Stuart Klugman wrote:
>
> Wang's solution is almost correct. The probabilities in the table add to .2552. If you add the probability of no payments (e^-.3)=.7408, you see that only .996 of the probability has been accounted for. Left off is the probability that three or more claims occur. However, adding the consequences of this only raised the answer to 2,344,000 which is still (D). My approach to dealing with this issue is to note that the probability of two or more claims is 1 - 1.3exp(-.3) since total claims is Poisson(.3). For the first two of these claims the probabilites are 1/9 for AA, 4/9 for MM, and 4/9 for one of each. That is because the probability a randomly selected claim is A is .1/(.1+.2) = 1/3.
>
> My solutions to all 30 problems are available for download on our departments home page www.drake.edu/cbpa/acts (all lower case) in PDF format (3 pages). They are not official, and may not even be correct. Corrections and improved solutions are welcome.
>
> ------------------------------------------------------------------------------------------------
> Stuart Klugman, FSA, PhD
> Principal Financial Group Professor of Actuarial Science
> Drake University
> 2507 University Avenue
> Des Moines, IA 50311 USA
> ph: 515-271-4097
> e-mail: Stuart.Klugman@drake.edu
> Drake Act. Sci. web site: www.drake.edu/cbpa/acts