SOCIETY OF ACTUARIES/CASUALTY ACTUARIAL SOCIETY EXAM P PROBABILITY P SAMPLE EXAM SOLUTIONS

SOCIETY OF ACTUARIES/CASUALTY ACTUARIAL SOCIETY EXAM P PROBABILITY P SAMPLE EXAM SOLUTIONS Copyright 9 by the Society of Actuaries and the Casualty Actuarial Society Some of the questions in this study note are taken from past SOA/CAS eaminations. PRINTED IN U.S.A. of 66

. Solution: D Let G event that a viewer watched gymnastics B event that a viewer watched baseball S event that a viewer watched soccer Then we want to find c Pr ( G B S) Pr ( G B S ) Pr G + Pr B + Pr S Pr G B Pr G S Pr B S + Pr G B S.8 +.9 +.9.4.. +.8.48.5 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ). Solution: A Let R event of referral to a specialist L event of lab work We want to find P[R L] P[R] + P[L] P[R L] P[R] + P[L] + P[~(R L)] P[R] + P[L] + P[~R ~L]. +.4 +.5.5.. Solution: D First note P[ A B] P[ A] + P[ B] P[ A B] P[ A B' ] P[ A] + P[ B' ] P[ A B' ] Then add these two equations to get P A B + P A B' P A + P B + P B' P A B + P A B' [ ] [ ] [ ] ( [ ] [ ]) ( [ ] [ ]) + P[ A] + P ( A B) ( A B ) P[ A] + P[ A] [ ].6.7.9 '.6 P A of 66

4. Solution: A For i,, let Ri event that a red ball is drawn form urn i Bi event that a blue ball is drawn from urn i. Then if is the number of blue balls in urn,.44 Pr[ R R B B ] Pr[ R R ] + Pr B B Pr[ R] Pr[ R] + Pr[ B] Pr[ B] 4 6 6 + + 6 + 6 Therefore, +. + + 6 + 6 + 6.+ 5. +.8. 4 ( ) ( ) [ ] 5. Solution: D Let N(C) denote the number of policyholders in classification C. Then N(Young Female Single) N(Young Female) N(Young Female Married) N(Young) N(Young Male) [N(Young Married) N(Young Married Male)] (4 6) 88. 6. Solution: B Let H event that a death is due to heart disease F event that at least one parent suffered from heart disease Then based on the medical records, c 8 P H F 97 97 c 97 65 P F 97 97 c P H F c 8 65 8 and P H F.7 c P F 97 97 65 of 66

7. Solution: D Let A event that a policyholder has an auto policy H event that a policyholder has a homeowners policy Then based on the information given, Pr A H.5 ( ) ( A c H ) ( A) ( A H) c ( A H) ( H) ( A H) Pr Pr Pr.65.5.5 Pr Pr Pr.5.5.5 and the portion of policyholders that will renew at least one policy is given by c c.4 Pr A H +.6 Pr A H +.8 Pr A H ( ) ( ) ( ) (.4)(.5) (.6)(.5) (.8)(.5).5 ( 5% ) + + 9 B-9 8. Solution: D Let C event that patient visits a chiropractor T event that patient visits a physical therapist We are given that Pr[ C] Pr[ T] +.4 Pr C T. ( ) c c ( C T ) Pr. Therefore, c c.88 Pr C T Pr C T Pr C + Pr T Pr C T Pr[ T] +.4 + Pr[ T]. Pr[ T ].8 or Pr[ T ] (.88 +.8).48 [ ] [ ] [ ] [ ] 4 of 66

9. Solution: B Let M event that customer insures more than one car S event that customer insures a sports car Then applying DeMorgan s Law, we may compute the desired probability as follows: c c c Pr M S Pr M S Pr M S Pr M + Pr S Pr M S Pr M Pr S + Pr S M Pr M.7. +.5.7.5 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ). Solution: C Consider the following events about a randomly selected auto insurance customer: A customer insures more than one car B customer insures a sports car We want to find the probability of the complement of A intersecting the complement of B (eactly one car, non-sports). But P ( A c B c ) P (A B) And, by the Additive Law, P ( A B ) P ( A) + P ( B ) P ( A B ). By the Multiplicative Law, P ( A B ) P ( B A ) P (A).5 *.64.96 It follows that P ( A B ).64 +..96.744 and P (A c B c ).744.56. Solution: B Let C Event that a policyholder buys collision coverage D Event that a policyholder buys disability coverage Then we are given that P[C] P[D] and P[C D].5. By the independence of C and D, it therefore follows that.5 P[C D] P[C] P[D] P[D] P[D] (P[D]) (P[D]).5/.75 P[D].75 and P[C] P[D].75 Now the independence of C and D also implies the independence of C C and D C. As a result, we see that P[C C D C ] P[C C ] P[D C ] ( P[C]) ( P[D]) (.75 ) (.75 ).. 5 of 66

. Solution: E Boed numbers in the table below were computed. High BP Low BP Norm BP Total Regular heartbeat.9..56.85 Irregular heartbeat.5..8.5 Total.4..64. From the table, we can see that % of patients have a regular heartbeat and low blood pressure.. Solution: C The Venn diagram below summarizes the unconditional probabilities described in the problem. In addition, we are told that P[ A B C] P[ A B C A B] P A B +. [ ] It follows that ( +.) +.4.4.6 Now we want to find c P ( A B C) c c P ( A B C) A c P A P[ A B C] P A [ ] ( ) ( ) ( )...6...6.8.467.6 6 of 66

4. Solution: A p k pk pk pk... p k 5 55 5 5 5 5 k p 5 pk p p k k 5 4 5 p 4/5. Therefore, P[N > ] P[N ] (4/5 + 4/5 /5) 4/5 /5.4. 5. Solution: C A Venn diagram for this situation looks like: k We want to find w ( + y+ z) 5 We have + y, + z, y+ z 4 Adding these three equations gives 5 ( + y) + ( + z) + ( y+ z) + + 4 + y+ z ( ) + y+ z w ( + y+ z) Alternatively the three equations can be solved to give /, y /6, z /4 again leading to w + + 6 4 7 of 66

6. Solution: D Let N and N denote the number of claims during weeks one and two, respectively. Then since N and N are independent, 7 Pr N + N 7 Pr N n Pr N 7 n [ ] [ ] [ n ] 7 n+ 8 n n 7 n 9 8 9 6 64 7. Solution: D Let O Event of operating room charges E Event of emergency room charges Then.85 Pr ( O E) Pr ( O) + Pr ( E) Pr ( O E) Pr ( O) + Pr ( E) Pr ( O) Pr ( E) ( Independence) Since Pr ( E c ).5 Pr ( E), it follows Pr ( E ).75..85 Pr O +.75 Pr O.75 So ( ) ( )( ) Pr ( O)(.75). Pr ( O ).4 8. Solution: D Let X and X denote the measurement errors of the less and more accurate instruments, respectively. If N(μ,σ) denotes a normal random variable with mean μ and standard deviation σ, then we are given X is N(,.56h), X is N(,.44h) and X, X are X+ X.56 h +.44 h independent. It follows that Y is N (, ) N(, 4.56h). Therefore, P[.5h Y.5h] P[Y.5h] P[Y.5h] P[Y.5h] P[Y.5h].5h P[Y.5h] P Z.56h P[Z.4] (.99).84. 8 of 66

