î# oe \$ . The interpretation of this conditional T ÒÖ#×Ó &quot;Î'probability is that if we know that event F has occurred, then the toss must be 2, 4 or 6. Since the original 6 possible tosses of a die were equally likely, if we are given the additional information that the toss is 2, 4 or 6, it seems reasonable that each of those is equally, each with a probability of &quot; . Then within the reduced probability space F , the (conditional) probability that a event E \$ occurs is the probability, in the reduced space, of tossing a 2; this is &quot; . \$ The conditional probability of E given G is T ÒElGÓ oe &quot;. To say that G has occurred means that the toss is 1 or 2. It is then guaranteed that event E has occurred, since G § E. The conditional probability of F given G is T ÒFlGÓ oe &quot; . #© ACTEX 2008SOA Exam P/CAS Exam 1 - Probability&amp; ( Example 2-2: If T ÒEÓ oe &quot; and T ÒFÓ oe &quot;# , and T ÒElFÓ  T ÒFlEÓ oe &quot;! , 'find T ÒE  FÓ . Solution: T ÒFlEÓ oeT ÒEFÓ T ÒEÓoe 'T ÒE  FÓ and T ÒElFÓ oeT ÒEFÓ T ÒFÓoe &quot;# T ÒE  FÓ &amp;( &quot; p Ð'  &quot;# Ñ  T ÒE  FÓ oe &quot;! p T ÒE  FÓ oe &quot;2 Þ &amp;IMPORTANT NOTE: The following manipulation of event probabilities arises from time to time: T ÒEÓ oe T ÒElFÓ  T ÐFÑ  T ÒElF w Ó  T ÐF w Ñ . This relationship is a version of the Law of Total Probability. This relationship is valid since for any events F and E, we have T ÒEÓ oe T ÒE  FÓ  T ÒE  F w Ó . We then use the relationships T ÒE  FÓ oe T ÒElFÓ  T ÐFÑ and T ÒE  F w Ó oe T ÒElF w Ó  T ÐF w Ñ . If we know the conditional probabilities for event E given some other event F and if we also know the conditional probability of E given the complement F w , and if we are given the (unconditional) probability of event F , then we can find the (unconditional) probability of event E. An application of this concept occurs when an experiment has two (or more) steps. The following example illustrates this idea.Example 2-3: Urn I contains 2 white and 2 black balls and Urn II contains 3 white 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 white. Solution: Let F be the event that Urn I is chosen and F w 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 &quot;an Urn is chosen at random&quot;). Therefore, T ÒFÓ oe &quot; and T ÒF w Ó oe &quot; . Let E be the event that the # # ball chosen in white. If we know that Urn I was chosen, then there is &quot; probability of choosing a # white ball (2 white out of 4 balls, it is assumed that each ball has the same chance of being chosen); this can be described as T ÒElFÓ oe &quot; . In a similar way, if Urn II is chosen, then # T ÒElF w Ó oe \$ (3 white out of 5 balls). We can now apply the relationship described prior to this &amp; example. T ÒE  FÓ oe T ÒElFÓ  T ÒFÓ oe Ð &quot; ÑÐ &quot; Ñ oe &quot; , and # # %\$ T ÒE  F w Ó oe T ÒElF w Ó  T ÒF w Ó oe Ð \$ ÑÐ &quot; Ñ oe &quot;! . Finally, &amp; # \$ T ÒEÓ oe T ÒE  