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**Problem 21A**

Ten percent of the policyholders for an auto insurance company are considered high risk and ninety percent of its policyholders are considered low risk. The number of claims made by a policyholder in a calendar year follows a Poisson distribution with mean .

For high risk policyholders, . For low risk policyholders, . An actuary selects one policyholder at random. If by the end of one year, this policyholder has made exactly one claim, what is the probability that this policyholder is a high risk policyholder?

**Problem 21B**

Use the same information from Problem 21A. If by the end of one year, the selected policyholder has made at least one claims, what is the probability that the policyholder is a high risk policyholder?

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**Answers**

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for 21A, I used Bay’s Theorem to solve. here is my solution: ( 0.1*0.9e^(-0.9) ) / (0.1*0.9e^(-0.9) + 0.1*0.1e^(-0.1)) = 0.8 your answer is 0.31. I really can’t figure out what I did wrong. Could you please give me some hints? Thank you so much.

The thing that you did wrong is in the denominator. Remember that 90% of the risks are in the low risk group.