EXAM P SAMPLE SOLUTIONS

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22. 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 1 Now we are given that Pr ⎡⎣DL⎤ ⎦ = 2 Pr ⎡⎣DN⎤⎦ and Pr ⎡⎣DL⎤ ⎦ = Pr ⎡DH 2 ⎣ ⎤⎦ 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] ( ) 2Pr⎡DL⎤ 0.2 0.4 = ⎣ ⎦ = = 0.42 1 Pr⎡DL⎤ 0.25 0.3 0.4 ( 0.5) + Pr⎡DL⎤ ( 0.3) + 2Pr⎡DL⎤( 0.2 + + ) 2 ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ -------------------------------------------------------------------------------------------------------- 23. 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 [ ] (0.08)(0.16) = = 0.22 . (0.15)(0.08) + (0.08)(0.16) + (0.04)(0.45) + (0.05)(0.31) -------------------------------------------------------------------------------------------------------- 24. Solution: B Observe Pr 1≤ Pr ⎡⎣N ≥1N ≤4⎤ ⎦ = Pr ≤4 ≤ 4 ⎡1 1 1 1 ⎤ = ⎢ + + + ⎣6 12 20 30⎥ ⎦ ⎡1 1 1 1 1 ⎤ ⎢ + + + + ⎣2 6 12 20 30⎥ ⎦ [ N ] [ N ] 10 + 5 + 3+ 2 20 2 = = = 30 + 10 + 5 + 3+ 2 50 5 Page 10 of 55
25. Solution: B Let Y = positive test result D = disease is present (and ~D = not D) Using Baye’s theorem: PY [ | DPD ] [ ] (0.95)(0.01) PY [ | DPD ] [ ] PY [ |~ DP ] [~ D ] (0.95)(0.01) (0.005)(0.99) = P[D|Y] = = 0.657 . + + -------------------------------------------------------------------------------------------------------- 26. Solution: C Let: S = Event of a smoker C = Event of a circulation problem Then we are given that P[C] = 0.25 and P[S⏐C] = 2 P[S⏐C C ] PSCPC [ ] [ ] Now applying Bayes’ Theorem, we find that P[C⏐S] = C C PSCPC [ ] [ ] + PSC [ ]( PC [ ] ) C 2 PSC [ ] PC [ ] 2(0.25) 2 2 = = = = . C C 2 PSC [ ] PC [ ] + PSC [ ](1 −PC [ ]) 2(0.25) + 0.75 2 + 3 5 -------------------------------------------------------------------------------------------------------- 27. Solution: D Use Baye’s Theorem with A = the event of an accident in one of the years 1997, 1998 or 1999. PA [ 1997] P[1997] P[1997|A] = PA [ 1997][ P[1997] + PA [ 1998] P[1998] + PA [ 1999] P[1999] (0.05)(0.16) = = 0.45 . (0.05)(0.16) + (0.02)(0.18) + (0.03)(0.20) -------------------------------------------------------------------------------------------------------- Page 11 of 55
Page 1 and 2: SOCIETY OF ACTUARIES/CASUALTY ACTUA
Page 3 and 4: 4. Solution: A For i = 1, 2, let Ri
Page 5 and 6: 9. Solution: B Let M = event that c
Page 7 and 8: 14. Solution: A 1 11 1 1 1 ⎛1⎞
Page 9: 19. Solution: B Apply Bayes’ Form
Page 13 and 14: 31. Solution: D Let X denote the nu
Page 15 and 16: 36. Solution: B To determine k, not
Page 17 and 18: 41. Solution: E Let X = number of g
Page 19 and 20: 44. Solution: C If k is the number
Page 21 and 22: 5 ( ) = ∑ ( ) Pr[ = ] E⎡⎣f X
Page 23 and 24: 54. Solution: B Let Y denote the cl
Page 25 and 26: 59. Solution: B The distribution fu
Page 27 and 28: 64. Solution: A Let X denote claim
Page 29 and 30: 68. Solution: C Note that X has an
Page 31 and 32: 74. Solution: E First note R = 10/T
Page 33 and 34: ( X < ) ∪ ( Y < ) Pr ⎡⎣ 1 1
Page 35 and 36: 40 − 3n −2≥ n To find such an
Page 37 and 38: -----------------------------------
Page 39 and 40: 92. Solution: B Let X and Y denote
Page 41 and 42: Consequently, 1 2 1 2 ( , ) 34 f t
Page 43 and 44: 99. Solution: C We use the relation
Page 45 and 46: 105. Solution: A The calculation re
Page 47 and 48: 2 [ X Y] E⎡( X Y) ⎤ ( E[ X Y] )
Page 49 and 50: 111. Solution: E 3 f Pr ⎡ ⎣1< Y
Page 51 and 52: 116. Solution: D Denote the number
Page 53 and 54: 120. Solution: A We are given that
Page 55: 124. Solution: A − x −2 y Becau

EXAM P SAMPLE SOLUTIONS

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