Answers to Sections 9 and 10 of Marcel B. Finan's "A Probability Course for the Actuaries"
Dr. Finan's study guide is an excellent resource for those preparing to take Actuarial Exam 1/P on probability.
Problems that require numerical answers are answered here, but it is still the responsibility of the student to provide his or her own work for these problems. These answers are meant to enable students to independently verify the correctness of their reasoning by checking to see if the end result they obtained is correct. Questions from the study guide that require proofs or diagrams are not addressed here, as the end result of those questions is known in advance, and it is the responsibility of the student to provide the procedure for getting there.
Section 9.
Answer 9.1. 108/625=0.1728
Answer 9.2. 0.205
Answer 9.3. 7/15= about 0.46666666666667
Answer 9.4. ½=0.5
Answer 9.5a. 19/100= 0.19
Answer 9.5b. 3/5=0.6
Answer 9.5c. 31/100=0.31
Answer 9.5d. 19/60= about 0.31666667
Answer 9.5e. 19/31= about 0.6129032
Answer 9.6. 5/33= about 0.15151515151515
Answer 9.7. 2/15= about 0.133333333333
Answer 9.8. 45 of every 46 parts that make it through the inspection machine are good, but 1 of every 46 parts that make it through the inspection machine is defective.
Answer 9.9. 72/73= about 0.9863013699
Answer 9.10. 7/1912= about 0.0036610879
Answer 9.11a. 1/221= about 0.0045248869
Answer 9.11b. 1/169= about 0.0059171598
Answer 9.12. 1/114= about 0.0087719298
Answer 9.13. 80.20202020202%
Answer 9.14. 0.476
Answer 9.15a. 0.48
Answer 9.15b. 0.625
Answer 9.15c. 0.1923
Answer 9.16a. about 0.0821
Answer 9.16b. about 0.1268
Section 10.
Answer 10.1a. 0.26
Answer 10.1b. 6/13 or about 0.4615
Answer 10.2. 16/101 or about 0.1584
Answer 10.3. 1/71 or about 0.01408
Answer 10.4. 45/154 or about 0.2922
Answer 10.5. 8/19= about 0.42105
Answer 10.6. 128/583= about 0.2196
Answer 10.7. 190/289= about 0.6574
Answer 10.8. 2/5=0.4
Answer 10.9. 5/11 or about 0.4545
Answer 10.10. 0.66= 33/50
Answer 10.11a. 0.22 = 11/50
Answer 10.11b. 15/22= about 0.68181818181818
Answer 10.12a. 0.56= 14/25
Answer 10.12b. 4/7 = about 0.5714285714
Answer 10.13. 5/14= about 0.3571428571
Answer 10.14. 1/3
Answer 10.15. 15/31 = about 0.48387096777
Answer 10.16a. 17/140
Answer 10.16b. 7/17
See a list of all the answer keys to Dr. Finan's study guide for Exam 1/P here.
Published by G. Stolyarov II
G. Stolyarov II is a science fiction novelist, independent essayist, poet, amateur mathematician, composer, author, and actuary. View profile
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6 Comments
Post a CommentWhat say about answer 10.11 (b). It is mentioned that "suppose I come back with injury". This indicates he did not play soccer, so the probability should be .5*.3/(.5*.3 + .3*.1) = 5/6
Hi Morrison,
Your method is correct but you are puting wrong data. The second term in the summation should be 0.3*0.1 so the probability would be: 0.5*0.3 + 0.3*0.1 + 0.2*0.2 = 0.22.
I have the same answers as Charlie for 9.14. Oddly enough, my copy of the handbook seems to be missing problem 9.16. Perhaps Dr. Finan revised?
I think ans of 10.11(a) would be:
0.5*0.3 + 0.2*0.1 + 0.2*0.2
= 0.21
Any suggestion??
I believe the answers to 9.14 are:
a) ~.021
b) from Firm A, .238; from Firm B, .286; from Firm C, .476
I think the answer to 9.14 is incorrect. Perhaps written in the wrong spot. First, question 9.14 has two parts, a) and b). Second, there is no way 47% of all cords are defective since the defective rates for the shipments are only 5%, 2%, and 1%.