9. Solution: B Apply Bayes Formula. Let A Event of an accident B Event the driver s age is in the range 6- B Event the driver s age is in the range - B Event the driver s age is in the range -65 B 4 Event the driver s age is in the range 66-99 Then Pr ( AB) Pr ( B) Pr ( B A) Pr AB Pr B + Pr AB Pr B + Pr AB Pr B + Pr AB Pr B ( ) ( ) ( ) ( ) ( ) ( ) ( 4) ( 4) (.6)(.8) (.6)(.8) + (.)(.5) + (.)(.49) + (.4)(.8).584. Solution: D Let S Event of a standard policy F Event of a preferred policy U Event of an ultra-preferred policy D Event that a policyholder dies Then P[ D U] P[ U] PU [ D] P D S P S + P D F P F + P D U P U [ ] [ ] [ ] [ ] [ ] [ ] (.)(.) (.)(.5) + (.5)(.4) + (.)(.).4. Solution: B Apply Baye s Formula: Pr Seri. Surv. Pr Surv. Seri. Pr[ Seri. ] Pr Surv. Crit. Pr[ Crit. ] + Pr Surv. Seri. Pr[ Seri. ] + Pr Surv. Stab. Pr [ Stab. ] (.9)(.).9 (.6)(.) + (.9)(.) + (.99)(.6) 9 of 66

. Solution: D Let H Event of a heavy smoker L Event of a light smoker N Event of a non-smoker D Event of a death within five-year period Now we are given that Pr DL Pr DN and Pr DL Pr DH Therefore, upon applying Bayes Formula, we find that Pr DH Pr[ H] Pr HD Pr DN Pr[ N] + Pr DL Pr[ L] + Pr DH Pr[ H] ( ) Pr DL..4.4 Pr DL.5..4 (.5) + Pr DL (.) + Pr DL (. + + ). Solution: D Let C Event of a collision T Event of a teen driver Y Event of a young adult driver M Event of a midlife driver S Event of a senior driver Then using Bayes Theorem, we see that PCY [ ] PY [ ] P[Y C] PCT [ ] PT [ ] + PCY [ ] PY [ ] + PCM [ ] PM [ ] + PCS [ ] PS [ ] (.8)(.6).. (.5)(.8) + (.8)(.6) + (.4)(.45) + (.5)(.) 4. Solution: B Observe [ N ] [ N ] Pr 4 Pr N N 4 + + + + + + + Pr 4 6 6 + 5 + + + + 5 + + 5 5 of 66

5. Solution: B Let Y positive test result D disease is present (and ~D not D) Using Baye s theorem: PY [ DPD ] [ ] (.95)(.) P[D Y] PY [ DPD ] [ ] + PY [ ~ DP ] [~ D ] (.95)(.) + (.5)(.99).657. 6. Solution: C Let: S Event of a smoker C Event of a circulation problem Then we are given that P[C].5 and P[S C] P[S C C ] PSCPC [ ] [ ] Now applying Bayes Theorem, we find that P[C S] C C PSCPC [ ] [ ] + PSC [ ]( PC [ ]) C PSC [ ] PC [ ] (.5) C C PSC [ ] PC [ ] + PSC [ ]( PC [ ]) (.5) +.75 + 5. 7. Solution: D Use Baye s Theorem with A the event of an accident in one of the years 997, 998 or 999. PA [ 997] P[997] P[997 A] PA [ 997][ P[997] + PA [ 998] P[998] + PA [ 999] P[999] (.5)(.6).45. (.5)(.6) + (.)(.8) + (.)(.) of 66

8. Solution: A Let C Event that shipment came from Company X I Event that one of the vaccine vials tested is ineffective P[ I C] P[ C] Then by Bayes Formula, PC [ I] c c P[ I C] P[ C] + P I C P C Now [ ] PC 5 c 4 P C P[ C] 5 5 P I..9.4 [ C] ( )( )( 9 ) c C ( )( )( ) P I..98.4 Therefore, (.4)( / 5) PC [ I ].4 / 5 +.4 4 / 5.96 9 ( )( ) ( )( ) 9. Solution: C Let T denote the number of days that elapse before a high-risk driver is involved in an accident. Then T is eponentially distributed with unknown parameter λ. Now we are given that 5 λt λt 5. P[T 5] λe dt e e 5λ Therefore, e 5λ.7 or λ (/5) ln(.7) 8 λt λt 8 It follows that P[T 8] λe dt e e 8λ e (8/5) ln(.7) (.7) 8/5.45.. Solution: D Let N be the number of claims filed. We are given P[N ] 4]4 λ 6 λ 4 λ 4 λ Therefore, Var[N] λ. λ λ 4 e λ e λ P[N! 4! of 66

. Solution: D Let X denote the number of employees that achieve the high performance level. Then X follows a binomial distribution with parameters n and p.. Now we want to determine such that Pr[ X > ]. or, equivalently,.99 Pr[ ] ( )(.) k (.98) X k k k The following table summarizes the selection process for : Pr X Pr X [ ] [ ] ( ) 9 ( )( ) 8 ( ) ( ) 5.99.98.668.668..98.7.94 9..98. Consequently, there is less than a % chance that more than two employees will achieve the high performance level. We conclude that we should choose the payment amount C such that C, or C 6,. Solution: D Let X number of low-risk drivers insured Y number of moderate-risk drivers insured Z number of high-risk drivers insured f(, y, z) probability function of X, Y, and Z Then f is a trinomial probability function, so Pr z + f,, 4 + f,, + f,, + f,, [ ] ( ) ( ) ( ) ( ) + + 4! +!!.488 (.) 4(.5)(.) 4(.)(.) (.) (.) 4 of 66

. Solution: B Note that Pr[ X > ].5( t) dt.5 t t.5 4 +.5 + where < <. Therefore, Pr[ X > 6] ( 6) + ( 6) 8 Pr X > 6 X > 8 Pr[ X > 8] ( 8) + ( 8) 7 9 4. Solution: C We know the density has the form C( ) + for < < 4 (equals zero otherwise). First, determine the proportionality constant C from the condition f ( d ) : 4 4 C C C( + ) d C( + ) C 5 5 so C 5, or.5. Then, calculate the probability over the interval (, 6): 6 6.5 ( + ) d ( + ) (.5).47 6. 5. Solution: C Let the random variable T be the future lifetime of a -year-old. We know that the density of T has the form f () C( + ) for < < 4 (and it is equal to zero otherwise). First, determine the proportionality constant C from the condition 4 f ( ) d : 4 4 f ( d ) C( + ) C 5 so that C 5.5. Then, calculate P(T < 5) by integrating f ().5 ( + ) over the interval (.5). 4 4 of 66

6. Solution: B To determine k, note that 4 k 5 ( ) ( ) k k y dy y 5 5 k 5 We net need to find P[V >,] P[, Y >,] P[Y >.] 4 5. (.9) 5.59 and P[V > 4,]. 4 5 5.4.4 5 ( y ) dy ( y ) (.6) 5.78. P[, Y > 4,] P[Y >.4] ( ) ( ) It now follows that P[V > 4, V >,] PV [ > 4, V >, ] PV [ > 4, ].78.. PV [ >, ] PV [ >, ].59 7. Solution: D Let T denote printer lifetime. Then f(t) ½ e t/, t Note that t/ t/ P[T ] e dt e e /.9 P[ T ] t/ t/ e dt e e / e.9 Net, denote refunds for the printers sold by independent and identically distributed random variables Y,..., Y where with probability.9 Y i with probability.9 i,..., with probability.68 Now E[Y i ] (.9) + (.9).56 Therefore, Epected Refunds EY [ ] i (.56),56. i 5 of 66

8. Solution: A Let F denote the distribution function of f. Then 4 ( ) Pr[ ] F X t dt t [ X < X ] Using this result, we see Pr.5 Pr Pr.5 Pr.5 ( X < ) ( X ) [ X < ] [ X ] Pr[ X.5] Pr[ X.5] ( ) F(.5) (.5) ( ).578 F (.5) (.5) 4 F 9. Solution: E Let X be the number of hurricanes over the -year period. The conditions of the problem give is a binomial distribution with n and p.5. It follows that P[X < ] (.95) (.5) + (.95) 9 (.5) + 9(.95) 8 (.5).58 +.77 +.89.95. 4. Solution: B Denote the insurance payment by the random variable Y. Then if < X C Y X C if C < X < Now we are given that.5.5+ C + C ( Y ) ( X C) d ( C).64 Pr <.5 Pr < <.5 +.5 + Therefore, solving for C, we find C ±.8.5 Finally, since < C <, we conclude that C. 6 of 66

