MHB Calculating Probability of Filing >3 Claims Under Insurance Policy

AI Thread Summary
The discussion focuses on calculating the probability of a policyholder filing more than three claims under an insurance policy that allows a maximum of five claims per year. Given the conditions that probabilities decrease with increasing claims and that 40% of policyholders file fewer than two claims, the solution reveals that all probabilities are equal at 0.2 each. If the assumption about filing fewer claims is adjusted to two or fewer, the probabilities change to 1/15 for zero claims, 2/15 for one claim, and so on. The calculations hinge on the relationships established by the observations made by the actuary. Ultimately, the correct interpretation of the constraints significantly affects the resulting probabilities.
Jason123
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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.
 
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Jason123 said:
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 -:)
 
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