EXAM P SAMPLE SOLUTIONS

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16. Solution: D Let 1 denote the number of claims during weeks one and two, respectively. Then since are independent, and N 2 N N N 1 and 2 [ N1+ N2 = 7 ] = ∑ n= 0 [ N1 = n] [ N2 = −n] 7 = ∑ ⎛ ⎜ 1 ⎞⎛ n+ 1⎟⎜ 1 ⎞ 8−n ⎟ Pr 7 Pr Pr 7 n= 0 ⎝2 7 1 = ∑ n= 0 9 2 ⎠⎝2 ⎠ = 8 = 1 = 1 9 6 2 2 64 -------------------------------------------------------------------------------------------------------- 17. Solution: D Let O = Event of operating room charges E = Event of emergency room charges Then 0.85 = Pr ( O ∪ E) = Pr ( O) + Pr ( E) −Pr ( O∩E) = Pr ( O) + Pr ( E) −Pr( O) Pr ( E) ( Independence) Since c Pr ( E ) = 0.25 = 1− Pr ( E) , it follows Pr ( E ) = 0.75 . So 0.85 = Pr O + 0.75 −PrO 0.75 ( ) ( )( ) Pr ( O)( 1− 0.75) = 0.10 Pr ( O ) = 0.40 -------------------------------------------------------------------------------------------------------- 18. Solution: D Let X 1 and X 2 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 X1 is N(0, 0.0056h), X 2 is N(0, 0.0044h) and X 1, X2 are X1+ X2 independent. It follows that Y = 2 is N (0, 2 2 2 2 0.0056 h + 0.0044 h ) = N(0, 4 0.00356h) . Therefore, P[−0.005h ≤ Y ≤ 0.005h] = P[Y ≤ 0.005h] – P[Y ≤ −0.005h] = P[Y ≤ 0.005h] – P[Y ≥ 0.005h] ⎡ 0.005h ⎤ = 2P[Y ≤ 0.005h] – 1 = 2P ⎢ Z ≤ ⎣ 0.00356h ⎦ ⎥ − 1 = 2P[Z ≤ 1.4] – 1 = 2(0.9192) – 1 = 0.84. Page 8 of 55
19. Solution: B Apply Bayes’ Formula. Let A = Event of an accident B 1 = Event the driver’s age is in the range 16-20 B 2 = Event the driver’s age is in the range 21-30 B = Event the driver’s age is in the range 30-65 Then Pr 3 B = 4 ( B1A) Event the driver’s age is in the range 66-99 Pr ( AB1) Pr ( B1) ( A B1) ( B1) + ( A B2) ( B2) + ( A B3) ( B3) + ( A B4) ( B4) = Pr Pr Pr Pr Pr Pr Pr Pr ( 0.06)( 0.08) ( 0.06)( 0.08) + ( 0.03)( 0.15) + ( 0.02)( 0.49) + ( 0.04)( 0.28) = = 0.1584 -------------------------------------------------------------------------------------------------------- 20. 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 = [ ] [ ] [ ] [ ] [ ] [ ] ( 0.001)( 0.10) ( 0.01)( 0.50) + ( 0.005)( 0.40) + ( 0.001)( 0.10) = 0.0141 -------------------------------------------------------------------------------------------------------- 21. 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. ] ( 0.9)( 0.3) 0.29 ( 0.6)( 0.1) + ( 0.9)( 0.3) + ( 0.99)( 0.6) = = Page 9 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: 14. Solution: A 1 11 1 1 1 ⎛1⎞
Page 11 and 12: 25. Solution: B Let Y = positive te
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

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