4. Solution: E Let X number of group participants that complete the study. Y number of group participants that complete the study. Now we are given that X and Y are independent. Therefore, P{ ( X 9) ( Y < 9) ( X < 9) ( Y 9) } P ( X 9) ( Y < 9) + P ( X < 9) ( Y 9) P ( X 9) ( Y < 9 ) (due to symmetry) P[ X 9] P[ Y < 9] P X 9 P X < 9 (again due to symmetry) [ ] [ ] [ ]( P[ X ]) P X 9 9 ( 9 )( )(.8) + ( )(.8) ( 9 )(.)(.8) ( )(.8) 9 9..76.76.469 [ ][ ] 4. Solution: D Let I A Event that Company A makes a claim I B Event that Company B makes a claim X A Epense paid to Company A if claims are made X B Epense paid to Company B if claims are made Then we want to find C Pr{ I I ( ) ( ) } A B I I X < X A B A B C Pr IA I B + Pr ( IA IB) ( X A < XB) C Pr IA Pr[ IB] + Pr[ IA] Pr[ IB] Pr [ X A < XB] (independence) (.6)(.) + (.4)(.) Pr[ XB X A ].8 +.Pr[ XB X A ] Now X B X A is a linear combination of independent normal random variables. Therefore, X B X A is also a normal random variable with mean M E X X E X E X [ ] [ ] [ ] 9,,, B A B A and standard deviation ( X ) ( X ) ( ) ( ) It follows that σ Var + Var + B A 7 of 66

Finally, Pr[ XB X A ] Pr Z ( Z is standard normal) Pr Z Pr Z < Pr <.54 [ Z ].68.6 C { IA I B ( IA IB) ( X A < XB) } + ( )( ) Pr.8..6. 4. Solution: D If a month with one or more accidents is regarded as success and k the number of failures before the fourth success, then k follows a negative binomial distribution and the requested probability is 4 k + k Pr[ k 4] Pr[ k ] ( k ) k 5 5 4 4 5 6 ( ) + ( ) + ( ) + ( ) 5 5 5 5 5 4 8 8 + + + 5 5 5 5.898 Alternatively the solution is 4 4 4 4 4 5 6 + ( ) + ( ) + ( ).898 5 5 5 5 5 5 5 which can be derived directly or by regarding the problem as a negative binomial distribution with i) success taken as a month with no accidents ii) k the number of failures before the fourth success, and Pr k iii) calculating [ ] 8 of 66

44. Solution: C If k is the number of days of hospitalization, then the insurance payment g(k) is k for k,, g(k) { + 5( k ) for k 4, 5. Thus, the epected payment is 5 gk ( ) pk p+ p + p+ 5 p4 + 4 p5 k ( 5 + 4 + + 5 + 4 ) 5 45. Solution: D Note that ( ) 4 4 8 64 56 8 E X d + d + + 5 46. Solution: D The density function of T is t / f () t e, < t< Therefore, E[ X] E ma ( T, ) t/ t t/ e dt+ e dt t/ t/ t/ e te + e dt e + + e e + e / / t / / 9 of 66

47. Solution: D Let T be the time from purchase until failure of the equipment. We are given that T is eponentially distributed with parameter λ since E[T] λ. Net define the payment for T P under the insurance contract by P for < T for T > We want to find such that E[P] e t/ dt + e t/ t/ t/ dt e e e / + (/) e / + (/) e / ( ½ e / ½ e / ).77. We conclude that 5644. 48. Solution: E Let X and Y denote the year the device fails and the benefit amount, respectively. Then the density function of X is given by f ( ) (.6) (.4 ),,,... and ( 5 ) if,,,4 It follows that y if > 4 [ ] 4(.4) (.6)(.4) (.6) (.4) (.6) (.4) EY + + + 694 49. Solution: D Define f ( X ) to be hospitalization payments made by the insurance policy. Then X if X,, f( X) + 5( X ) if X 4,5 and of 66

5 ( ) ( ) Pr[ ] E f X f k X k k 5 4 + + + 5 + 5 5 5 5 5 5 64 [ + 6 + 8 + + 7]. 5. Solution: C Let N be the number of major snowstorms per year, and let P be the amount paid to / (/ ) n e the company under the policy. Then Pr[N n], n,,,... and n! for N P., ( N ) for N n / (/ ) e Now observe that E[P],( n ) n! n, e / n / (/ ) e +,( n ), e / + E[, (N )] n n!, e / + E[,N] E[,], e / +, (/), 7,. 5. Solution: C Let Y denote the manufacturer s retained annual losses. for.6 < Then Y for >.5.5.5.5.5(.6).5(.6).5(.6) (.6) and E[Y] d d d.5 +.5.5.5.6.6.5.5.5.5.5.5(.6) (.6).5(.6).5(.6) (.6) + + +.94..5.6.5.5.5.5.5 ().5().5(.6) of 66

5. Solution: A Let us first determine K. Observe that 6 + + + 5 + 7 K + + + + K K 4 5 6 6 6 K 7 It then follows that Pr N n Pr N n Insured Suffers a Loss Pr Insured Suffers a Loss 6 (.5 ), N,...,5 7N 7N P E X where [ ] [ ] Now because of the deductible of, the net annual premium [ ], if N X N, if N > Then, 5 P E[ X] ( N ) ().4 N + + 7N 7 7( 4) 7( 5) 5. Solution: D y for < y Let W denote claim payments. Then W for y It follows that E[W] y dy+ dy y / + /.9. y y y of 66

54. Solution: B Let Y denote the claim payment made by the insurance company. Then with probability.94 Y Ma (, ) with probability.4 4 with probability. and 5 / EY.94 +.4.5 e d+. 4 [ ] ( )( ) ( )( ) ( ) ( )( ) 5 5 / / (.) e d e d +.8 5 5 / 5 / /.8 + (.) e + e d e d 5 7.5.5 /.8 + (.) e + e + e d 7.5.5 / 5.8 + (.) e + e e 7.5.5 7.5.5.8 +. e + e e + e ( )( ) 7.5.5 ( )( e e ).8 +. + 4 ( )( ).8 +..48.8 (in thousands) It follows that the epected claim payment is 8. 55. Solution: C k The pdf of is given by f(), < <. To find k, note 4 ( + ) k k k d 4 ( + ) ( + ) k It then follows that E[] d and substituting u +, du d, we see 4 ( + ) 4 u u ( u ) E[] du 4 u ( u u ) du / ½. of 66

56. Solution: C Let Y represent the payment made to the policyholder for a loss subject to a deductible D. for X D That is Y D for D < X Then since E[X] 5, we want to choose D so that ( D) ( D) 5 ( ) 4 D d D D ( D) /4 5 5 D ± 5 D 5 (or D 5 which is etraneous). 57. Solution: B We are given that M (t) for the claim size X in a certain class of accidents. 4 ( 5 t) ( 4)( 5), First, compute M (t) 5 5 ( 5 t) ( 5 t) (, )( 5)( 5) 5,, M (t) 6 6 ( 5 t) ( 5 t) Then E[X] M (), E[X ] M () 5,, Var[X] E[X ] {E[X]} 5,, (,) 5,, Var[ X ] 5,. 58. Solution: E Let X J, X K, and X L represent annual losses for cities J, K, and L, respectively. Then X X J + X K + X L and due to independence t ( J + K + L) t t J Kt Lt M(t) E e E e E e E e E e M J (t) M K (t) M L (t) ( t) ( t).5 ( t) 4.5 ( t) Therefore, M (t) ( t) M (t) 44( t) M (t),56( t) E[X ] M (),56 4 of 66

