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Thread: Random Variable

  1. MHB Apprentice

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    #1
    Under an insurance policy, a maximum of of 5 claims may be filed per year by a policy holder. Let
    p(n) be the probability that a policyholder files n claims during a given year, where n= 0,1,2,3,4,5. An actuary makes the following observations:
    1) p (n) >= p (n+1) for 0<=n <=4
    2) the difference between p (n) and p (n+1) is the same for 0 <= n <= 4
    3) exactly 40% of policyholders file fewer than 2 claims during a given year.

    Calculate the probability that a random policyholder will file more than three claims during a given year.

    Can't seem to figure this one out. Any help would be great.

  2. MHB Apprentice

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    #2
    Quote Originally Posted by Jason123 View Post
    Under an insurance policy, a maximum of of 5 claims may be filed per year by a policy holder. Let
    p(n) be the probability that a policyholder files n claims during a given year, where n= 0,1,2,3,4,5. An actuary makes the following observations:
    1) p (n) >= p (n+1) for 0<=n <=4
    2) the difference between p (n) and p (n+1) is the same for 0 <= n <= 4
    3) exactly 40% of policyholders file fewer than 2 claims during a given year.

    Calculate the probability that a random policyholder will file more than three claims during a given year.

    Can't seem to figure this one out. Any help would be great.
    The answer is all probabilities are .2 each. If a = prob of 0 claims then given your constraints the probabilities are a , a+d, a + 2*d, a+ 3*d and a + 4*d (diff is equal by assumption 2 and is d here). So 5*a + 10*d = 1. Also by 3, a + a + d = 2*a + d = .4. The only soln to these two is d = 0 and a =.2.

    If however there was a typo and assumption 3 reads 2 or fewer claims then the second eqn is 3*a + 3*d = .4. Then the solution is a = 1/15 and d= 1/15 so the probs are 1/15, 2/15 ... 5/15. Take your pick -
    Last edited by vkalyan; October 19th, 2016 at 23:42.

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