Answers to Sections 5 and 6 of Marcel B. Finan's "A Probability Course for the Actuaries"

These answer keys are meant to assist students using Marcel B. Finan's A Probability Course for the Actuaries. With Dr. Finan's permission, Mr. Stolyarov wrote solutions for the problems in his study guide and is endeavoring to make the answer keys to each section publicly available. You can see his full List of Answer Keys here. Do the problems at the end of each section and then check your answers with these keys.

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 5.

Answer 5.1. 19,958,400 rearrangements

Answer 5.2. 9,777,287,520 ways

Answer 5.3. 5.364473777*10²⁸ ways

Answer 5.4. 4,504,501 ways

Answer 5.5. 12,600 possible outcomes

Answer 5.6. 1,260 possible outcomes

Answer 5.7. If all the money needs to be invested, 1771 ways. If not all the money needs to be invested, 10,626 ways.

Answer 5.8. 21 solutions

Answer 5.9a. 28 distinct terms

Answer 5.9b. 60

Answer 5.9c. 729

Answer 5.10. 211,876 solutions

Answer 5.11. About 3.605238748*10¹⁶ ways

Answer 5.12a. 136 solutions

Answer 5.12b. C(15, 2)C(8, 2) = 2940 solutions

Section 6.

Answer 6.1a. S={1, 2, 3, 4, 5, 6}

Answer 6.1b. {2, 4, 6}

Answer 6.2. S={(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,H), (2,T), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,H), (4,T), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,H), (6,T)}

Answer 6.3. ½= 0.5

Answer 6.4a. S= {(1, 1), (1,2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (3, 4), (4, 1), (4, 2), (4, 3), (4, 4)}

Answer 6.4b. E^c= {(1, 1), (1,2), (2, 1)}