59. Solution: B The distribution function of X is given by.5.5.5.5( ) ( ) ( ) F( ) dt.5.5.5 t t, > th Therefore, the p percentile p of X is given by.5 p ( ) F( p ).5 ( ) (.p) 5.5.p.5 p 5 (.p) p p It follows that p (.) (.7) 7 5 5 9.6 6. Solution: E Let X and Y denote the annual cost of maintaining and repairing a car before and after the % ta, respectively. Then Y.X and Var[Y] Var[.X] (.) Var[X] (.) (6) 74. 6. Solution: A z The first quartile, Q, is found by ¾ f() d. That is, ¾ (/Q).5 or Q Q (4/).4 4.4. Similarly, the third quartile, Q, is given by Q (4).4 48.. The interquartile range is the difference Q Q. 5 of 66

6. Solution: C First note that the density function of X is given by if f ( ) if < < otherwise Then E ( X ) + ( ) d + ( ) d + 8 4 7 4 + + 4 E ( X ) + ( ) ( ) d + d + 4 6 8 7 7 + + 4 4 4 4 6 5 Var ( X ) E ( X ) E ( X ) 9 6 6. Solution: C X if X 4 Note Y 4 if 4 < X 5 Therefore, 4 54 4 4 5 EY [ ] d+ 4 5 d + 4 5 5 6 6 8 4 + + 5 5 5 5 5 4 5 6 4 6 5 E Y d+ d 4 5 + 4 5 5 5 64 8 64 64 6 64 48 + + + 5 5 5 5 5 5 5 5 Var[ Y] E Y ( E[ Y] ).7 5 5 6 of 66

64. Solution: A Let X denote claim size. Then E[X] [(.5) + (.) + 4(.5) + 5(.) + 6(.) + 7(.) + 8(.)] ( + + + + 6 + 7 + 4) 55 E[X ] 4(.5) + 9(.) + 6(.5) + 5(.) + 6(.) + 49(.) + 64(.) 6 + 9 + 8 + 5 + 6 + 49 + 9 5 Var[X] E[X ] (E[X]) 5 5 475 and Var[ X ].79. Now the range of claims within one standard deviation of the mean is given by [55..79, 55. +.79] [., 76.79] Therefore, the proportion of claims within one standard deviation is.5 +. +. +..45. 65. Solution: B Let X and Y denote repair cost and insurance payment, respectively, in the event the auto is damaged. Then if 5 Y 5 if > 5 and 5 [ ] ( ) ( ) 5 5 EY 5 d 5 5 5 5 5 5 5 5 E Y ( 5) d ( 5) 5 44,8 5 5 45 45 Var[ Y] E Y { E[ Y] } 44,8 ( 5) Var Y 4 [ ] 66. Solution: E Let X, X, X, and X 4 denote the four independent bids with common distribution function F. Then if we define Y ma (X, X, X, X 4 ), the distribution function G of Y is given by G( y) Pr[ Y y] Pr ( X y) ( X y) ( X y) ( X4 y) Pr X y Pr X y Pr X y Pr X y [ ] [ ] [ ] [ 4 ] 4 ( ) F y 4 5 ( ) + sin π y, y 6 It then follows that the density function g of Y is given by 7 of 66

Finally, EY ( ) '( ) g y G y ( + sin πy) ( π cos πy) 4 π 5 cosπy( + sin πy), y 4 5/ [ ] ( ) yg ydy / 5/π cosπ ( sinπ ) / y y + y dy 4 67. Solution: B The amount of money the insurance company will have to pay is defined by the random variable if < Y if where is a Poisson random variable with mean.6. The probability function for X is.6 k e (.6) p( ) k,,, and k! k.6.6.6 EY [ ] + (.6) e + e k k! k.6.6.6.6.6 (.6) e + e e (.6) e k k! k.6 (.6 ).6.6.6.6 e e (.6) e e 6e k k! 57 k.6.6.6 E Y ( ) (.6) e + ( ) e k k! k.6.6.6.6 ( ) e ( ) e ( ) ( ) (.6) e k k!.6.6 ( ) ( ) e ( ) ( ) (.6) e 86,89 [ Y] E Y { E[ Y] } ( ) [ Y ] Var 86,89 57 Var 699 488,564 8 of 66

68. Solution: C Note that X has an eponential distribution. Therefore, c.4. Now let Y denote the for < 5 claim benefits paid. Then Y and we want to find m such that.5 5 for 5 m m.4.4.4e d e e.4m This condition implies e.4m.5 m 5 ln 7.9. 69. Solution: D The distribution function of an eponential random variable t θ T with parameter θ is given by F( t) e, t > Since we are told that T has a median of four hours, we may determine θ as follows: 4 θ F( 4) e 4 θ e 4 ln ( ) θ 4 θ ln ( ) Therefore, ( ) ( ) ( ) 5ln 5 θ 4 5 4 Pr T 5 F 5 e e.4 7. Solution: E Let X denote actual losses incurred. We are given that X follows an eponential distribution with mean, and we are asked to find the 95 th percentile of all claims that eceed. Consequently, we want to find p 95 such that Pr[ < < p95] F( p95 ) F().95 where F() is the distribution function of X. PX [ > ] F() Now F() e /. p95 / / / p95 / e ( e ) e e / p95 / Therefore,.95 e e / / ( e ) e p 95 / e.5 e / p 95 ln(.5 e / ) 999 9 of 66

7. Solution: A The distribution function of Y is given by G y Pr T y Pr T y F y 4 y ( ) ( ) ( ) ( ) for y > 4. Differentiate to obtain the density function g( y) 4y Alternate solution: Differentiate F( t ) to obtain f ( t) 8t and set y t. Then t y and d g( y) f ( t( y) ) dt dy f ( y) ( y) 8y y 4y dt 7. Solution: E We are given that R is uniform on the interval (.4,.8 ) and V,e R Therefore, the distribution function of V is given by R F( v) Pr[ V v] Pr,e v Pr R ln ( v) ln (,) ln( v) ln(,).4.4.4 ( v) ( ) ln ln,.4 ( ) ( ) dr r 5ln v 5ln, v 5 ln.4, 7. Solution: E F y Y y X y X e Y 4 ( Y ) Therefore, f ( y) F ( y) e 54 8 ( ) ( ) [ ] ( ).8 8 Y Pr Pr Pr Y 8 of 66

74. Solution: E First note R /T. Then F R (r) P[R r] P r P T FT T r. Differentiating with respect to r r f R (r) F R (r) d/dr d F T FT () t r dt r d F T() t f T() t since T is uniformly distributed on [8, ]. dt 4 5 Therefore f R (r). 4 r r 75. Solution: A Let X and Y be the monthly profits of Company I and Company II, respectively. We are given that the pdf of X is f. Let us also take g to be the pdf of Y and take F and G to be the distribution functions corresponding to f and g. Then G(y) Pr[Y y] P[X y] P[X y/] F(y/) and g(y) G (y) d/dy F(y/) ½ F (y/) ½ f(y/). 76. Solution: A First, observe that the distribution function of X is given by F( ) dt 4, > t t Net, let X, X, and X denote the three claims made that have this distribution. Then if Y denotes the largest of these three claims, it follows that the distribution function of Y is given by G y Pr X y Pr X y Pr X y ( ) [ ] [ ] [ ], y > y while the density function of Y is given by 9 g( y) G' ( y) 4 4 y y y y, y> Therefore, of 66

9 9 E [ Y ] dy dy + 6 y y y y y 9 8 9 9 8 9 dy 6 9 5 8 y y y y 5y 8y + + 9 +.5 (in thousands) 5 8 77. Solution: D Prob. ( ) 8 + yddy.65 Note c { } ( X ) ( Y ) ( X > ) ( Y > ) Pr Pr (De Morgan's Law) Pr ( ) ( ) ( ) ( ) X > Y > + y ddy 8 8 + y dy 64 7 7 8 6 + + + + + 48 48 8.65 48 48 ( y ) ( y ) dy ( y ) ( y ) ( ) 78. Solution: B That the device fails within the first hour means the joint density function must be integrated over the shaded region shown below. This evaluation is more easily performed by integrating over the unshaded region and subtracting from. of 66

