SOCIETY OF ACTUARIES EXAM P PROBABILITY EXAM P SAMPLE SOLUTIONS

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1 SOCIETY OF ACTUARIES EXAM P PROBABILITY EXAM P SAMPLE SOLUTIONS Copyright 8 by the Society of Actuaries Some of the questions in this study note are taken from past examinations. Some of the questions have been reformatted from previous versions of this note. Questions were added in October 4. Questions 56-6 were added January 5. Questions 7-37 were added April 5. Questions 38-4 were added May 5. Questions 4-4 were added November 5. Questions were added September 6. Question 37 was added January 8. Page of 9

2 . Solution: D Let G = viewer watched gymnastics, B = viewer watched baseball, S = 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) = =.48=.5 ( ). Solution: A Let R = referral to a specialist and L = lab work. Then c PR [ L] = PR [ ] + PL [ ] PR [ L] = PR [ ] + PL [ ] + P[( R L) ] c = pr [ ] + PL [ ] + PR [ L ] = = Solution: D First note P A B = P A + P B P A B [ ] [ ] [ ] [ ] [ ] c c c P A B = P A + P B P A B Then add these two equations to get [ ] [ ] [ ] [ ] P A =.6 [ ] P[ A] ( ) ( [ ] ) c c c P A B + P A B = P A + P B + P B P A B + P A B c = P[ A] + P ( A B) ( A B ).6 = P A + Page of 9

3 4. Solution: A For i =,, let Ri = event that a red ball is drawn from urn i and let Bi = event that a blue ball is drawn from urn i. Then, if x is the number of blue balls in urn,.44 = Pr[ ( R R) ( B B) ] = Pr[ R R] + Pr[ B B] = Pr R Pr R + Pr B Pr B [ ] [ ] [ ] [ ] x = + x+ 6 x+ 6 Therefore, 3 3x 3x+ 3. = + = x+ 6 x+ 6 x+ 6.x+ 35. = 3x+ 3.8x = 3. x = 4 5. Solution: D Let N(C) denote the number of policyholders in classification C. Then N(Young and Female and Single) = N(Young and Female) N(Young and Female and Married) = N(Young) N(Young and Male) [N(Young and Married) N(Young and Married and Male)] = 3 3 (4 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 = = c P F = = c P H F c and P H F = = = =.73 c P F Page 3 of 9

4 7. Solution: D Let A = event that a policyholder has an auto policy and H = event that a policyholder has a homeowners policy. Then, 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 =.35 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)(.35) (.8)(.5).53 ( 53% ) = + + = = 8. Solution: D Let C = event that patient visits a chiropractor and 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] Pr[ T]. = Pr[ T ].8 or PT = + = [ ] (.88.8) Solution: B [ ] [ ] [ ] [ ] Let M = event that customer insures more than one car and S = event that customer insurers a sports car. Then applying DeMorgan s Law, compute the desired probability as: c c c Pr ( M S ) = Pr ( M S) = Pr ( M S) = Pr ( M) + Pr ( S) Pr ( M S) = Pr M Pr S + Pr SM Pr M = =.5 ( ) ( ) ( ) ( ) ( )( ). This question duplicates Question 9 and has been deleted Page 4 of 9

5 . Solution: B Let C = Event that a policyholder buys collision coverage and D = Event that a policyholder buys disability coverage. Then we are given that PC [ ] = PD [ ] and PC [ D] =.5. By the independence of C and D,.5 = [ C D] = PCPD [ ] [ ] = PD [ ] PD [ ] =.5 / =.75 PD [ ] =.75, PC [ ] =.75. Independence of C and D implies independence of C c and D c. Then PC [ c D c ] = PC [ c ] PD [ c ] = (.75)(.75) =.33.. Solution: E Boxed numbers in the table below were computed. High BP Low BP Norm BP Total Regular heartbeat Irregular heartbeat Total From the table, % of patients have a regular heartbeat and low blood pressure. 3. Solution: C Let x be the probability of having all three risk factors. P[ A B C] x = P[ A B C A B] = = 3 P[ A B] x+. It follows that x= ( x+.) = x x =.4 3 x =.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 [ ] ( ) ( ) ( ) = = = Page 5 of 9

6 4. Solution: A pk = pk = pk = pk 3 = = p k k p 5 = = pk = p = = p, p = 4/5 k= k= Therefore, P[N > ] = P[N ] = (4/5 + 4/5 x /5) = 4/5 = /5 = Solution: C k Let x be the probability of choosing A and B, but not C, y the probability of choosing A and C, but not B, z the probability of choosing B and C, but not A. We want to find w = (x + y + z). We have x + y = /4, x + z = /3, y + z = 5/. Adding these three equations gives 5 ( x+ y) + ( x+ z) + ( y+ z) = ( x+ y+ z) = x+ y+ z = w= ( x+ y+ z) = =. Alternatively the three equations can be solved to give x = /, y = /6, z =/4 again leading to w = + + = Solution: D Let N and N denote the number of claims during weeks one and two, respectively. Then since they are independent, 7 P N + N = 7 = P N = n Pr N = 7 n [ ] [ ] [ n= ] 7 = n+ 8 n n= 7 = n= 9 8 = = = Page 6 of 9

7 7. Solution: D Let O = event of operating room charges and E = event of emergency room charges. Then.85 = P O E = P O + P E P O E ( ) ( ) ( ) ( ) ( ) ( ) ( ) P ( ) ( Independence) = P O + P E P O E P E =.5 = P E, PE ( ) =.75, c Because ( ) ( ) ( ) P( O)( ).85 = P O PO ( )(.75) = =. PO ( ) =. /.5 = Solution: D Let X and X denote the measurement errors of the less and more accurate instruments, respectively. If N ( µσ, ) denotes a normal random variable then X N(,.56 h), X N(, 44 h) and they are independent. It follows that X+ X.56 h +.44 h Y = N(, =.356 h). Therefore, 4.5h.5h P(.5h Y.5 h) = P Z.356h.356h = P(.4 Z.4) = PZ (.4) [ PZ (.4)] = (.99) = 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 -3 B 3 = Event the driver s age is in the range 3-65 B 4 = Event the driver s age is in the range Then P( AB) P( B) P( B A) = P AB P B + P AB P B + P AB P B + P AB P B ( ) ( ) ( ) ( ) ( 3) ( 3) ( 4) ( 4) (.6)(.8) (.6)(.8) + (.3)(.5) + (.)(.49) + (.4)(.8) = =.584 Page 7 of 9