FÓ  T ÒE  F w Ó oe &quot;  &quot;! oe &quot;&quot; . % #! The order of calculations can be summarized in the following table E F 1. T ÐE  FÑ oe T ÒElFÓ  T ÒFÓFw2. T ÐE  F w Ñ oe T ÒElF w Ó  T ÒF w Ó 3. T ÐEÑ oe T ÐE  FÑ  T ÐE  F w Ñ© ACTEX 2008SOA Exam P/CAS Exam 1 - ProbabilityBayes' rule and Bayes' Theorem (basic form): For any events E and F with T ÒEÓ  !, T ÒFlEÓ oeT ÒFEÓ T ÒEÓoeT ÒElFÓT ÒFÓ T ÒEÓ.The usual way that this is applied is in the case that we are given the values of T ÒFÓ , T ÒElFÓ and T ÒElF w Ó (from T ÒFÓ we get T ÒF w Ó oe &quot;  T ÒFÓ), and we are asked to find T ÒFlEÓ (in other words, we are asked to &quot;turn around&quot; the conditioning of the events E and F ). We can summarize this process by calculating the quantities in the following table in the order indicated numerically (1-2-3-4) (other entries in the table are not necessary in this calculation, but might be needed in related calculations). E F ß T ÒFÓ Fw ß T ÒF w Ó &quot;  T ÒFÓ T ÒElF w Ó 2. T ÒE  F w Ó oe T ÒElF w Ó  T ÒF w Ó 3. T ÒEÓ oe T ÒE  FÓ  T ÒE  F w Ó T ÒF w Ó oe &quot;  T ÒFÓ Step 4: T ÒFlEÓ oeT ÒFEÓ T ÒEÓEwT ÒElFÓ 1. T ÒE  FÓ oe T ÒElFÓ  T ÒFÓT ÒEw lFÓ oe &quot;  T ÒElFÓ T ÒEw  FÓ oe T ÒEw lFÓ  T ÒFÓT ÒEw lF w Ó oe &quot;  T ÒElF w Ó T ÒEw  F w Ó oe T ÒEw lF w Ó  T ÒF w Ó T ÒEw Ó oe &quot;  T ÒEÓ.This can also be summarized in the following sequence of calculations. T ÒFÓ , T ÒElFÓ , given Ì T ÒE  FÓ oe T ÒElFÓ  T ÒFÓ Ì T ÒEÓ oe T ÒE  FÓ  T ÒE  F w Ó Algebraically, we have done the following calculation: T ÒFlEÓ oeT ÒFEÓ T ÒEÓT ÒF w Ó oe &quot;  T ÒFÓ , T ÒElF w Ó , given Ì T ÒE  F w Ó oe T ÒElF w Ó  T ÒF w Óoe T ÒEFÓT ÒEF w Ó oe T ÒElFÓT ÒFÓT ÒElF w ÓT ÒF w Ó ,T ÒEFÓT ÒElFÓT ÒFÓwhere all the factors in the final expression were originally known. Note that the numerator is one of the components of the denominator.© ACTEX 2008SOA Exam P/CAS Exam 1 - ProbabilityExam questions on this topic (and the extended form of Bayes' rule reviewed below) have occurred quite regularly. The key starting point is identifying and labeling unconditional events and conditional events and their probabilities in an efficient way.Example 2-4: Urn I contains 2 white and 2 black balls and Urn II contains 3 white 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 white. Find the probability that the ball transferred from Urn I to Urn II was white. Solution: Let F denote the event that the ball transferred from Urn I to Urn II was white and let E denote the event that the ball chosen from Urn II is white. We are asked to find T ÒFlEÓ . From the simple nature of the situation (and the usual assumption of uniformity in such a situation, meaning that all balls are equally likely to be chosen from Urn I in the first step), we have T ÒFÓ oe &quot; (2 of the 4 balls in Urn I are white), and T ÒF w Ó oe &quot; . # # If the ball transferred is white, then Urn II has 4 white and 2 black balls, and the probability of choosing a white ball out of Urn II is # ; this is T ÒElFÓ oe # . \$ \$ If the ball transferred is black, then Urn II has 3 white and 3 black balls, and the probability of choosing a white ball out of Urn II is &quot; ; this is T ÒElF w Ó oe &quot; . # # All of the information needed has been identified. From the table described above, we do the calculations in the following order: 1. 