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. Originally Posted by Jason123
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 -