8 . 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[ DU ] PU [ ] PU [ D] = P D S P S + P D F P F + P D U P U = [ ] [ ] [ ] [ ] [ ] [ ] (.)(.) (.)(.5) + (.5)(.4) + (.)(.) =.4. Solution: B P Seri. Surv. = P Surv. Seri. P[ Seri. ] P Surv. Crit. P[ Crit. ] + P Surv. Seri. P[ Seri. ] + P Surv. Stab. P[ Stab. ] (.9)(.3).9 (.6)(.) + (.9)(.3) + (.99)(.6) = =. Solution: D Let H = heavy smoker, L = light smoker, N = non-smoker, D = death within five-year period. We are given that PD [ L] = PD [ N] and PD [ L] = PD [ H] Therefore, P DH P[ H] P HD = P D N P[ N] + P D L P[ L] + P D H P[ H] P DL (.).4 = = =.4 P DL (.5) + P DL (.3) + P DL (. + + ) Page 8 of 9

9 3. 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, PCY [ ] PY [ ] PY [ C] = PCT [ ] PT [ ] + PCY [ ] PY [ ] + PCM [ ] PM [ ] + PCS [ ] PS [ ] (.8)(.6) = =.. (.5)(.8) + (.8)(.6) + (.4)(.45) + (.5)(.3) 4. Solution: B [ N ] P[ N 4] P 4 P N N 4 = = = = = Solution: B Let Y = positive test result D = disease is present Then, PY [ DPD ] [ ] (.95)(.) PD [ Y] = = =.657. c c PY [ DPD ] [ ] + PY [ D ] PD [ ] (.95)(.) + (.5)(.99) 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 ] Then,, PS [ CPC ] [ ] PC [ S] = c c PS [ CPC ] [ ] + PS [ C ] PC [ ] c PS [ C ] PC [ ] (.5) = = = = c c PS [ C ] PC [ ] + PS [ C ]( PC [ ]) (.5) Page 9 of 9

10 7. Solution: D Let B, C, and D be the events of an accident occurring in 4, 3, and, respectively. Let A= B C D. PABPB [ ] [ ] PB [ A] = PAB [ ][ PB [ ] + PACPC [ ] [ ] + PADPD [ ] [ ] Use Bayes Theorem (.5)(.6) = =.45. (.5)(.6) + (.)(.8) + (.3)(.) 8. Solution: A Let C = Event that shipment came from Company X I = Event that one of the vaccine vials tested is ineffective PI [ CPC ] [ ] Then, PC [ I] =. c c P I C P C + P I C P C Now [ ] PC = 5 [ ] [ ] c 4 PC = PC [ ] = = 5 5 P I C =..9 = [ ] ( )( )( ) c 3 ( )( )( ) P I C =..98 =.334 Therefore, PC [ I ] = (.4)( / 5).4 / / 5 = Solution: C ( )( ) ( )( ) 9 Let T denote the number of days that elapse before a high-risk driver is involved in an accident. Then T is exponentially distributed with unknown parameter λ. We are given that 5 λ t λt 5 5λ λ.3 = P[ T 5] = e dt = e = e. Therefore, e Then, = 5λ =.7 and λ = (/ 5) ln(.7). 8 λ λt λt 8 8λ (8/5)ln(.7) 8/5 P[ T 8] = e dt = e = e = e 8λ = e =.7 =.435. Page of 9

11 3. Solution: D λ λ 4 e λ e λ Let N be the number of claims filed. We are given PN [ = ] = = 3 PN [ = 4] = 3.! 4! Then, 3 4 λ = λ or λ = 4 or λ =, which is the variance of N Solution: D Let X denote the number of employees who achieve the high performance level. Then X follows a binomial distribution with parameters n = and p =.. Now we want to determine x such that P[X > x] <. or equivalently.99 [ ] x ( )(.) k (.98) P X x = k k = k The first three probabilities (at,, and ) are.668,.7, and.53. The total is.993 and so the smallest x that has the probability exceed.99 is. Thus C = / = 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(x, y, z) = probability function of X, Y, and Z Then f is a trinomial probability function, so P z x+ = f,, 4 + f,,3 + f,,3 + f,, [ ] ( ) ( ) ( ) ( ) 33. Solution: B = + + 4! +!! =.488 (.) 4(.5)(.) 4(.3)(.) (.3) (.) P[ X > x] =.5( t) dt =.5 x t t x =.5 4 x+ x =.5 x+ x where < x <. Therefore, P[ X > 6] ( 6) + ( 6) 8 P 6 8 X > X > = = = =. P[ X > 8] ( 8) + ( 8) 7 9 Page of 9

12 34. Solution: C + for < x < 4 (equals zero otherwise). First, determine the proportionality constant C. 4 4 C C = C ( + x) dx = C( + x) = = C 5 5 So C = 5/ or.5. Then, calculate the probability over the interval (, 6): ( + x) dx =.5( + x) =.5.47 = 6. We know the density has the form C( x) 35. This question duplicates Question 34 and has been deleted 36. Solution: B To determine k, 4 k 5 k, so k = = k ( y) dy ( y) = = = We next need to find P[V >,] = P[, Y >,] = P[Y >.], which is 4 5 ( ) ( ) 5 5 y dy = y =.9 =.59 and P[V > 4,] which is ( y) dy ( y) 5 = =.6 =.78. Then,.4 PV [ > 4, V >, ] PV [ > 4, ].78 PV [ > 4, V >, ] = = = =.3. PV [ >, ] PV [ >, ] Solution: D / Let Τ denote printer lifetime. The distribution function is Ft () = e t. The probability of failure in the first year is F() =.3935 and the probability of failure in the second year is F() F() = =.386. Of printers, the expected number of failures is and 3.86 for the two periods. The total expected cost is (39.35) + (3.86) =, Solution: A The distribution function is x 4 3 x 3 F ( x ) = P [ X x ] = 3 t dt = t = x [ ] P[ X.5] Pr[ X.5] P ( X < ) and ( X.5) PX [ < ] Pr[X <.5] PX [ < X.5] = = 3 3 F() F(.5) ( ) (.5 ) / / 7 37 = = = = = F(.5) (.5 ) 8 / Then, Page of 9

