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  • P Question

    I was wondering how to solve this question:

    An insurance company issues policies to two groups of insured A and B. For a certain type of coverage, the number of claims per year for each group follows the Poisson distribution

    For group A the mean is 1, and for group B the mean is 1.5.

    If in calendar year 2003, 2/5 of this company's policyholders are of group A, what is the probability that the combined number of claims from both groups is no more that two.

    A. 0.627

    B. 0.857

    C. 0.544

    D. 0.287

    E. 0.879

    Thanks!

  • #2
    I'm having trouble with this one... I did a variant of it that doesn't involve the 2/5ths of the company being A.

    In general when you get a problem like this my strategy is to just add up the probabilities of all possible scenarios, in this case it would be:

    P(A=0) * P(B=0) +
    P(A=1) * P(B=0) +
    P(A=0) * P(B=1) +
    P(A=1) * P(B=1) +
    P(A=2) * P(B=0) +
    P(A=0) * P(B=2)

    This gets you .544, or C) The problem in this case is, I never used the 2/5. And when I throw it in term-by-term (along with weighting B at 3/5) I get an answer of .13 or so, which isn't on the list.

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    • #3
      This gets you .544, or C) The problem in this case is, I never used the 2/5. And when I throw it in term-by-term (along with weighting B at 3/5) I get an answer of .13 or so, which isn't on the list.
      I tried the same method :geek:- Apparently the Correct Answer is B however. :unsure:

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      • #4
        Is this problem in the ASM manual? If it gives you the answer surely it gives you a solution as well?

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        • #5
          You should probably find a way to incorporate that 2/5..

          What if I told 99.9999999999% were from A, would you still give me the same answer?

          Comment


          • #6
            Originally posted by NoMoreExams View Post
            You should probably find a way to incorporate that 2/5..

            What if I told 99.9999999999% were from A, would you still give me the same answer?
            If this were the case, we could ignore B correct? In which case we would just have A = 0, 1 or 2. Then we'd just calculate those probabilities and add them.

            Comment


            • #7
              Originally posted by Lucretius View Post
              If this were the case, we could ignore B correct? In which case we would just have A = 0, 1 or 2. Then we'd just calculate those probabilities and add them.
              Why would you ignore it? Fine, let's assume the split is 1/3 and 2/3 or 1/5 and 4/5. Are you saying that regardless of how many policyholders are in each bucket, you will get the same answer?

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              • #8
                Well my original thought had been to multiply the probability for A by 2/5, and the probability for B by 3/5 - and do this for each term in the sum. I mean, it obviously has to play an important role. But I'm all out of ideas!

                Comment


                • #9
                  Originally posted by Lucretius View Post
                  Is this problem in the ASM manual? If it gives you the answer surely it gives you a solution as well?
                  I was just searching around on the internet, and it came from one of those Be An Actuary http://www.beanactuary.org/exams/OnlineExam/index.cfm exams. But it seems to be a random bank of questions, and they don't provide a full solution, only the correct answer :wacko:

                  Comment


                  • #10
                    If A and B are poisson with means 1 and 1.5 and if C = (2/5)A + (3/5)B, then the mean of C (which is Poisson) is (2/5)(1) + (3/5)(1.5). The question really is what is the probability that C is at most 2. Exam questions never contain arbitrary information.
                    Last edited by WyoFire; January 20 2011, 10:34 AM.

                    Comment


                    • #11
                      Originally posted by WyoFire View Post
                      If A and B are poisson with means 1 and 1.5 and if C = (2/5)A + (3/5)B, then the mean of C (which is Poisson) is (2/5)(1) + (3/5)(1.5). The question really is what is the probability that C is at most 2. Exam questions never contain arbitrary information.
                      Hmm and how would you modify for a different distribution where you can't scale the parameter(s)?

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                      • #12
                        Mixed distribution aah... I can't believe I worked the same thing out for the normal distribution to answer some other question here but didn't think to do it for the Poisson. Funny thing is, I woke up this morning with that question on my mind, and my brain had put it together for me. I just made that new variable (f_c = (1/x!)[(2/5)e^-1 + (3/5)e^-1.5*1.5^x] and used x=0,1,2. Gets me .853 ~.857 ~ B)

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                        • #13
                          Originally posted by NoMoreExams View Post
                          Hmm and how would you modify for a different distribution where you can't scale the parameter(s)?
                          Wouldn't you simply do a transformation using, for instance, U = (2/5)A + (3/5)B and V = (3/5)B then find P(U is at most 2)?

                          Comment


                          • #14
                            Originally posted by WyoFire View Post
                            Wouldn't you simply do a transformation using, for instance, U = (2/5)A + (3/5)B and V = (3/5)B then find P(U is at most 2)?
                            How would that work if your distribution has more than one parameter? Would you scale both? You sure that works?

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