EXAM P SAMPLE SOLUTIONS

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89. Solution: B We are given that ⎧ 6 ⎪ (50 − x− y) for 0 < x< 50 − y< 50 f( x, y) = ⎨125,000 ⎪⎩ 0 otherwise and we want to determine P[X > 20 ∩ Y > 20] . In order to determine integration limits, consider the following diagram: y 50 x>20 y>20 (20, 30) (30, 20) x 50 30 50− x 6 We conclude that P[X > 20 ∩ Y > 20] = (50 − x − y) 125,000 ∫∫ dy dx . -------------------------------------------------------------------------------------------------------- 90. Solution: C Let T1 be the time until the next Basic Policy claim, and let T2 be the time until the next Deluxe policy claim. Then the joint pdf of T 1 and T 2 is ⎛1 ⎞⎛1 ⎞ 1 f( t1, t2) = ⎜ e ⎟⎜ e ⎟= e e ⎝2 ⎠⎝3 ⎠ 6 P[T < T ] = 2 1 −t1/2 −t2/3 −t1/2 −t2/3 20 20 ∞ t1 ∞ 1 −t1/2 −t2/3 ⎡ 1 −t1/2 −t2/3⎤ t1 ∫∫ e e dt2dt1 = e e dt1 6 ∫⎢ − 2 ⎥ 0 0 0⎣ ⎦0 , 0 < t 1 < ∞ , 0 < t 2 < ∞ and we need to find ∞ ∞ ∞ ⎡1 −t1/2 1 −t1/2 −t1/3 ⎤ ⎡1 −t1/2 1 −5 t1/6⎤ ⎡ −t1/2 3 −5 t1/6⎤ 3 2 = ∫⎢ e − e e dt1 = e − e d 1 2 2 ⎥ ∫⎢ t 2 2 0⎣ ⎦ 0⎣ ⎥ = e e 1 ⎦ ⎢ − + = − = ⎣ 5 ⎥ ⎦05 5 = 0.4 . -------------------------------------------------------------------------------------------------------- 91. Solution: D We want to find P[X + Y > 1] . To this end, note that P[X + Y > 1] = 1 2 1 2 ⎡2x+ 2− y⎤ ⎡1 1 1 2 ⎤ ∫∫⎢ dydx xy y y dx 4 ⎥ = ∫ + − 2 2 8 01−x ⎣ ⎦ ⎢ ⎥ 0⎣ ⎦ 1−x 1 ⎡ 1 1 1 1 2 ⎤ = x + − − x −x − − x + −x dx ∫ 0 ⎢ 1 (1 ) (1 ) (1 ) ⎣ 2 2 2 8 ⎥ ⎦ 1 ⎡5 2 3 1⎤ = x + x+ dx ∫ 0 ⎢ ⎣8 4 8⎥ ⎦ 1 3 2 x + x + x 0 ⎡ 5 3 1 ⎤ = ⎢ ⎣24 8 8 ⎥ = ⎦ Page 38 of 55 = 1 ∫ 0 ⎡ 1 1 1 1 ⎤ ⎢ + + − + ⎣ 2 8 4 8 ⎥ ⎦ 2 2 x x x x dx 5 3 1 17 + + = 24 8 8 24
92. 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 20: Based on the graph and the uniform distribution: 1 − ⋅ Shaded Region Area Pr 20 2 ⎣⎡ X − Y < ⎤ ⎦ = = 2 2 200 ( 2200 − 2000) ( ) 2 200 2 180 2 180 2 = 1− = 1− 2 ( 0.9) = 0.19 200 More formally (still using symmetry) Pr ⎡⎣ X − Y < 20⎤ ⎦ = 1−Pr⎡⎣ X −Y ≥20⎤ ⎦ = 1−2Pr[ X −Y ≥20] 2200 x−20 1 2200 1 = 1− 2∫ 1 2 2020 ∫ dydx = − 2000 2 2020 2 200 ∫ 200 x−20 y 2000 dx 2 = 1− 2 200 ∫ 2020 x−20 − 2000 1 dx= 1− 2 200 x−2020 2 ⎛180 ⎞ = 1− ⎜ ⎟ = 0.19 ⎝200 ⎠ ( ) ( ) 2200 2 2200 2020 Page 39 of 55 2
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 and 10: 19. Solution: B Apply Bayes’ Form
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: -----------------------------------
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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