For P(B|d)>P(A|d) it would be (d-10)^2 > 8/3 *ln(2) => |d-10| > (8/3
*ln(2))^.5 and
for P(C|d) > P(A|d) I get (d-10)^2 < 32/3 * ln (2) => |d-10| < 4*(2/3
ln(2))^.5 answer c
Thanks for the solutions. Do you have any estimates of the passing scores
for the last three exams?
-----Original Message-----
From: Stuart Klugman <stuart.klugman@drake.edu>
To: Study group 4B <studygroup4B@lists.casact.org>
Date: Friday, May 07, 1999 1:33 PM
Subject: #8
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.
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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