Answers to Sections 3 and 4 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 3.
Answer 3.1a. 100 ways
Answer 3.1b. 900 ways
Answer 3.1c. 5040 ways
Answer 3.1d. 90000 ways
Answer 3.2a. 336 finishing orders
Answer 3.2b. 6 finishing orders
Answer 3.3. 6 choices of outfits
Answer 3.4. 90 ways
Answer 3.6. 36 ways
Answer 3.7. 380 ways
Answer 3.8. 6,497,400 ways
Answer 3.9. 5040 ways
Answer 3.10. 3840 ways
Section 4.
Answer 4.1. m=9 and n=3
Answer 4.2a. 456,976 words
Answer 4.2b. 358,800 words
Answer 4.3a. 15,600,000 possible license plates
Answer 4.3b. 11,232,000 possible license plates
Answer 4.4a. 64,000 possible combinations
Answer 4.4b. 59,280 possible combinations
Answer 4.5a. 479,001,600 arrangements
Answer 4.5b. 604,800 arrangements
Answer 4.6a. 5
Answer 4.6b. 20
Answer 4.6c. 60
Answer 4.6d. 120
Answer 4.7. 20 ways
Answer 4.8a. 362,880 ways
Answer 4.8b. 15,600 ways
Answer 4.9. m=13; n=1 or n=12
Answer 4.10. 11,480 possible choices
Answer 4.11. 300 handshakes
Answer 4.12. 10 ways
Answer 4.13. 28 ways
Answer 4.14. 4060 ways
Answer 4.15. The number of permutations of a set of objects is usually greater.
Answer 4.16. 125,970 possible juries
Answer 4.17. 14/2!= 43,589,145,600 ways
Answer 4.18. 144 ways
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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3 Comments
Post a CommentThe correction to the answer to Problem 4.17 has now been made. Thank you, Joe and Frank, for your feedback.
I agree with Frank. The answer should be 14!/2 because the circular band can be simply flipped over. Both of these combinations of pearls would be the same and this rule would apply for all 14! of the combinations. So 14!/2 is how many different combinations there actually are.
On question 4.17 I think you have overlooked a fundamental
symmetry in your solution. This is of course a matter of
interpretation but given the physical reality of the circumstance I
think it is reasonable to assume that the action of picking up the
band and flipping it over does not structurally alter it in any
fundamental way. Therefore any two outcomes which would be reflections
of one another along any axis of symmetry should be considered
fundamentally the same in this example. As such rather than having our
outcome invariant under action of the cyclic group, it is invariant
under action of the dihedral group D_{30}. So we have the following
short exact sequence.
0-->D_{30}-->S_{15}-->X-->0
Where X is our space of outcomes gifted with an inheirited, but
irrelevant, group structure from S_{15}.
In any case |X|=|S_{15}|/|D_{30}|= 15!/30=14!/2
This is a nice generalization to the circle permutations.