( X < ) ( Y < ) Pr + y + y d dy dy ( 9 6y y) dy 7 + 54 54 ( ) ( ) ( ) 54 y dy y y 54 54 54 7 8 + 4 8 + 4 + 8 8.4 79. Solution: E The domain of s and t is pictured below. Note that the shaded region is the portion of the domain of s and t over which the device fails sometime during the first half hour. Therefore, / / Pr S T f ( s, t) dsdt + f ( s, t) dsdt / (where the first integral covers A and the second integral covers B). 8. Solution: C By the central limit theorem, the total contributions are approimately normally nμ 5 5 6,8,5 and standard deviation distributed with mean ( )( ) σ n 5 5, 5. From the tables, the 9 th percentile for a standard normal random variable is.8. Letting p be the 9 th percentile for total contributions, p nμ.8, and so p nμ+.8σ n 6,8,5 + (.8)(, 5) 6,4,548. σ n of 66

8. Solution: C Let X,..., X 5 denote the 5 collision claims, and let X (X +... +X 5 ). We are 5 given that each X i (i,..., 5) follows a normal distribution with mean 9,4 and standard deviation 5. As a result X also follows a normal distribution with mean 9,4 and standard deviation (5). We conclude that P[ X >,] 5 X 9, 4, 9, 4 X 9, 4 P P > >.6 Φ(.6).757.74. 8. Solution: B Let X,..., X 5 be the number of claims filed by each of the 5 policyholders. We are given that each X i follows a Poisson distribution with mean. It follows that E[X i ] Var[X i ]. Now we are interested in the random variable S X +... + X 5. Assuming that the random variables are independent, we may conclude that S has an approimate normal distribution with E[S] Var[S] ()(5) 5. Therefore P[45 < S < 6] 45 5 S 5 6 5 S 5 P < < P 5 5 5 < < 5 S 5 S 5 P < P < 5 5 S 5 Then using the normal approimation with Z, we have P[45 < S < 6] 5 P[Z < ] P[Z > ] P[Z < ] + P[Z < ].977 +.84.886. 8. Solution: B Let X,, X n denote the life spans of the n light bulbs purchased. Since these random variables are independent and normally distributed with mean and variance, the random variable S X + + X n is also normally distributed with mean μ n and standard deviation σ n Now we want to choose the smallest value for n such that S n 4 n.977 Pr[ S > 4] Pr > n n This implies that n should satisfy the following inequality: 4 of 66

4 n n To find such an n, let s solve the corresponding equation for n: 4 n n n 4 n n n 4 ( n )( n ) + 4 n 4 n 6 84. Solution: B Observe that E X + Y E X + E Y 5 + 7 [ ] [ ] [ ] [ ] [ ] [ ] [ ] Var X + Y Var X + Var Y + Cov X, Y 5 + + for a randomly selected person. It then follows from the Central Limit Theorem that T is approimately normal with mean ET 7 7 [ ] ( ) and variance Var T [ ] ( ) Therefore, T 7 7 7 Pr[ T < 7] Pr < Pr <.84 [ Z ] where Z is a standard normal random variable. 5 of 66

85. Solution: B Denote the policy premium by P. Since is eponential with parameter, it follows from what we are given that E[X], Var[X],,, Var[ X ] and P + E[X],. Now if policies are sold, then Total Premium Collected (,), Moreover, if we denote total claims by S, and assume the claims of each policy are independent of the others then E[S] E[X] ()() and Var[S] Var[X] ()(,,). It follows from the Central Limit Theorem that S is approimately normally distributed with mean, and standard deviation,. Therefore,,, P[S,] P[S,] P Z, P[Z ].84.59. 86. Solution: E Let X,..., X denote the number of pensions that will be provided to each new recruit. Now under the assumptions given, with probability.4.6 X i with probability (.4)(.5). with probability (.4)(.75). for i,...,. Therefore, E X.6 +. +..7, [ ] ( )( ) ( )( ) ( )( ) i ( ) ( ) ( ) ( ) ( ) ( ) E X i.6 +. +.., and Var[ Xi] E X i { E[ Xi] }. (.7).8 Since X,..., X are assumed by the consulting actuary to be independent, the Central Limit Theorem then implies that S X +... + X is approimately normally distributed with mean E[ S] E[ X] +... + E[ X ] (.7) 7 and variance Var[ S] Var [ X] +... + Var[ X] (.8) 8 Consequently, S 7 9.5 7 Pr[ S 9.5] Pr 9 9 Pr[ Z.8].99 6 of 66

87. Solution: D Let X denote the difference between true and reported age. We are given X is uniformly distributed on (.5,.5). That is, X has pdf f() /5,.5 < <.5. It follows that μ E[X] σ Var[X] E[X ].5.5 (.5) d.5.8 5 5 5.5 σ.44 Now X 48, the difference between the means of the true and rounded ages, has a distribution that is approimately normal with mean and standard deviation.44 48.8. Therefore,.5.5 P X48 P Z 4 4.8.8 P[. Z.] P[Z.] P[Z.] P[Z.] + P[Z.] P[Z.] (.8849).77. 88. Solution: C Let X denote the waiting time for a first claim from a good driver, and let Y denote the waiting time for a first claim from a bad driver. The problem statement implies that the respective distribution functions for X and Y are F e, > and /6 ( ) y / ( ) G y e, y > Therefore, Pr ( X ) ( Y ) Pr[ X ] Pr[ Y ] F G e e e e + e ( ) ( ) ( )( ) / / / / 7/6 7 of 66

89. Solution: B We are given that 6 (5 y) for < < 5 y< 5 f(, y) 5, otherwise and we want to determine P[X > Y > ]. In order to determine integration limits, consider the following diagram: y 5 We conclude that P[X > Y > ] > y> (, ) (, ) 5 5 6 (5 y) 5, dy d. 9. Solution: C Let T be the time until the net Basic Policy claim, and let T be the time until the net Delue policy claim. Then the joint pdf of T and T is t/ t/ t/ t/ f( t, t) e e e e, < t <, < t < and we need to find 6 P[T < T ] t t/ t/ t/ t/ t e e dtdt e e dt 6 t/ t/ t/ t/ 5 t/6 e e e dt e e dt / 5 /6 + t t e e.4. 5 5 5 9. Solution: D We want to find P[X + Y > ]. To this end, note that P[X + Y > ] + y dyd y y y d 4 + 8 + ( ) ( ) + ( ) d 8 5 + + d 8 4 8 5 + + 4 8 8 5 + + 7 4 8 8 4 + + + 8 4 8 d 8 of 66

9. Solution: B Let X and Y denote the two bids. Then the graph below illustrates the region over which X and Y differ by less than : Based on the graph and the uniform distribution: Shaded Region Area Pr X Y < ( ) ( ) 8 8 (.9).9 More formally (still using symmetry) Pr X Y < Pr X Y Pr[ X Y ] dyd y d d 8.9 ( ) ( ) 9 of 66

9. Solution: C Define X and Y to be loss amounts covered by the policies having deductibles of and, respectively. The shaded portion of the graph below shows the region over which the total benefit paid to the family does not eceed 5: We can also infer from the graph that the uniform random variables X and Y have joint density function f (, y), < <, < y< We could integrate f over the shaded region in order to determine the desired probability. However, since X and Y are uniform random variables, it is simpler to determine the portion of the square that is shaded in the graph above. That is, Pr Total Benefit Paid Does not Eceed 5 ( ) Pr ( X 6, Y ) Pr ( X, Y 7) Pr ( X 6, Y 8 X) ( )( ) ()( ) ( )( )( ) < < < < + < < < < + < < < < 6 5 5 5 5.5 + + + +.95 94. Solution: C Let f ( t, t ) denote the joint density function of T and T. The domain of f is pictured below: Now the area of this domain is given by A 6 6 4 6 4 ( ) 4 of 66

