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View Full Version : CAS Exam 3 Spring 2005 #17



M2010
October 13th 2010, 04:42 PM
An insurer selects risks from a population that consists of 3 indep groups.

The claims generation process for each group is poisson.

The 1st group consists of 50% of the population. These individuals are expected to generate one claim per year.

The 2nd group consists of 35% of the population. These individuals are expected to generate two claims per year.

Individuals in the 3rd group are expected to generate three claims per year.

A certain insured has two claims in year 1.

What is the probability that this insured has more than two claims in year 2?

A. Less than 21%
B. At least 21%, but less than 25%
C. At least 25%, but less than 29%
D. At least 29%, but less than 33%
E. 33% or more

I got the answer was 23.98%, which is B. However, the answer sheet says the correct one is C.

I calculated the probabilities of the insured has more than two claims in year 2 for each group, then times the each groups' proportion of the population. I did not use the condition that the insured has two claims in year 1, since I think year 2 is indep with year 1. Is there anything wrong?

Thank you very much

alekhine4149
October 13th 2010, 09:20 PM
You should use the condition that the insured has 2 claims in year 1. This helps identify which group he is in. The technique to do so is called Bayes' Theorem.

M2010
October 14th 2010, 01:19 PM
I think I got it. Please correct me if I am wrong.
Assume A = 2 claims in year 1, B = >2 claims in year 2, I,II and III = Group 1,2 and 3, respectively.
I first find p(A|I),p(A|II),p(A|III) and p(B|I),p(B|II),p(B|III), use Bayes' Theorem find p(B), then I find p(A&B|I)=p(A|I)*p(B|I), and p(A&B|II), p(A&B|III) as well.
p(A&B)=p(A&B|I)*p(I)+p(A&B|II)*p(II)+p(A&B|III)*p(III)
finally, p(A&B)/p(B)=26.05%

alekhine4149
October 14th 2010, 05:04 PM
I think I got it. Please correct me if I am wrong.
Assume A = 2 claims in year 1, B = >2 claims in year 2, I,II and III = Group 1,2 and 3, respectively.
I first find p(A|I),p(A|II),p(A|III) and p(B|I),p(B|II),p(B|III), use Bayes' Theorem find p(B), then I find p(A&B|I)=p(A|I)*p(B|I), and p(A&B|II), p(A&B|III) as well.
p(A&B)=p(A&B|I)*p(I)+p(A&B|II)*p(II)+p(A&B|III)*p(III)
finally, p(A&B)/p(B)=26.05%

Hi M2010, you have the right answer choice, wrong answer. That's progress! :) You need to find the probability of being in each group. Bayes theorem should be set up as follows:

Pr (Group I | 2 claims in a year) = Pr (2 claims in a year | Group I) * Pr (Group I) = ((e^-1)/2)*(0.5) = 0.092.

Do this for all 3 groups. Sum up the 3 probabilities. Divide each probability by the total probability to get a relative probability. These 3 numbers should add up to 1, and they are the conditional probability of being in each group.

Finally, find the probability of having more than 2 claims in a year for each group. Multiply these 3 probabilities by each group's conditional probability, and you should end up with about 0.2689.

This problem is calculation intensive, but it's absolutely essential to know how to do. Good luck!