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 low-risk 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

• 0.2857
• 0.3214
• 0.3333
• 0.3571
• 0.4000

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Solution: A

The probability that Rahul examines exactly n policies is 0.1(0.9) n−1 . The probability that Toby examines more than n policies is 0.8n . The required probability is thus

[[math]] \sum_{n=1}^{\infty} 0.1 (0.9)^{n-1}(0.8)^n = \frac{1}{9} \sum_{n=1}^{\infty} 0.72^n = \frac{0.72}{9(1-0.72)} = 0.2857. [[/math]]

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:

1. Both find a claim. The probability is 0.02.
2. Rahul finds a claim and Toby does not. The probability is 0.08.
3. Toby finds a claim and Rahul does not. The probability is 0.18
4. Neither finds a claim. The probability is 0.72.

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 0.08/(0.02+ 0.08 + 0.18) = 8/28 = 0.2857.

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.