Consequently, f ( t t ), < t < 6, < t < 6, t+ t <, 4 elsewhere and ET+ T ET + ET ET (due to symmetry) [ ] [ ] [ ] [ ] 4 6 6 t 4 t 6 6 t t t dt dt + 4 t 4 dt dt 4 t 4 dt + t 4 4 dt 4t 6 t 4 6 dt + ( t t ) dt 4 + 5t t 7 4 4 4 4 4 64 + 8 7 8 + 5.7 7 4 95. Solution: E tw + tz t( X+ Y) + t( Y X) ( t t) X ( t+ t) Y M ( t, t) E e E e E e e ( ) ( ) ( t t) ( t+ t) ( t tt+ t) ( t + tt+ t) E e E e e e e e e t t X t+ t Y t + t 96. Solution: E Observe that the bus driver collect 5 5 for the tickets he sells. However, he may be required to refund to one passenger if all ticket holders show up. Since passengers show up or do not show up independently of one another, the probability that..98.65. Therefore, the tour all passengers will show up is ( ) ( ) operator s epected revenue is 5 ( )(.65) 985. 4 of 66

97. Solution: C We are given f(t, t ) /L, t t L. L t Therefore, E[T + T ] ( t + t ) dt dt L L t L t t t + t dt t + dt L L 4 L 4 t t dt L L L L t ( L, L) t 98. Solution: A Let g(y) be the probability function for Y X X X. Note that Y if and only if X X X. Otherwise, Y. Since P[Y ] P[X X X ] P[X ] P[X ] P[X ] (/) 8/7. 9 for y 7 8 We conclude that g( y) for y 7 otherwise y 9 8 t t and M(t) E e + e 7 7 4 of 66

99. Solution: C We use the relationships Var ax + b a Var X, Cov ax, by ab Cov X, Y, and ( ) ( ) ( ) ( ) ( X + Y) ( X) + ( Y) + ( X Y) ( X+ Y) + + ( XY) ( XY) ( X + ) + Y ar ( X +.Y) + [ X Y] X ( ) Y ( X Y) Var Var Var Cov,. First we observe 7, Var 5, Cov,, and so Cov,. We want to find Var. V Var +. Var + Var. + Cov,. Var X +. VarY +. Cov X, Y 5 +, + 9,. ( ) ( ) ( ). Solution: B Note P(X ) /6 P(X ) / + /6 / P(X ) / + / + /6 7/. E[X] ()(/6) + ()(/) + ()(7/) 7/ E[X ] () (/6) + () (/) + () (7/) / Var[X] / (7/).58.. Solution: D Note that due to the independence of X and Y Var(Z) Var(X Y 5) Var(X) + Var(Y) Var(X) + Var(Y) 9() +.. Solution: E Let X and Y denote the times that the two backup generators can operate. Now the variance of an eponential random variable with mean β is β. Therefore, Var[ X] Var[ Y] Then assuming that X and Y are independent, we see Var X+Y Var X + Var Y + [ ] [ ] [ ] 4 of 66

. Solution: E Let X, X, and X denote annual loss due to storm, fire, and theft, respectively. In addition, let Y Ma( X, X, X). Then Pr Y > Pr Y Pr X Pr X Pr X [ ] [ ] [ ] [ ] [ ] ( e )( e )( e ) ( e )( e )( 5 e ).44 * Uses that if X has an eponential distribution with mean μ Pr ( X ) Pr( X ) e dt ( e ) e μ * t μ t μ μ 4. Solution: B Let us first determine k: k k kddy k dy dy k Then E[ X] dyd d E [ Y ] y ddy ydy y E[ XY] yddy y dy ydy y 6 6 Cov [ XY, ] EXY [ ] EX [ ] EY [ ] (Alternative Solution) Define g() k and h(y). Then f(,y) g()h() In other words, f(,y) can be written as the product of a function of alone and a function of y alone. It follows that X and Y are independent. Therefore, Cov[X, Y]. 44 of 66

5. Solution: A The calculation requires integrating over the indicated region. 4 5 8 4 4 4 4 E ( X ) y dy d ( 4 ) 4 y d d d 5 5 4 5 8 8 8 56 56 56 E ( Y ) y dy d ( 8 ) y dy d d d 9 9 9 45 45 5 8 8 8 56 56 8 E( XY) y dyd y d ( 8 ) d d 9 9 9 54 7 8 56 4 Cov ( XY, ) E( XY) E( X) EY ( ).4 7 45 5 6. Solution: C The joint pdf of X and Y is f(,y) f (y ) f () (/)(/), < y <, < <. Therefore, y E[X] dyd d d 4 E[Y] E[XY] 6 y y 44 dyd d d 4 4 48 48 y y () dyd d d 4 4 4 7 7 Cov(X,Y) E[XY] E[X]E[Y] 4 ()(6) 4 8 6. 45 of 66

7. Solution: A Cov ( C, C) Cov ( X + Y, X +.Y) Cov X, X + Cov Y, X + Cov X,.Y + Cov Y,.Y Var X + Cov ( X, Y) +.Cov ( X, Y) +.VarY Var X +. Cov X, Y +.VarY ( ) ( ) ( ) ( ) ( ) ( ( )) ( ) ( E Y ) ( ) Var X E X E X 7.4 5.4 VarY E Y ( ) 5.4 7.4 Var ( X + Y) Var X + VarY + Cov ( X, Y) Cov ( XY, ) ( Var ( X+ Y) Var X VarY) ( 8.4.4).6 Cov C, C.4 +..6 +..4 8.8 ( ) ( ) ( ) 7. Alternate solution: We are given the following information: C X + Y C X +.Y E[ X] 5 E X 7.4 EY [ ] 7 E Y 5.4 Var[ X + Y] 8 Now we want to calculate Cov ( C, C) Cov ( X + Y, X +.Y) E ( X + Y)( X +.Y) E[ X + Y] ie[ X +.Y] E X +.XY +.Y E[ X] + E[ Y] E X +.E Y E X +.E[ XY] +.E Y ( 5 + 7) ( 5 + (.) 7) 7. 4 +.E XY +. 5.4.4 ( )( [ ] [ ]) [ ] ( ) ( )( ).E[ XY] 7.7 Therefore, we need to calculate E[ XY ] first. To this end, observe 46 of 66

( ) [ X Y] E ( X Y) E[ X Y] 8 Var + + + + + + ( [ ] [ ]) E X XY Y E X E Y E X + E[ XY] + E Y ( 5 + 7) 7.4 + E[ XY] + 5.4 44 E[ XY] 65. E[ XY] ( 8 + 65.) 6.6 Finally, Cov( CC ).( 6.6) 7.7 8.8, 8. Solution: A The joint density of T and T is given by t t f ( t, t) e e, t >, t > Therefore, Pr X Pr T + T [ ] [ ] ( t ) ( t ) t t t t e e dtdt e e dt + t t t t e e dt e e e dt e + e e e + e e + e t t e + e e e + e, > It follows that the density of X is given by d g( ) e + e e e, > d 47 of 66

9. Solution: B Let u be annual claims, v be annual premiums, g(u, v) be the joint density function of U and V, f() be the density function of X, and F() be the distribution function of X. Then since U and V are independent, u v/ u v/ g( u, v) ( e ) e e e, < u<, < v< and u F( ) Pr[ X ] Pr Pr[ U V] v v v u v/ g ( u, v) dudv e e dudv u v/ v v v/ v/ e e dv e e e dv + v ( + / ) v / e e dv + v ( + /) v / e e + + + Finally, f ( ) F' ( ) + ( ). Solution: C Note that the conditional density function f (, y) f y,, < y< f ( ) 6 f 4( ) ydy 8ydy 4y 9 f y 9 f, y 9 y, y < < 6 It follows that ( ) Consequently, 9 9 Pr Y < X X ydy y 4 4 48 of 66

