Krzys’ Ostaszewski: http://www.krzysio.net
Author of the BTDT Manual (the “Been There Done That!” manual) for Course P/1
http://smartURL.it/krzysioP (paper) or http://smartURL.it/krzysioPe (electronic)
Instructor of online P/1 seminar: http://www.math.ilstu.edu/actuary/prepcourses.html If
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P Sample Exam Questions, Problem No. 152, also Dr. Ostaszewski’s online exercise
posted August 21, 2010
Automobile policies are separated into two groups: low-risk and high-risk. Actuary Rahul
examines low-risk policies, continuing until a policy with a claim is found and then
stopping. Actuary Toby follows the same procedure with high-risk policies. Each lowrisk policy has a 10% probability of having a claim. Each high-risk policy has a 20%
probability of having a claim. The claim statuses of polices are mutually independent.
Calculate the probability that Actuary Rahul examines fewer policies than Actuary Toby.
A. 0.2857
B. 0.3214
C. 0.3333
D. 0.3571
E. 0.4000
Solution.
For a positive integer n, the probability that Actuary Rahul examines exactly n policies is
0.9 n−1 ⋅ 0.1. The probability that Actuary Toby examines more than n policies is 0.8n.
Since they are independent, the probability that those two events happen simultaneously
is 0.9 n−1 ⋅ 0.1⋅ 0.8n. The probability that Actuary Rahul examines fewer policies than
Actuary Toby is the sum of that probability for each possible value of n, where n = 1, 2,
3, …, and thus the probability sought is
+∞
+∞
0.1
0.1 +∞
1 0.72
n−1
n
n
n
∑ 0.9 ⋅ 0.1⋅ 0.8 = ∑ 0.9 ⋅ 0.9 ⋅ 0.8 = 0.9 ⋅ ∑ 0.72n = 9 ⋅ 1 − 0.72 ≈ 0.2857.
n=1
n=1
n=1
Answer A.
© Copyright 2010 by Krzysztof Ostaszewski.
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