Exam P Practice Problem 36 – Number of Claims

This post has no alternate problem. It has one problem with 2 parts.

Problem 36

A claim examiner of an insurer reviews the claim history of two independent insureds. Let X be the annual number of claims of the first insured (Insured # 1). Let Y be the annual number of claims of the second insured (Insured # 2). The claim examiner finds tht X follows a Poisson distribution with mean 1, and that Y follows a distribution with the following probability function.

      \displaystyle P(Y=y)=\frac{3!}{y! (3-y)!} \ \biggl(\frac{1}{3}\biggr)^y \ \biggl(\frac{2}{3}\biggr)^{3-y} \ \ \ \ \ \ y=0,1,2,3
  1. Between these two insureds, what is the probability that one of the insureds has two more claims than the other insured in a year?
  2. Given that one of the insured has two more claims than the other insured, what is the probability that Insured # 1 has more claims than Insured # 2?

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Answers

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