2. 3. 4. T ÒE  FÓ oe T ÒElFÓ  T ÒFÓ oe Ð # ÑÐ &quot; Ñ oe &quot; \$ # \$ T ÒE  F w Ó oe T ÒElF w Ó  T ÒF w Ó oe Ð &quot; ÑÐ &quot; Ñ oe &quot; # # %( T ÒEÓ oe T ÒE  FÓ  T ÒE  F w Ó oe &quot;  &quot; oe &quot;# \$ % T ÒFEÓ &quot;Î\$ % T ÒFlEÓ oe T ÒEÓ oe (Î&quot;# oe ( .Example 2-5: Identical twins come from the same egg and hence are of the same sex. Fraternal twins have a 50-50 chance of being the same sex. Among twins, the probability of a fraternal set is : and an identical set is ; oe &quot;  : . If the next set of twins are of the same sex, what is the probability that they are identical?© ACTEX 2008SOA Exam P/CAS Exam 1 - ProbabilitySolution: Let E be the event &quot;the next set of twins are of the same sex&quot;, and let F be the event &quot;the next sets of twins are identical&quot;. We are given T ÒElFÓ oe &quot; ß T ÒElF w Ó oe Þ&amp; T ÒFÓ oe ; ß T ÒF w Ó oe : oe &quot;  ; . Then T ÒFlEÓ oeT ÒEFÓ T ÒEÓis the probability we are asked tofind. But T ÒE  FÓ oe T ÒElFÓ  T ÒFÓ oe ; , and T ÒE  F w Ó oe T ÒElF w Ó  T ÒF w Ó oe Þ&amp;: . Thus, T ÒEÓ oe T ÒE  FÓ  T ÒE  F w Ó oe ;  Þ&amp;: oe ;  Þ&amp;Ð&quot;  ;Ñ oe Þ&amp;Ð&quot;  ;Ñ , and T ÒFlEÓ oe Þ&amp;Ð&quot;;Ñ . This can be summarized in the following table E oe Same sex F oe identical prob. ; T ÒElFÓ oe &quot; (given) , T ÒE  FÓ oe T ÒElFÓ  T ÒFÓ oe ; T ÒElF w Ó oe Þ&amp; (given) , T ÒE  F w Ó oe T ÒElF w Ó  T ÒF w Ó oe Þ&amp;Ð&quot;  ;Ñ Ì T ÒEÓ oe T ÒE  FÓ  T ÒE  F w Ó oe ;  Þ&amp;Ð&quot;  ;Ñ oe Þ&amp;Ð&quot;  ;Ñ . T ÒFlEÓ oeT ÒEFÓ T ÒEÓ ;Ew oe not same sex T ÒEw lFÓ oe &quot;  T ÒElFÓ oe !F w oe fraternal prob. : oe &quot;  ;T ÒEw lF w Ó oe &quot;  T ÒEw lF w Ó oe Þ&amp;oe Þ&amp;Ð&quot;;Ñ .;Bayes' rule and Bayes' Theorem (extended form): If F&quot; ß F# ß ÞÞÞß F8 form a partition of the entire probability space W , then T ÒF4 lEÓ oeT ÒEF4 Ó T ÒEÓoeT ÒEF4 Ó8oeT ÒElF4 ÓT ÒF4 Ó8for each 4 oe &quot;ß #ß ÞÞÞß 8.T ÒEF3 Ó3oe&quot;T ÒElF3 ÓT ÒF3 Ó3oe&quot;For example, if the F 's form a partition of 8 oe \$ events, we have T ÒF&quot; lEÓ oeT ÒF&quot; EÓ T ÒEÓ&quot; &quot; oe T ÒEF ÓT ÒEF ÓT ÒEF Ó &quot; # \$T ÒElF ÓT ÒF Óoe T ÒElF ÓT ÒF ÓT ÒElF&quot; ÓT ÒF&quot; ÓT ÒElF ÓT ÒF Ó &quot; &quot; # # \$ \$8T ÒElF ÓT ÒF ÓThe relationship in the denominator, T ÒEÓ oeT ÒElF3 Ó  T ÒF3 Ó is the general Law of Total3oe&quot;Probability. The values of T ÒF4 Ó are called prior probabilities, and the value of T ÒF4 lEÓ is called a posterior probability. The basic form of Bayes' rule is just the case in which the partition consists of two events, F and F w . The main application of Bayes' rule occurs in the situation in which the T ÒF4 Ó probabilities are known and the T ÒElF4 Ó probabilities are known, and we are asked to find T ÒF3 lEÓ for one of the 3's. The series of calculations can be summarized in a table as in the basic form of Bayes' rule. This is illustrated in the following example.