**Problem 71-A**

An auto insurer issued policies to a large group of drivers under the age of 40. These drivers are classified into five distinct groups by age. These groups are equal in size.

The annual claim count distribution for any driver being insured by this insurer is assumed to be a binomial distribution. The following table shows more information about these drivers.

An insured driver is randomly selected from this large pool of insured and is observed to have one claim in the last year.

What is the probability that the mean number of claims in a year for this insured driver is greater than 1.5?

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**Problem 71-B**

An auto insurer issued policies to a large group of drivers under the age of 40. These drivers are classified into five distinct groups by age. These groups are equal in size.

The annual claim count distribution for any driver being insured by this insurer is assumed to be a geometric distribution. The following table shows more information about these drivers.

An insured driver is randomly selected from this large pool of insured and is observed to have one claim in the last year.

What is the probability that the mean number of claims in a year for this insured driver is greater than 2.5?

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Hello Dan. Thanks for posting these problems. I’ve gotten a lot of great insights from working on them. For this problem, having some trouble. For 71A, I take it to mean that each age group has it’s own binomial parameters. From the means and variances given, it looks like all age groups have p=0.5 and q=0.5, with age group 16-17 having an n of 5, 18-24 having an n of 4, and so on until age 35-39 has an n of 1. Now, we are given that a randomly selected driver has had 1 accident. So what I did was use the binomial distribution of each group with x=1, so for example for age group 16-17 was P(X=1|group 16-17) = 5C1(.5)(.5)^4 = .15625

P(X=1|group 18-24) = 4C1(.5)(.5)^3 = .25

so on until P(X=1|group 35-39) = .5

Now, this gives a total of 57/32. I don’t even know if the Bayesian denominator can be greater than 1 (seems like it could be), but if we want to know what the probability of the random driver having a mean claims per year greater than 1.5, then we are talking about the first two groups. This gives me (13/32)/(57/32) = 13/57. So I’d appreciate if you might be able to tell me where I got off track on this one. Thanks!