13 39. Solution: E The number of hurricanes has a binomial distribution with n = and p =.5. Then 9 8 PX< [ 3] =.95 + (.95) (.5) + 9(.95) (.5) = Solution: B Denote the insurance payment by the random variable Y. Then if < X C Y = X C if C < X < We are given that.5+ C.5+ C.64 = P [ Y <.5] = P [ < X <.5 + C ] = x dx = x = (.5 + C ). The quadratic equation has roots at C =.3 and.3. Because C must be between and, the solution is C = Solution: E The number completing the study in a single group is binomial (,.8). For a single group the 9 probability that at least nine complete the study is ( 9 )(.8) (.) + ( )(.8) =.376 The probability that this happens for one group but not the other is.376(.64) +.64(.376) = Solution: D There are two situations where Company B s total exceeds Company A s. First, Company B has at least one claim and Company A has no claims. This probability is.3(.6) =.8. Second, both have claims. This probability is.3(.4) =.. Given that both have claims, the distribution of B s claims minus A s claims is normal with mean 9,, =, and standard deviation, +, =, The probability that the difference exceeds zero is the probability that a standard normal variable exceeds (,) =.354. The,88.43 probability is.638 =.36. The probability of the desired event is.8 +.(.36) =.3. Page 3 of 9

14 43. Solution: D One way to view this event is that in the first seven months there must be at least four with no accidents. These are binomial probabilities: ( 4) ( 5) ( 6) ( 7) = =.897. Alternatively, consider a negative binomial distribution where K is the number of failures before the fourth success (no accidents). Then PK< [ 4] = = Solution: C ( ) ( ) ( 3) The probabilities of,, 3, 4, and 5 days of hospitalization are 5/5, 4/5, 3/5, /5, and /5 respectively. The expected payments are,, 3, 35, and 4 respectively. The expected value is [(5) + (4) + 3(3) + 35() + 4()]/5 =. 45. Solution: D x 4 x x x E ( X ) = x dx + x dx = + = + = = Solution: D The density function of T is t/3 f( t) = e, < t < 3 Therefore, t/3 t t/3 E [ X ] = E max ( T, ) = e dt + 3 e dt 3 t/3 t/3 t/3 /3 /3 t/3 = e te + e dt = e + + e 3e /3 = + 3e Alternatively, with probability e /3 the device fails in the first two years and contributes to the expected value. With the remaining probability the expected value is + 3 = 5 (employing /3 /3 /3 the memoryless property). The unconditional expected value is ( e ) + ( e )5 = + 3 e. Page 4 of 9

15 47. Solution: D We want to find x such that = E[P] = x e t/ dt + 3 x e t/ dt = 3 t/ t/ = x(.) e dt +.5 x(.) e dt t/ t/ 3 = xe.5xe / 3/ / = xe + x.5xe +.5xe =.77 x. Thus x = Solution: E EY = [ ] 4(.4) 3(.6)(.4) (.6) (.4) (.6) (.4) = This question duplicates Question 44 and has been deleted 5. Solution: C The expected payment is n 3/ n 3/ (3 / ) e (3 / ) e, ( n ) =, ( n ), ( ) e n= n! n= n! = + e = 3/, (.5 ), 7, Solution: C The expected payment is (.6).5(.6).5 x x x dx dx.5(.6) x + x = = + + = (.6) / Page 5 of 9

16 5. Solution: A First, determine K = K = K = K K = 37 Then, after applying the deductible, the expected payment is.5[(3 ) PN ( = 3) + (4 ) PN ( = 4) + (5 ) PN ( = 5)] =.5(6 / 37)[(/ 3) + (/ 4) + 3(/ 5)] = Solution: D The expected payment is 3 3 y dy + dy = = + + =.9. y y y y 54. Solution: B The expected payment is (in thousands) (.94)() + (.)(5 ) +.4 ( ).53 5 ( ) 7.5 x/ e ( e ) 5 ( ) ( )( ) x/ x e dx 5 x/ 5 x/ =.8 + (.) e ( x ) + e dx =.8 +. (4) + 4 =.8 +. e (4) 4e + 4e = = Solution: C k k k = dx = = and so k = ( + x) 3 ( + x) 3 The expected value is (where the substitution u = + x is used x dx = 3( u ) u du = 3 u / ( ) 3 u / ( 3) = 3 / = /. 4 ( + x) Integration by parts may also be used. Page 6 of 9

17 56. Solution: C With no deductible, the expected payment is 5. With the deductible it is to be 5. Let d be the deductible. Then, ( x d) 5 = ( x d)(.) dx = (.) =.5 ( d) d d 5, = ( d) 5 = d d = Solution: B This is the moment generating function of a gamma distribution with parameters 4 and,5. The standard deviation is the square root of the shape parameter times the scale parameter, or (,5) = 5,. But such recognition is not necessary. 5 M ( t) = 4(5)( 5 t) 6 M ( t) = (5) ( 5 t) EX ( ) = M () =, EX ( ) = M () = 5,, Var( X ) = 5,,, = 5,, SD( X ) = 5, 58. Solution: E Because the losses are independent, the mgf of their sum is the product of the individual mgfs, which is ( t) -. The third moment can be determined by evaluating the third derivative at zero. This is ()()()()()()( ()) -3 =,56. Page 7 of 9

18 59. Solution: B The distribution function of X is.5.5 x.5 x.5() () () F( x) = dt = =, x > t t x The pth percentile xp of X is given by.5 p () = F( xp ) =.5 x ().p = x.4 p = (. ).5 p.5 x p p xp =.4 (. p) It follows that x7 x3 = = (.3) (.7) 6. Solution: E Let X and Y denote the annual cost of maintaining and repairing a car before and after the % tax, respectively. Then Y =.X and Var( Y ) = Var(. X ) =.44 Var( X ) =.44(6) = This question duplicates Question 59 and has been deleted 6. Solution: C First note that the distribution function jumps ½ at x =, so there is discrete probability at that point. From to, the density function is the derivative of the distribution function, x. Then, 3 x x E( X ) = () + x( x ) dx = + = + + = 4 3 x x = + = + = + + = E( X ) () x ( x ) dx Var( X) = EX ( ) [ EX ( )] = = = Page 8 of 9

19 63. Solution: C E[ Y ] = x(.) dx + 4(.) dx =.x +.8 = E[ Y ] = x (.) dx + 4 (.) dx = (. / 3) x + 3. = ( EY ) Var[Y] = EY [ ] [ ] = = Solution: A The mean is (.5) + 3(.) + 4(.5) + 5(.) + 6(.) + 7(.) + 8(.3) = 55. The second moment is 4(.5) + 9(.) + 6(.5) + 5(.) + 36(.) + 49(.) + 64(.3) = 35. The variance is = 475. The standard deviation is.79. The range within one standard deviation of the mean is 33. to 76.79, which includes the values 4, 5, 6, and 7. The sum of the probabilities for those values is = Solution: B Let Y be the amount of the insurance payment E[ Y ] = ( x 5) dx = ( x 5) = = E[ Y ] = ( x 5) dx = ( x 5) = = 434,8 Var[Y] = 434, 8 (5) = 6,587 SD[ Y ] = DELETED 67. Solution: B The expected amount paid is (where N is the number of consecutive days of rain).6 e PN [ = ] + PN [ > ] = + ( e e.6) = 573.! The second moment is.6 e PN [ = ] + PN [ > ] = + ( e e.6) = 86,893.! The variance is 86, = 488,564 and the standard deviation is 699. Page 9 of 9

