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Thread: CAS Exam 3 Spring 2005 #17

  1. #1
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    CAS Exam 3 Spring 2005 #17

    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

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    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.

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    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%

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    Quote Originally Posted by M2010 View Post
    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!

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