. Solution: E f Pr < Y < X dy f ( ) (, y) ( ) ( ) f (, y) y y 4 4 f ( ) y dy y 4 4 y dy 8 9 9 Finally, Pr < Y < X y. Solution: D We are given that the joint pdf of X and Y is f(,y) (+y), < y < <. Now f () 4 (+ y) dy y+ y +, < < f ( y, ) ( + y) y so f(y ) ( ) + f, < y < y f(y.) [ ] + + y.., < y <..5 P[Y <.5 X.].5 5 [ + ydy ] y y + +.467.. Solution: E Let W event that wife survives at least years H event that husband survives at least years B benefit paid P profit from selling policies c Then Pr[ H] P[ H W] + Pr H W.96 +..97 and Pr[ W H].96 Pr[ W H].9897 Pr H.97 [ ] c Pr H W c. Pr W H. Pr.97 [ H ] 49 of 66

It follows that c { } [ ] [ ] [ ] ( ) [ ] ( ),(.) 897 E P E B E B Pr W H +, Pr W H 4. Solution: C Note that PX (, Y ) PX (, Y ).5 P(Y X ) PX ( ) PX (, Y ) + PX (, Y ).5+.5.86 P(Y X) P(Y X ).86.74 Therefore, E(Y X ) () P(Y X ) + () P(Y X ) ()(.74).74 E(Y X ) () P(Y X ) + () P(Y X ).74 Var(Y X ) E(Y X ) [E(Y X )].74 (.74). 5. Solution: A Let f () denote the marginal density function of X. Then + + f ( ) dy y ( + ), < < Consequently, f ( y ) f (, y) if: < y< + f ( ) otherwise + + EY [ X] ydy y ( + ) + + + + + E Y X y dy y ( + ) + + + + + Var [ ] { [ ]} Y X E Y X E Y X + + 4 + + + 5 of 66

6. Solution: D Denote the number of tornadoes in counties P and Q by N P and N Q, respectively. Then E[N Q N P ] [()(.) + ()(.6) + ()(.5) + (.)] / [. +.6 +.5 +.].88 E[N Q N P ] [() (.) + () (.6) + () (.5) + () (.)] / [. +.6 +.5 +.].76 and Var[N Q N P ] E[N Q N P ] {E[N Q N P ]}.76 (.88).9856. 7. Solution: C The domain of X and Y is pictured below. The shaded region is the portion of the domain over which X<.. Now observe.. X y dyd y y y d + [ < ] ( ) Pr. 6 6.. 6 ( ) ( ) ( ) ( ) 6 d d.. 6 ( ) d ( ) (.8) +.488 8. Solution: E The shaded portion of the graph below shows the region over which f (, ) y is nonzero: We can infer from the graph that the marginal density function of Y is given by y y ( ) ( ) ( ) g y 5yd 5y 5y y y 5y y, < y< y y 5 of 66

or more precisely, g( y) ( ) 5y y, < y< otherwise 9. Solution: D The diagram below illustrates the domain of the joint density ( ) f y, of Xand Y. We are told that the marginal density function of X is f ( ),< < while f ( y), < y< + y It follows that f (, y) f ( ) f ( y ) Therefore, if < <, < y< + y otherwise [ ] [ ] Pr Y >.5 Pr Y.5 dyd 7 y d d + 4 8 8 [Note since the density is constant over the shaded parallelogram in the figure the solution is also obtained as the ratio of the area of the portion of the parallelogram above y.5 to the entire shaded area.] 5 of 66

. Solution: A We are given that X denotes loss. In addition, denote the time required to process a claim by T., < t <, Then the joint pdf of X and T is f(, t) 8 8, otherwise. Now we can find P[T ] 4 4 4 4 6 7 ddt 8 6 dt t dt t / t / 6 64 6 64 4 6 64 t /64.7. t t 4 t. Solution: C The marginal density of X is given by y f ( ) ( y ) dy y 64 64 64 Then EX ( ) f ( ) d d 64 5 8 64 9 9 5.778 5 64 9 5 of 66

. Solution: D The marginal distribution of Y is given by f (y) 6 e y e y + 6e y 6 e y 6 e y, < y < y 6 y y (6ye 6 ye ) Therefore, E(Y) y f (y) dy dy 6 6 6 ye y dy y e y dy But y e y dy and y e e e y d 6 e y d y ye dy 6 y e y dy y e y dy are equivalent to the means of eponential random variables with parameters / and /, respectively. In other words, and y e y dy / y e y dy /. We conclude that E(Y) (6/) (/) (6/) (/) / / 9/6 4/6 5/6.8.. Solution: C Observe Pr 4 < S < 8 Pr 4 < S < 8 N Pr N + Pr 4 < S < 8 N > Pr N > 8 ( ) ( ) 4 5 5 e e + e e * 6. *Uses that if X has an eponential distribution with mean μ a b t μ t μ μ Pr ( a X b) Pr ( X a) Pr ( X b) e dt e dt e e μ μ μ [ ] [ ] [ ] a b 54 of 66

4. Solution: A y Because f(,y) can be written as f ( ) f ( y) e e and the support of f(,y) is a cross product, X and Y are independent. Thus, the condition on X can be ignored and it suffices y to just consider f ( y) e. Because of the memoryless property of the eponential distribution, the conditional density of Y is the same as the unconditional density of Y+. Because a location shift does not affect the variance, the conditional variance of Y is equal to the unconditional variance of Y. Because the mean of Y is.5 and the variance of an eponential distribution is always equal to the square of its mean, the requested variance is.5. ------------- 5. Solution: E The support of (X,Y) is < y < <. f X, Y (, y) f ( y ) f X ( ) on that support. It is clear geometrically (a flat joint density over the triangular region < y < < ) that when Y y we have X ~ U(y, ) so that f ( y) for y < <. y By computation: f Y f X, Y (, y) ( y) d y f ( y) for y < < y f ( y) y y Y 55 of 66

6. Solution: C Using the notation of the problem, we know that p + p and 5 p + p+ p + p+ p4 + p5. Let pn pn+ c for all n 4. Then pn p nc for n 5. Thus p + ( p c) + ( p c) +... + ( p 5c) 6 p 5c. 6p 5c Also p + p p + ( p c) p c. Solving simultaneously 5 p c 5 6 6p c 5 5 5 6p + 5c. So c and p +. Thus p. 6 5 6 6 c 5 7 5 We want p4 + p5 ( p 4c) + ( p 5c) +.67. --- 7. Solution: D Because the number of payouts (including payouts of zero when the loss is below the deductible) is large, we can apply the central limit theorem and assume the total payout S is normal. For one loss there is no payout with probability.5 and otherwise the payout is U(, 5). So, E [ X ].5* +.75*75 565, 5 E [ X ].5* +.75*(75 + Var ( X ) E[ X ] E[ X ] 4,69,75. ) 56,5,, so the variance of one claim is Applying the CLT, S ()(565) P [,, < S <,,] P.78746 < <. 6944968 ()(4,69,75) which interpolates to.8575-(-.966).8 from the provided table. 56 of 66

8. Key: B Let H be the percentage of clients with homeowners insurance and R be the percentage of clients with renters insurance. Because 6% of clients do not have auto insurance and none have both homeowners and renters insurance, we calculate that 8% (6% 7% %) must have renters insurance, but not auto insurance. (H )% have both homeowners and auto insurance, (R 8)% have both renters and auto insurance, and none have both homeowners and renters insurance, so (H + R 9)% must equal 5%. Because H R, R must be 8%, which implies that % have both renters and auto insurance. 9. Key: B The reimbursement is positive if health care costs are greater than, and because of the memoryless property of the eponential distribution, the conditional distribution of health care costs greater than is the same as the unconditional distribution of health care costs. We observe that a reimbursement of 5 corresponds to health care costs of 5 (% ( ) + 5% (5 )), which is greater than the deductible of. Therefore, G(5) F() e.77.. Key: C E X [ (.5) ] E (.5) X ( ln.5) [ ] E e X X [ ] M ( ln.5) 4. 9 ln.5 57 of 66