20 68. Solution: C X has an exponential distribution. Therefore, c =.4 and the distribution function is.4x F( x) = e. For the moment, ignore the maximum benefit. The median is the solution to.4m.5 = Fm ( ) = e, which is m = 5ln(.5) = Because this is below the maximum benefit, it is the median regardless of the existence of the maximum. Note that had the question asked for a percentile such that the solution without the maximum exceeds 5, then the answer is Solution: D t θ The distribution function of an exponential random variable, T, is Ft ( ) = e, t>. With a 4/ median of four hours,.5 = F(4) = e θ and so θ = 4 / ln(.5). The probability the 5ln(.5)/4 5/4 component works for at least five hours is PT [ 5] = F(5) = + e =.5 = Solution: E This is a conditional probability. The solution is P[ X p] F( p) F().95 = PX [ p X> ] = = PX [ > ] F() p/3 /3 /3 p/3 e + e e e p = = = e /3 /3 + e e ( p )/3.5 = e.9957 = ( p ) / 3 p = Solution: A ( )/3 The distribution function of Y is given by G( y) = PT ( y) = PT ( y) = F( y) = 4 y for y > 4. Differentiate to obtain the density function g( y) = 4y. 3 Alternatively, the density function of T f() t = F () t = 8t.5.5. We have t = y and dt =.5y dy Then g( y) = f ( y ) dt dy = 8( y ) (.5 y ) = 4y. Page of 9

21 7. Solution: E The distribution function of V is given by R F( v) = PV [ v] = P,e v = P R ln( v) ln(,) ln( v) ln(,).4 [ ] ln( v) ln(,) r = dr = = 5ln( v) 5ln(, ) v = 5 ln.4., 73. Solution: E F( y) = PY [ y] = P[ X y] = Pr X (. y) = e Therefore, (. y). f( y) F ( y).5(. y) e.5.5 (. y) = =. 74. Solution: E First note R = /T. Then, F( r) = P[ R r] = P[ / T r] = P[ T / r] =.5dt =.5( / r). The density function is f( r) = F ( r) =.5 / r. Alternatively, t = /r and dt = /r dr. Then fr( r) = ft( / r) dt / dr =.5( / r ) =.5 / r. 75. Solution: A Let Y be the profit for Company II, so Y = X. G( y) = P[ Y y] = P[ X y] = P[ X y/ ] = F( y/ ) g( y) = G ( y) = F ( y/ ) = (/ ) f( y/ ). Alternatively, X = Y/ and dx = dy/. Then, g( y) = f ( y / ) dx / dy = f ( y / )(/ ). / r Page of 9

22 76. Solution: A The distribution function of X is given by x 3 3 x 3 F( x) = dt = t = x, x > 4 t Next, let X, X, and X3 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 3 3 Gy ( ) = PX [ ypx ] [ ypx ] [ 3 y] = ( y ). The density function of Y is given by 3 4 g( y) = G ( y) = 3( y ) (3 y ). Therefore, E[ Y ] = y3( y ) 3y dy = 9 y y + y dy y / y /5 y /8 9 / /5 /8 [ ] = + = + =.5 (in thousands). 77. Solution: D The probability it works for at least one hour is the probability that both components work for more than one hour. This probability is x + y.5x + xy.5 + y.5y +.5y dxdy = dy = dy 8 = = 8 8 The probability of failing within one hour is the complement, This question duplicates Question 77 and has been deleted 79. Solution: E Let s be on the horizontal axis and t be on the vertical axis. The event in question covers all but the upper right quarter of the unit square. The probability is the integral over the other three quarters. Answer (A) is the lower left quarter. Answer B is the left half. Answer (C) is the upper right quarter. Answer (D) is the lower half plus the left half, so the lower left quarter is counted twice. Answer (E) is the lower right corner plus the left half, which is the correct region. For this question, the regions don t actually need to be identified. The area is.75 while the five answer choices integrate over regions of area.5,.5,.5,, and.75 respectively. So only (E) can be correct. Page of 9

23 8. Solution: C The mean and standard deviation for the 5 contributions are 5(35) = 6,38,5 and 45(5) =,5. By the central limit theorem, the total contributions are approximately normally distributed. The 9 th percentile is the mean plus.8 standard deviations or 6,38,5 +.8(,5) = 6,34, Solution: C The average has the same mean as a single claim, 9,4. The standard deviation is that for a single claim divided by the square root of the sample size, 5,/5 =,. The probability of exceeding, is the probability that a standard normal variable exceeds (, 9,4)/, =.6. From the tables, this is.757 = Solution: B A single policy has a mean and variance of claims. For 5 polices the mean and variance of the total are both 5. The standard deviation is the square root, or 5. The approximate probability of being between 45 and 6 is the same as a standard normal random variable being between (45 5)/5 = and (6 5)/5 =. From the tables, the probability is.977 (.843) = Solution: B Let n be the number of bulbs purchased. The mean lifetime is 3n and the variance is n. From the normal tables, a probability of.977 is standard deviations below the mean. Hence 4 = 3n sqrt(n). Let m be the square root of n. The quadratic equation is 3m m 4. The roots are 4 and /3. So n is either 6 or /9. At 6 the mean is 48 and the standard deviation is 4, which works. At /9 the mean is /3 and the standard deviation is /3. In this case 4 is two standard deviations above the mean, and so is not appropriate. Thus 6 is the correct choice. 84. Solution: B Observe that (where Z is total hours for a randomly selected person) EZ [ ] = EX [ + Y] = EX [ ] + EY [ ] = 5 + = 7 Var[ Z] = Var[ X + Y ] = Var[ X ] + Var[ Y ] + Cov[ X, Y ] = =. It then follows from the Central Limit Theorem that T is approximately normal with mean (7) = 7 and variance () =, and standard deviation. The probability of being less than 7 is the probability that a standard normal variable is less than (7 7)/ =. From the tables, this is.843. Page 3 of 9