© ACTEX 2008SOA Exam P/CAS Exam 1 - ProbabilityExample 2-6: Three dice have the following probabilities of throwing a &quot;six&quot;: : ß ; ß &lt; ß respectively. One of the dice is chosen at random and thrown (each is equally likely to be chosen). A &quot;six&quot; appeared. What is the probability that the die chosen was the first one? Solution: The event &quot; a 6 is thrown&quot; is denoted by &quot;6&quot; T Òdie &quot;l&quot;'&quot;Ó oeT ÒÐdie &quot;ÑÐ&quot;'&quot;ÑÓ T Ò&quot;'&quot; Ó T Ò&quot;'&quot;ldie &quot;ÓT Òdie &quot;Ó T Ò&quot;'&quot;Ó\$ oe T Ò&quot;'&quot;Ó .oe: &quot;But T Ò&quot;'&quot;Ó oe T ÒÐ&quot;'&quot;Ñ  Ðdie &quot;ÑÓ  T ÒÐ&quot;'&quot;Ñ  Ðdie #ÑÓ  T ÒÐ&quot;'&quot;Ñ  Ðdie \$ÑÓ oe T Ò&quot;'&quot;ldie &quot;Ó  T Òdie &quot;Ó  T Ò&quot;'&quot;ldie #Ó  T Òdie #Ó  T Ò&quot;'&quot;ldie \$Ó  T Òdie \$Ó oe :&quot; \$;&quot; \$&lt;&quot; \$oe:;&lt; \$\$ \$ Ê T Òdie &quot;l&quot;'&quot;Ó oe T Ò&quot;'&quot;Ó oe Ð:;&lt;Ñ &quot; oe :;&lt; . \$: &quot;: &quot;:These calculations can be summarized in the following table. Die 1 , T Ðdie 1) oe &quot; \$ (given) T Ò&quot;6&quot;ldie 1Ó oe : (given) T Ò&quot;6&quot;  die 1Ó oe T Ò&quot;6&quot;ldie 1Ó  T Òdie 1Ó oe: &quot; \$ Die 2 , T Ðdie 2) oe &quot; \$ (given) T Ò&quot;6&quot;ldie 2Ó oe ; (given) T Ò&quot;6&quot;  die 2Ó oe T Ò&quot;6&quot;ldie 2Ó  T Òdie 2Ó oe; &quot; \$ Die 3 , T Ðdie 3) oe &quot; \$ (given) T Ò&quot;6&quot;ldie 3Ó oe &lt; (given) T Ò&quot;6&quot;  die 3Ó oe T Ò&quot;6&quot;ldie 3Ó  T Òdie 3Ó oe&lt; &quot; \$Toss &quot;6&quot;T Ò&quot;6&quot;Ó oe :  &quot;  ;  &quot;  &lt;  &quot; oe &quot; Ð:  ;  &lt;Ñ . \$ \$ \$ \$ In terms of Venn diagrams, the conditional probability is the ratio of the shaded area probability in the first diagram to the shaded area probability in the second diagram.© ACTEX 2008SOA Exam P/CAS Exam 1 - ProbabilityIn Example 2-6 there is a certain symmetry to the situation and general reasoning can provide a shortened solution. In the conditional probability T Òdie &quot;l&quot;'&quot;Ó oeT ÒÐdie &quot;ÑÐ&quot;'&quot;ÑÓ T Ò&quot;'&quot; Ó, we can thinkof the denominator as the combination of the three possible ways a &quot;6&quot; can occur, :  ;  &lt; , and we can think of the numerator as the &quot;6&quot; occurring from die 1, with probability :. Then the conditional probability is the fraction :;&lt; . The symmetry involved here is in the assumption that each die was equally likely to be chosen, so there is a &quot; chance of any one die being chosen. \$: \$ This factor of &quot; cancels in the numerator and denominator of Ð:;&lt;Ñ &quot; . If we had not had this \$ \$&quot;:symmetry, we would have to apply different &quot;weights&quot; to the three dice. Another example of this sort of symmetry is a variation on Example 2-3 above. Suppose that Urn I has 2 white and 3 black balls and Urn II has 4 white and 1 black balls. An Urn is chosen at random and a ball is chosen. The reader should verify using the usual conditional probability' rules that the probability of choosing a white is &quot;! . This can also be seen by noting that if weconsider the 10 balls together, 6 of them are white, so that the chance of picking a white out of' the 20 is &quot;! . This worked because of two aspects of symmetry, equal chance for picking eachUrn, and same number of balls in each Urn.Independent events E and F : If events E and F satisfy the relationship T ÒE  FÓ oe T ÒEÓ  T ÒFÓ , 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 ÒElFÓ oe T ÒEÓ , and also is equivalent to T ÒFlEÓ oe T ÒFÓ .© ACTEX 2008SOA Exam P/CAS Exam 1 - ProbabilityExample 2-1 continued: A fair six-sided die is tossed. E oe &quot;the number tossed is Y \$&quot; oe Ö&quot;ß #ß \$× , F oe &quot;the number tossed is even&quot; oe Ö#ß %ß '× , G oe &quot;the number tossed is a &quot; or a #&quot; oe Ö&quot;ß #× , H oe &quot;the number tossed doesn't start with the letters 'f' or 't'&quot; oe Ö&quot;ß '× . We have the following probabilities: T ÒEÓ oe &quot; ß T ÒFÓ oe &quot; ß T ÒGÓ oe &quot; ß T ÒHÓ oe &quot; Þ # # \$ \$ Events E and F are not independent since &quot; oe T ÒE  FÓ Á T ÒEÓ  T ÒFÓ oe &quot;  &quot; oe &quot; . We also ' # # % &quot; &quot; see that E and F are not independent because T ÒElFÓ oe \$ Á # oe T ÒEÓ Þ Also, E and G are not independent, since T ÒE  GÓ oe &quot; Á &quot;  &quot; oe T ÒEÓ  T ÒGÓ (also since \$ # \$ T ÒElGÓ oe &quot; Á &quot; oe T ÒEÓ). # Events F and G are independent, since T ÒF  GÓ oe &quot; oe &quot;  &quot; oe T ÒFÓ  T ÒGÓ (alternatively, ' # \$ &quot; T ÒFlGÓ oe # oe T ÒFÓ , so that F and G are independent). The reader should check that both E and F are independent of H.Mutually independent events E&quot; ß E# ß ÞÞÞß E8 : If events E&quot; ß E# ß ÞÞÞß E8 satisfy the relationship T ÒE&quot;  E#  â  E8 Ó oe T ÒE&quot; Ó  T ÒE# ÓâT ÒE8 Ó oe # T ÒE3 Ó , then8 3oe&quot;the events are said to be mutually independent.Some rules concerning conditional probability and independence are: (i) T ÒE  FÓ oe T ÒFlEÓ  T ÒEÓ oe T ÒElFÓ  T ÒFÓ for any events E and F (ii) If T ÒE&quot;  E#  â  E8&quot; Ó  ! , then T ÒE&quot;  E#  â  E8 Ó oe T ÒE&quot; Ó  T ÒE# lE&quot; Ó  T ÒE\$ lE&quot;  E# ÓâT ÒE8 lE&quot;  E#  â  E8&quot;Ó (iii) T ÒEw lFÓ oe &quot;  T ÒElFÓ (iv) T ÒE  FlGÓ oe T ÒElGÓ  T ÒFlGÓ  T ÒE  FlGÓ ; properties (iv) and (v) are the same properties satisfied by unconditional events (v) if E § F then T ÒElFÓ oeT ÒEFÓ T ÒFÓoe T ÒFÓ , and T ÒFlEÓ oe &quot;T ÒEÓ(vi) if E and F are independent events then Ew and F are