. Solution: E First, find the conditional probability function of N given n where p ( n ) is the marginal probability function of N : ( n n ) ( n, n ) p ( n ) p p, N. To find the latter, sum the joint probability function over all possible values of N obtaining n n n n n p ( n ) p( n, n ) e ( e ), n 4 4 n 4 n n n n since ( e ) 4 e as the sum of the probabilities of a geometric random variable. The conditional probability function is ( n n ) p ( n, n ) p ( n ) n ( e ) n n p e, n which is the probability function of a geometric random variable with parameter p e. The n n mean of this distribution is / p / e e, and becomes e when n.. Solution: C The number of defective modems is % + 8% 5. 7 The probability that eactly two of a random sample of five are defective is.. 8 5. Solution: B Pr(man dies before age 5) Pr(T < 5 T > 4) Pr(4 < T < 5) Pr( T > 4) e 4. 5. e e 4. e F(5) F(4) F(4) 4 5 (.. ).696 Epected Benefit 5 Pr(man dies before age 5) (5) (.696) 47.96 58 of 66

4. Solutions: C Letting t denote the relative frequency with which twin-sized mattresses are sold, we have that the relative frequency with which king-sized mattresses are sold is t and the relative frequency with which queen-sized mattresses are sold is (t+t)/4, or t. Thus, t. since t + t + t. The probability we seek is t + t.8. 59 of 66

5. Key: E Var (N) E [ Var ( N λ )] + Var [ E ( N λ )] E (λ) + Var (λ).5 +.75.5 6. Key: D X follows a geometric distribution with p. Y implies the first roll is not a 6 and the 6 second roll is a 6. This means a 5 is obtained for the first time on the first roll (probability %) or a 5 is obtained for the first time on the third or later roll (probability 8%). p [ X X ] + 6 + 8 E, so E [ X Y ].( ) +.8( 8) 6. 6 7. Key: E Because X and Y are independent and identically distributed, the moment generating function of X + Y equals K (t), where K(t) is the moment generating function common to X and Y. Thus, K(t).e -t +.4 +.e t. This is the moment generating function of a discrete random variable that assumes the values -,, and with respective probabilities.,.4, and.. The value we seek is thus.7. 6 of 66

8. Key: D Suppose the component represented by the random variable X fails last. This is represented by the triangle with vertices at (, ), (, ) and (5, 5). Because the density is uniform over this region, the mean value of X and thus the epected operational time of the machine is 5. By symmetry, if the component represented by the random variable Y fails last, the epected operational time of the machine is also 5. Thus, the unconditional epected operational time of the machine must be 5 as well. 9. Key: B The unconditional probabilities for the number of people in the car who are hospitalized are.49,.4 and.9 for, and, respectively. If the number of people hospitalized is or, then the total loss will be less than. However, if two people are hospitalized, the probability that the total loss will be less than is.5. Thus, the epected number of people in the car who are hospitalized, given that the total loss due to hospitalizations from the accident is less than is.49.4.9.5 + +.54.49 +.4 +.9.5.49 +.4 +.9.5.49 +.4 +.9.5 4. Key: B Let X equal the number of hurricanes it takes for two losses to occur. Then X is negative binomial with success probability p.4 and r successes needed. n n PX [ n] p( p) (.4) (.4) ( n )(.4) (.6) r r n r n n We need to maimize P[X n]. Note that the ratio, for n. n PX [ n+ ] n(.4) (.6) n n [ ] ( )(.4) (.6) PX n n n (.6). This ratio of consecutive probabilities is greater than when n and less than when n. Thus, P[X n] is maimized at n ; the mode is. 6 of 66

4. Key: C There are (5 choose ) ways to select the three columns in which the three items will appear. The row of the rightmost selected item can be chosen in any of si ways, the row of the leftmost selected item can then be chosen in any of five ways, and the row of the middle selected item can then be chosen in any of four ways. The answer is thus ()(6)(5)(4). Alternatively, there are ways to select the first item. Because there are squares in the row or column of the first selected item, there are ways to select the second item. Because there are 8 squares in the rows or columns of the first and second selected items, there are 8 ways to select the third item. The number of permutations of three qualifying items is ()()(). The number of combinations is thus ()()()/!. 4. Key: B The epected bonus for a high-risk driver is.8 (months) 5. 48. The epected bonus for a low-risk driver is.9 (months) 5. 54. The epected bonus payment from the insurer is 6 48 + 4 54 5, 4. 6 of 66

4. Key: E P(Pr Li) P(Pr) + P(Li Pr'). +.. Subtract from to get the answer. 44. Key: E The total time is less than 6 minutes, so if minutes are spent in the waiting room, less than 6 minutes are spent in the meeting itself. 45. Key: C f (.75, y) f y.75) ( f (.75, y) dy Thus, 4 / for < y <.5 f ( y.75), / for.5 < y < f (.75, y)..5 which leads to Var (Y X.75) /44.76. 46. Key: B C the set of TV watchers who watched CBS over the last year N the set of TV watchers who watched NBC over the last year A the set of TV watchers who watched ABC over the last year H the set of TV watchers who watched HGTV over the last year The number of TV watchers in the set C N A is 4 + 5+ 7 6 5 + 4 45. Because C N A and H are mutually eclusive, the number of TV watchers in the set C N A H is 45 + 8 6. The number of TV watchers in the complement of C N A H is thus 6 7. 6 of 66

47. Key: A Let X denote the amount of a claim before application of the deductible. Let Y denote the amount of a claim payment after application of the deductible. Let be the mean of X, which because X is eponential, implies that is the variance of X ande X. By the memoryless property of the eponential distribution, the conditional distribution of the portion of a claim above the deductible given that the claim eceeds the deductible is an eponential distribution with mean. Given that E Y.9, this implies that the probability of a claim eceeding the deductible is.9 and thus. Then, E Y.8.9.99 Var Y..9.8 48. Key: C Let N denote the number of hurricanes, which is Poisson distributed with mean and variance 4. Let X i denote the loss due to the i th hurricane, which is eponentially distributed with mean, and therefore variance (,),,. Let X denote the total loss due to the N hurricanes. This problem can be solved using the conditional variance formula. Note that independence is used to write the variance of a sum as the sum of the variances. X X N X N Var E X... X N E Var X... X N Var NE X E NVar X Var,N E,,N, Var N,,E N Var Var E E Var,, (4),, (4) 8,, 64 of 66

49. Key: B Let N denote the number of accidents, which is binomial with parameters 4 and and thus has mean 4 4 9 and variance. 4 4 6 Let X i denote the unreimbursed loss due to the i th accident, which is. times an eponentially distributed random variable with mean.8 and therefore variance (.8).64. Thus, X has mean.8(.).4 and variance.64(.).576. i Let X denote the total unreimbursed loss due to the N accidents. This problem can be solved using the conditional variance formula. Note that independence is used to write the variance of a sum as the sum of the variances. X X N X N Var E X... X N E Var X... X N Var NE X E NVar X Var.4N E.576N Var Var E E Var.4 Var N.576E N 9.576.576.756. 6 4 65 of 66

5. Key: B The 95 th percentile is in the range when an accident occurs. It is the 75 th percentile of the payout, given that an accident occurs, because (.95.8)/(.8).75. Letting be the 75 th percentile of the given eponential distribution, F( ) e.75, so 459. Subtracting the deductible of 5 gives 659 as the (unconditional) 95 th percentile of the insurance company payout. 5. Key: C The ratio of the probability that one of the damaged pieces is insured to the probability r 7 r that none of the damaged pieces are insured is 7 4 4r, where r is the total r 7 r 4 r 4 7 number of pieces insured. Setting this ratio equal to and solving yields r 8. The probability that two of the damaged pieces are insured is 4 r 7 r 8 9 (8)(7)(9)(8)(4)()()() 66 7 7 (7)(6)(5)(4)()()()() 975 4 4.7. 5. Key: A The probability that Rahul eamines eactly n policies is..9 n. The probability that Toby eamines more than n policies is.8 n. The required probability is thus n n n.7..9.8.7.857. 9 9.7 n n 66 of 66