24 85. Solution: B A single policy has an exponential distribution with mean and standard deviation. The premium is then + =. For policies, the total claims have mean () =, and standard deviation () =,. Total premiums are () =,. The probability of exceeding this number is the probability that a standard normal variable exceeds (,,)/, =. From the tables this probability is.843 = Solution: E For a single recruit, the probability of pensions is.6, of pension is.4(.5) =., and of pensions is.4(.75) =.3. The expected number of pensions is (.6) + (.) + (.3) =.7. The second moment is (.6) + (.) + 4(.3) =.3. The variance is.3.49 =.8. For recruits the mean is 7 and the variance is 8. The probability of providing at most 9 pensions is (with a continuity correction) the probability of being below 9.5. This is (9.5 7)/9 =.8 standard deviations above the mean. From the tables, this probability is Solution: D For one observation, the mean is and the variance is 5/ (for a uniform distribution the variance is the square of the range divided by ). For 48 observations, the average has a mean of and a variance of (5/)/48 =.434. The standard deviation is years is.5/.83 =. standard deviations from the mean. From the normal tables the probability of being within. standard deviations is.8849 (.8849) = Solution: C For a good driver, the probability is e 3/6 and for a bad driver, the probability is 3/6 /3 / /3 7/6 probability of both is the product, ( e )( e ) = e e + e. 89. Solution: B e /3. The The probability both variables exceed is represented by the triangle with vertices (,), (,3), and (3,). All the answer choices have x as the outer integral and x ranges from to 3, eliminating answers (A), (D), and (E). For a given value of x, the triangle runs from the base at y = to the diagonal line at y = 5 x. This is answer (B). Page 4 of 9

25 9. Solution: C Let B be the time until the next Basic Policy claim, and let D be the time until the next Deluxe policy claim. Then the joint pdf of B and D is b/ d/3 b/ d/3 f(, bd) = e e = e e. 3 6 The desired probability is b b b/ d/3 b/ d/3 P[ B > D] = e e dddb = e e db 6 b/ b/3 b/ 5 b/6 b/ 3 = e e + e db = e e = = Solution: D x + y xy + y.5y P[ X + Y ] = dydx = dx 4 4 x x 4x+ 4 x( x) + ( x).5( x) = 4 4.5x + 3x+.5.5 / 3+ 3 / +.5 = dx = = dx 9. Solution: B Because the distribution is uniform, the probability is the area of the event in question divided by the overall area of () = 4,. It is easier to get the complement as it comprises two triangles. One triangle has vertices at (,), (,), and (8,). The area is 8(8)/ = 6,. The other triangle has the same area for a total of 3,4. The area in question is 7,6 for a probability of Solution: C Because losses are uniformly distribution, the desired probability is the area of the event in question divided by the total area, in this case () =. Let X be the loss on the policy with a deductible of and Y the loss on the policy with a deductible of. The area where total payment is less than 5 can be broken down as follows: For < X <, there is no payment on that policy. Then Y must be less than 7. Area equals 7. For < X < 6, the payment on that policy is X. The other policy must have a loss of less than + 5 (X ) = 8 X. This is a trapezoid with a base of width 5 and a height that starts at 7 and decreases linearly to. The area is 5(7 + )/ =.5. The total area is 9.5 and so the probability is.95. Page 5 of 9

26 94. Solution: C First note that due to symmetry the two random variables have the same mean. Second note that probability is on the square from to 6 but the upper right corner is cut off by the diagonal line where the sum is. The integral must be split into two parts. The first runs horizontally from to 4 and vertically from to 6. The second runs horizontally from 4 to 6 and vertically from to t. The area of this total region is 4(6) + (6 + 4)/ = 34. So the constant density is /34. The mean is 6 t t 4 t 6 t E[ T] = t dtdt+ t 4 dtdt t dt t 4 dt 34 = t 6 3t 5 t t /3 dt ( t t ) dt = + = = = The expected sum is (.86) = 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 The tour operator collects x5 = 5 for the tickets sold. The probability that all passengers will show up is (.) = (.98) =.65. Therefore, the tour operator s expected revenue is 5 (65) = Solution: C The domain has area L / and so the density function is /L. Then, L t L t 3 + = + = ( ) L + L L L L = ( t /3 + t) dt = ( t /3) =. L L 3 E[ T T ] ( t t ) dt dt t /3 t t dt. Page 6 of 9

27 98. Solution: A The product of the three variables is only if all three are, so P[Y = ] = 8/7. The remaining probability of 9/7 is on the value. The mgf is ty t() t() 9 8 t M( t) = Ee [ ] = e (9 / 7) + e (8 / 7) = + e Solution: C First obtain the covariance of the two variables as (7, 5,,)/ =,. The requested variance is Var( X + +. Y ) = Var( X ) + Var(. Y ) + Cov( X,. Y ) = Var( X ) +. Var( Y ) + (.) Cov( X, Y ) = 5, +.(, ) +.(, ) = 9,3.. Solution: B P(X = ) = /6 P(X = ) = / + /6 = 3/ P(X = ) = / + /3 + /6 = 7/. E[X] = ()(/6) + ()(3/) + ()(7/) = 7/ E[X ] = () (/6) + () (3/) + () (7/) = 3/ Var[X] = 3/ (7/) =.58.. Solution: D Due to the independence of X and Y Var( Z) = Var(3X Y 5) = 3 Var( X ) + ( ) Var( Y ) = 9() + =.. Solution: E Let X and Y denote the times that the generators can operate. Now the variance of an exponential random variable is the square of them mean, so each generator has a variance of. Because they are independent, the variance of the sum is. 3. Solution: E Let S, F, and T be the losses due to storm, fire, and theft respectively. Let Y = max(s,f,t). Then, PY [ > 3] = PY [ 3] = P[max( S, F, T) 3] = PS [ 3] PF [ 3] PT [ 3] 3/ 3/.5 3/.4 = ( e )( e )( e ) =.44. Page 7 of 9

28 4. Solution: B First determine k:.5 = kxdxdy = kx dy =.5 kdy =.5 k k =. Then E[ X ] = x dydx = x dx = 3 E[ Y ] = y x dxdy = ydy = E[ XY ] = x ydxdy = ( / 3) ydy = 3 Cov( XY, ) = EXY [ ] EXEY [ ] [ ] = =. 3 3 Alternatively, note that the density function factors as the product of one term that depends only on x and one that depends only on y. Therefore, the two variables are independent and the covariance must be. 5. Solution: A Note that although the density function factors into expressions involving only x and y, the variables are not independent. An additional requirement is that the domain be a rectangle. x 4 5 x x x E[ X ] = x y dy dx = ( 4 ) 4 3 x y dx = x x x dx x dx x 3 = = = x x x x E[ Y ] = xy dy dx = ( 8 ) 3 xy dx = x x x dx x dx x 9 = = = x x ( ) x 3 9 x E[ XY ] = x y dy dx = x y dx = x 8x x dx = x dx = = Cov( XY, ) = EXY [ ] EXEY [ ] [ ] = = Page 8 of 9