independent events, E and F w are independent events, and Ew and F w are independent events (vii) since T ÒgÓ oe T Òg  EÓ oe ! oe T ÒgÓ  T ÒEÓ for any event E, it follows that g is independent of any event E© ACTEX 2008SOA Exam P/CAS Exam 1 - ProbabilityExample 2-7: Suppose that events E and F are independent. Find the probability, in terms of T ÒEÓ and T ÒFÓ, that exactly one of the events E and F occurs. Solution: T Òexactly one of E and FÓ oe T ÒÐE  F w Ñ  ÐEw  FÑÓ . Since E  F w and F  Ew are mutually exclusive, it follows that T Òexactly one of E and FÓ oe T ÒE  F w Ó  T ÒEw  FÓ . Since E and F are independent, it follows that E and F w are also independent, as are F and Ew . Then T ÒÐE  F w Ñ  ÐEw  FÑÓ oe T ÒEÓ  T ÒF w Ó  T ÒFÓ  T ÒEw Ó oe T ÒEÓÐ&quot;  T ÒFÓÑ  T ÒFÓÐ&quot;  T ÒEÓÑ oe T ÒEÓ  T ÒFÓ  #T ÒEÓ  T ÒFÓExample 2-8: A survey is made to determine the number of households having electric appliances in a certain city. It is found that 75% have radios (V ), 65% have irons (M ), 55% have electric toasters (X ), 50% have (MV ), 40% have (VX ), 30% have (MX ), and 20% have all three. Find the following proportions. (i) Of those households that have a toaster, find the proportion that also have a radio. (ii) Of those households that have a toaster but no iron, find the proportion that have a radio. Solution: This is a continuation of Example 1-3 given earlier in the study guide. The diagram below deconstructs the three events.(i) This is T ÒVlX Ó . The language &quot;of those households that have a toaster&quot; means, &quot;given that the household has a toaster&quot;, so we are being asked for a conditional probability. Then, T ÒVlX Ó oeT ÒVX Ó T ÒX Ó Þ% ) oe Þ&amp;&amp; oe &quot;&quot; Þ T ÒVX M w Ó T ÒX M w Ó Þ# % oe Þ#&amp; oe &amp; Þ(ii) This T ÒVlX  M w Ó oe© ACTEX 2008SOA Exam P/CAS Exam 1 - ProbabilityExample 2-9: In a survey of 94 students, the following data was obtained. 60 took English, 56 took Math, 42 took Chemistry, 34 took English and Math, 20 took Math and Chemistry, 16 took English and Chemistry, 6 took all three subjects. Find the following proportions. (i) Of those who took Math, the proportion who took neither English nor Chemistry, (ii) Of those who took English or Math, the proportion who also took Chemistry. Solution: The following diagram illustrates how the numbers of students can be deconstructed. We calculate proportion of the numbers in the various subsets.(i) 56 students took Math, and 8 of them took neither English nor Chemistry. T ÐI w  G w lQ Ñ oeT ÐI w G w Q Ñ T ÐQ Ñ ) &quot; oe &amp;' oe ( .(ii) 82 ( oe )  &quot;%  '  #)  &quot;'  &quot;! in I  Q Ñ students took English or Math (or both), and 30 of them ( oe &quot;%  '  &quot;! in ÐI  Q Ñ  G ) also took Chemistry . T ÐGlI  Q Ñ oeT ÒGÐIQ ÑÓ T ÐIQ Ñ \$! &quot;&amp; oe )# oe %&quot; Þ© ACTEX 2008SOA Exam P/CAS Exam 1 - Probability`

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