29 6. Solution: C The joint density function of X and Y is f ( x, y ) = f ( x ) f ( y x ) = =,<,. x x x < < y < x x x x y x x 44 E[ X ] = dydx = dx dx x = = = 4 4 x x y y x x 44 E[ Y ] = dydx = dx dx x = 4x = = x x 3 xy y x x 78 E[ XY ] = dydx = dx dx x = 4 = = Cov( X, Y ) = = Solution: A First obtain Var(X) = =.4, Var(Y) = =.4, Cov(X,Y) = (8.4.4)/ =.6. Then, Cov( C, C ) = Cov( X + Y, X +. Y ) = Cov( X, X ) +. Cov( X, Y ) + Cov( Y, X ) +. Cov( Y, Y ) = Var( X ) +. Var( Y ) +. Cov( X, Y ) =.4 +.(.4) +.(.6) = Solution: A t t The joint density of T and T is given by f( t, t) = e e, t >, t > Therefore, PX [ x] = P[ T+ T x] x t x t t/ t t/ x ( ) ( )/ ( ) x ( x t )/ x ( )/ x t t t t t x t e e dt dt e e dt e e dt = = = = e e e dt = e + e e = e + e + e x/ x. x/ x/ x x x/ = e + e The density function is the derivative, d x/ x x/ x e + e = e e. dx Page 9 of 9

30 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(x) be the density function of X= U/V, and F(x) be the distribution function of X. Then because U and V are independent, u v/ u v/ guv (, ) = e (.5 e ) =.5 e e, < uv, <. Then, F( x) = P[ X x] = PU [ / V x] = PU [ Vx] vx vx u v/ g( u, v) dudv.5 e e dudv u v/ vx vx v/ v/ vx v/ v/.5 = = = e e dv = e + e dv = e e = + = +. x+.5 x+.5 x+ Then, f( x) = F ( x) =. x +. Solution: C ( ) The calculations are: f( / 3, y) f( y x= / 3) =, < y< f ( / 3) 3 f X /3 X (/3) = 4(/3) ydy = 8(/3) /= 6/9 8y f( y x= / 3) = = 4.5 y, < y< 6 / 9 3 /3 [ < = / 3] = [ < / 3 = / 3] = 4.5 = 4.5(/ 3) / = / 4. PY X X PY X ydy Page 3 of 9

31 . Solution: E 3 f(, y) P[ < Y < 3 X = ] = dy f () f(, y) = y =.5y (4 ) ( ) 3 ( ) 3 f () = f (, y) dy =.5y dy =.5y =.5 X X 3 3.5y 3 P[ < Y < 3 X = ] = dy = y = / 9 + = 8 / Solution: D Because only those with the basic policy can purchase the supplemental policy, < y < x <. Then,.5 f(., y) P[ Y <.5 X =.] = dy f X (.) f(., y) = (. + y), < y< f (.) = f (., y) dy = (. + y) dy =.y + y =.3 X (. + y).y+ y.+.5 P[ Y <.5 X =.] = dy = = = Solution: E Because the husband has survived, the only possible claim payment is to the wife. So we need the probability that the wife dies within ten years given that the husband survives. The numerator of the conditional probability is the unique event that only the husband survives, with probability.. The denominator is the sum of two events, both survive (.96) and only the husband survives (.). The conditional probability is./(.96 +.) = /97. The expected claim payment is,/97 = 3 and the expected excess is, 3 = Solution: C PX ( =, Y= ) PX ( =, Y= ).5 PY [ = X= ] = = = =.86 PX ( = ) PX ( =, Y= ) + PX ( =, Y= ) PY [ = X= ] =.86 =.74. The conditional variable is Bernoulli with p =.74. The variance is (.74)(.86) =.4. Page 3 of 9

32 5. Solution: A f( y x) = f( xy, ) f ( x) x+ f ( x) = xdy = x X x X x f( y x) = =, x< y< x+. x The conditional variance is uniform on the interval (x, x + ). A uniform random variable on a unit interval has variance /. Alternatively the integrals can be done to obtain the mean of x +.5 and second moment of x + x + /3. The second moment minus the square of them mean gives the variance of /. 6. Solution: D With no tornadoes in County P the probabilities of,,, and 3 tornadoes in County Q are /5, 6/5, 5/5, and /5 respectively. The mean is ( )/5 = /5. The second moment is ( )/5 = 44/5. The variance is 44/5 (/5) = Solution: C The marginal density of X is x x f ( x) = 6( x y) dy = 6( y xy.5 y ) = 6[ x x( x).5( x) ] X = 6[ x x+ x.5 + x.5 x ] = 6(.5x x+.5), < x<. The requested probability is [.]. 6(.5.5) P X < = x x + dx = x x + x = = Solution: E y y 3 / ( ) = 5 = 5 = 5 ( ) = 5 ( ), < < y y g y y dx yx y y y y y y The limits are found by noting that x must be less than the square root of y and also must be greater than y. While not directly stated, the only values of x for which the square is smaller are < x <. This implies y is constrained to the same range and thus its square root must be larger, ensuring that the integral has the correct sign. Page 3 of 9

33 9. Solution: D The joint density is f( xy, ) = f ( xf ) ( y x) = () =, < x<, x< y< x+. The marginal density of Y is X min( y,) fy ( y) = dx = min( y,) max( y, ), < y <. Thus from max( y,) to the density is y and from to it is y. The requested probability is ydy + ydy =.5y +.5( y) = =.875 = 7 / Because the joint density is uniform, an alternative is to find the area represented by the event Y >.5.. Solution: A Let Y be the processing time. The joint density is 3x f( xy, ) = fx ( xf ) ( y x) =, x, x y x. 8 x min( y,) 3x 3 The marginal density of Y is fy ( y) = dx = [min( y,) y / 4], y 4. y/ P[ Y 3] = f ( ) [4 / 4] [4 /] 3 Y y dy = y dy = y y = [6 64 / + 7 /] = Solution: C The marginal density of X is given by 3 xy x f X ( x) = ( xy ) dy = y = Then, 3 x x x E[ X ] = x f ( ) X x dx = x dx = = + = Page 33 of 9

34 . Solution: D y 3 The marginal distribution of Y is ( ) 6 x y 6 x y 6 y 6 y f. Y y = e e dx= e e = e + e The expected value of Y can be found via integration by parts, or recognition, as illustrated here. 3y y 3y y E[ Y ] y( 6e 6 e ) dy y3e dt 3 = + = ye dt (/ 3) 3(/ ) 5 / = + = = With the constants factored out, the integrals are the expected value of exponential distributions with means /3 and /, respectively. 3. Solution: C From the Law of Total Probability: P[4 < S < 8] = P[4 < S < 8 N = ] P[ N = ] + P[4 < S < 8 N > ] P[ N > ] 4/5 8/5 4/8 8/8 = ( e e )(/ 3) + ( e e )(/ 6) =.. 4. Solution: A Because the domain is a rectangle and the density function factors, X and Y are independent. Also, Y has an exponential distribution with mean /. From the memoryless property, Y given Y > 3 has the same exponential distribution with 3 added. Adding a constant has no effect on the variance, so the answer is the square of the mean,.5. More formally, the conditional density is y f( y) e 6 y f( y Y > 3) = = = e, y > 3 6 PY ( > 3) e 6 y u 3 6 y u 3 E[ Y Y > 3] = y e dy = ( u + 3)e du = = 3.5 E[ Y Y > 3] = y e dy = ( u + 3) e du =.5 + 6(.5) + 9 =.5 Var[ Y Y > 3] = =.5. The second integrals use the transformation u = y 3 and then the integrals are moments of an exponential distribution. 5. Solution: E The joint density is f( xy, ) = f ( xf ) ( y x) = x(/ x) =,< y< x<. The marginal density of Y is The conditional density is X f ( y) = dx = ( y). Y y f( xy, ) f( x y) = = =. f ( y) ( y) y Y y Page 34 of 9

35 6. Solution: C Due to the equal spacing of probabilities, pn = p nc for c =,, 3, 4, 5. Also,.4 = p + p = p + p c= p c. Because the probabilities must sum to, = p + p c+ p c+ p 3c+ p 4c+ p 5c= 6p 5 c. This provides two equations in two unknowns. Multiplying the first equation by 5 gives 6 = 3 p 5 c. Subtracting the second equation gives 5 = 4 p p = 5 / 4. Inserting this in the first equation gives c = /6. The requested probability is p4 + p5 = 5 / 4 4 / / 4 5 / 6 = 3 / = Solution: D Because the number of payouts (including payouts of zero when the loss is below the deductible) is large, apply the central limit theorem and assume the total payout S is normal. For one loss, the mean, second moment, and variance of the payout are 5,, 5, 5,, dx x 5, 56, 5, 5, 65 = 4, 69,375., ( x 5, ) dx + ( 5, ) 5, 65, x = + =, 4, 5, 3, ( x 5, ) + ( 5, ) = + = 56, 5,,, 6, For losses the mean, variance, and standard deviation are,5,, 4,9,875,, and 7,56.,, is 5,/7,56 =.787 standard deviations below the mean and,, is 75,/7,56 =.69 standard deviations above the mean. From the standard normal tables, the probability of being between these values is.8575 (.966) = Solution: B Let H be the percentage of clients with homeowners insurance and R be the percentage of clients with renters insurance. Because 36% of clients do not have auto insurance and none have both homeowners and renters insurance, we calculate that 8% (36% 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 35%. Because H = R, R must be 8%, which implies that % have both renters and auto insurance. 5, Page 35 of 9

36 9. Solution: B Let Y be the reimbursement. Then, G(5) = P[Y < 5 X > ]. For Y to be 5, the costs must be above (up to accounts for a reimbursement of ). The extra 5 requires 3 in additional costs. Therefore, we need 5/ / PX [ 5] PX [ ] e + e PX [ 5 X> ] = = / PX [ > ] + e.5. e +.3 = = e =.77., e 3. Solution: C X E[(.5) ] = Ee [ ] = M X (ln.5) = = 4.9. ln.5 3. Solution: E X ln.5 The conditional probability function given claims in April is 3 n 3e 3 pn () = e ( e ) = = n = ( e ) 6 p(, n ) 3 n 6 n p( n N = ) = = e ( e ) = e ( e ). p () N This can be recognized as a geometric probability function and so the mean is / e = e. 3. Solution: C The number of defective modems is % x 3 + 8% x 5 =. 7 3 The probability that exactly two of a random sample of five are defective is = Page 36 of 9

37 33. Solution: B P(4 year old man dies before age 5) = P(T < 5 T > 4) Pr(4 < T < 5) F(5) F(4) = = Pr( T > 4) F(4) exp + exp exp exp = = exp exp = = Expected Benefit = 5(.696) = Solution: 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 3t and the relative frequency with which queen-sized mattresses are sold is (3t+t)/4, or t. Thus, t =. since t + 3t + t =. The probability we seek is 3t + t = Solution: E Var( N) = E[ Var( N λ)] + Var[ E( N λ)] = E[ λ] + Var( λ) = =.5. The variance of a uniform random variable is the square of the range divided by, in this case 3 / = Solution: D X follows a geometric distribution with p = /6 and Y = implies the first roll is not a 6 and the 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). EX [ X 3] = + = 6 + = 8., The expected value is.() +.8(8) = 6.6. p 37. Solution: 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, ( ).3 t t Kt = e e. This is the moment generating function of a discrete random variable that assumes the values,, and with respective probabilities.3,.4, and.3. The value we seek is thus =.7. Page 37 of 9

38 38. Solution: 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 expected operational time of the machine is 5. By symmetry, if the component represented by the random variable Y fails last, the expected operational time of the machine is also 5. Thus, the unconditional expected operational time of the machine must be 5 as well. 39. Solution: 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 expected number of people in the car who are hospitalized, given that the total loss due to hospitalizations from the accident is less than is () + () + () = Solution: 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 maximize P[X = n]. Note that the ratio, for n. n PX [ = n+ ] n(.4) (.6) n = n = PX [ = n] ( n )(.4) (.6) n (.6). This ratio of consecutive probabilities is greater than when n = and less than when n 3. Thus, P[X = n] is maximized at n = 3; the mode is 3. Alternatively, the first few probabilities could be calculated. Page 38 of 9

39 4. Solution: C There are (5 choose 3) 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 six 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 3 ways to select the first item. Because there are squares in the row or column of the first selected item, there are 3 = 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 3 8 = ways to select the third item. The number of permutations of three qualifying items is (3)()(). The number of combinations is thus (3)()()/3! =. 4. Solution: B The expected bonus for a high-risk driver is.8()(5) = 48. The expected bonus for a low-risk driver is.9()(5) = 54. The expected bonus payment from the insurer is 6(48) + 4(54) = 5, Solution: E Liability but not property =. (given) Liability and property =.4. =.3. Property but not liability =..3 =.7 Probability of neither =..3.7 = Solution: E The total time is to e less than 6 minutes, so if x minutes are spent in the waiting room (in the range to 6), from to 6 x minutes are spent in the meeting itself. Page 39 of 9

40 45. Solution: C f(.75, y) f(.75, y) f(.75, y) f( y x=.75) = = = f (.75, y) dy.5dy +.75dy.5 Thus,.5 /.5 = 4 / 3 for < y <.5 f( y x=.75) =,.75 /.5 = / 3 for.5 < y < Then,.5 EY ( X=.75) = y(4 / 3)dy + y( / 3)dy = (/ 8)(4 / 3) + (3 / 8)( / 3) = 5 /.5.5 EY ( X=.75) = y(4 / 3)dy + y( / 3)dy = (/ 4)(4 / 3) + (7 / 4)( / 3) = 8 / 7 Var Y.5 ( X =.75) = 8 / 7 (5 /) = /44 = Solution: 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 = 45. Because C N A and H are mutually exclusive, the number of TV watchers in the set C N A H is = 63. The number of TV watchers in the complement of C N A H is thus 63 = 37. Page 4 of 9

41 47. Solution: 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 λ be the mean of X, which because X is exponential, implies that λ is the variance of X and EX ( ) = λ. By the memoryless property of the exponential distribution, the conditional distribution of the portion of a claim above the deductible given that the claim exceeds the deductible is an exponential distribution with mean λ. Given that EY ( ) =.9λ, this implies that the probability of a claim exceeding the deductible is.9 and thus EY ( ) =.9( λ ) =.8λ. Then, Var( Y ) =.8 λ (.9 λ) =.99λ. The percentage reduction is %. 48. Solution: 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 exponentially 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. Var( X ) = Var[ E( X N)] + E[ Var( 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) =,, (4) +,, (4) = 8,,. Page 4 of 9

42 49. Solution: B Let N denote the number of accidents, which is binomial with parameters 3 and.5 and thus has mean 3(.5) =.75 and variance 3(.5)(.75) =.565. Let X i denote the unreimbursed loss due to the i th accident, which is.3 times an exponentially distributed random variable with mean.8 and therefore variance (.8) =.64. Thus, X i has mean.8(.3) =.4 and variance.64(.3) =.576. 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. Var( X ) = Var[ E( X N)] + E[ Var( X N)] = Var[ E( X+ + X N)] + E[ Var( X+ + X N)] = Var[ NE( X)] + E[ NVar( X)] = Var(.4 N) + E(.576 N) =.4 Var( N) E( N) =.576(.565) +.576(.75) = Solution: 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 x be the 75 th 3 percentile of the given exponential distribution, F( x) = =.75, so x = 459. Subtracting the deductible of 5 gives 3659 as the (unconditional) 95 th percentile of the insurance company payout. e x Page 4 of 9

43 5. Solution: C The ratio of the probability that one of the damaged pieces is insured to the probability that none of the damaged pieces are insured is r 7 r r =, r 7 r 4 r where r is the total 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 r 7 r 8 9 (8)(7)(9)(8)(4)(3)()() 66 = = = = (7)(6)(5)(4)()()()() Solution: A The probability that Rahul examines exactly n policies is.(.9) n. The probability that Toby examines more than n policies is.8 n. The required probability is thus n n n.7.(.9) (.8) =.7 = = n= n= ( ) An alternative solution begins by imagining Rahul and Toby examine policies simultaneously until at least one of the finds a claim. At each examination there are four possible outcomes:. Both find a claim. The probability is... Rahul finds a claim and Toby does not. The probability is Toby finds a claim and Rahul does not. The probability is.8 4. Neither finds a claim. The probability is.7. Conditioning on the examination at which the process ends, the probability that it ends with Rahul being the first to find a claim (and hence needing to examine fewer policies) is.8/( ) = 8/8 =.857. Page 43 of 9

44 53. Solution: E Let a be the mean and variance of X and b be the mean and variance of Y. The two facts are a = b 8 and a + a =.6(b + b ). Substituting the first equation into the second gives b 8 + ( b 8) =.6b+.6b b 8 + b 6b+ 64 =.6b+.6b b b+ = ± 5.6 4(.4)(56) 5.6 ±.4 b = = = 4 or 35. (.4).8 At b = 4, a is negative, so the answer is Solution: C Suppose there are N red sectors. Let w be the probability of a player winning the game. Then, w = the probability of a player missing all the red sectors and N w = Using the geometric series formula, N w = = = Thus we need 9 9. > w = + 9. > > 9 9 > 9 N N N N N > 45 9 ln(45) N > ln( / 9) Thus N must be the first integer greater than 4.767, or 5. N N Page 44 of 9

45 55. Solution: B The fourth moment of X is 4 5 x x dx = =. 5 The Y probabilities are / for Y = and, and / for Y =,,,9. EY = = [ ] ( 9 ) / / The absolute value of the difference is Solution: E Px= y= = Py= x= Px= = = (, ) ( ) ( ).3(.5).75 Px= y= = Py= x= Px= = = 3 (, ) ( ) ( ).5(.5).35 Px= y= = Py= x= Px= = = The total is.35. (, ) ( ) ( ).5(.5) Solution: C p x p E( X ) = x dx = ( p ) x dx p p x p ( p ) = = p p p = ( p ) = p 4 p = Solution: D The distribution function plot shows that X has a point mass at with probability.5. From to 3 it has a continuous distribution. The density function is the derivative, which is the constant (.5)/(3 ) =.5. The expected value is (.5) plus the integral from to 3 of.5x. The integral evaluates to.5, which is the answer. Alternatively, this is a 5-5 mixture of a point mass at and a uniform(,3) distribution. The mean is.5() +.5(.5) = Solution: E The dice rolls that satisfy this event are (,), (,), (,3), (,), (,), (,3), (,4), (3,), (3,), (3,3), (3,4), (3,5), (4,), (4,3), (4,4), (4,5), (4,6), (5,3), (5,4), (5,5), (5,6), (6,4), (6,5), and (6,6). They represent 4 of the 36 equally likely outcomes for a probability of /3. Page 45 of 9