Answers to Sections 13 and 14 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 13.

Answer 13.1a. Continuous

Answer 13.1b. Discrete

Answer 13.1c. Discrete

Answer 13.1d. Continuous

Answer 13.2. P(G then B)= 15/56; G then B implies X = 1.

P(G then G)= 3/28; G then G implies X = 0.

P(B then G)= 15/56; B then G implies X = 1.

P(B then B)= 5/14; B then B implies X = 2.

Answer 13.3. 5/36= about 0.1388888889

Answer 13.4. 0.85

Answer 13.5. (1/2)ⁿ

Answer 13.6. P(X = J)= 2/4= ½

Answer 13.7. 2/5= 0.4

Answer 13.8. 35/36= about 0.97222222222222

Answer 13.9a. X(s)= f:{HHH, HHM, HMH, HMM, MHH, MHM, MMH, MMM} → {0, 1, 2, 3}

Answer 13.9b. P(X=0)=9/100=0.09

Answer 13.9c. P(X=1)=9/25= 0.36

Answer 13.9d. P(X=2)=41/100= 0.41

Answer 13.9e. P(X=3)=7/50=0.14

Answer 13.10. P(X=0)=½; P(X=1)=1/6; P(X=2)=1/12; P(X=3)=1/4

Answer 13.11. P(0)= 1/(1+e)= about 0.2689414214

Section 14.

Answer 14.1a. The following table shows this probability mass function:

x:--------0-----1----2-----3

P(x):--1/8---3/8--3/8---1/8

Answer 14.2. F(x)=

0 for x

1/8 for 0≤x

½ for 1≤x

7/8 for 2≤x

1 for x≥3

Answer 14.3. Formula: P(x)= [(6-‌│7-x│)/36]*I_{2,3,4,5,6,7,8,9,10,11,12}(x)

Table:

x:--------2-----3-----4-----5-----6-----7-----8-----9-----10-----11-----12

P(x):--1/36--1/18--1/12-1/9--5/36-1/6--5/36-1/9---1/12---1/18---1/36

Answer 14.4. P(x)= (1/6)*I_1,2,3,4,5,6(x).

Answer 14.5. P(x) can be represented via the following table:
x:-------0----1----2

P(x):--1/4--1/2--1/4

P(x) can also be represented as an equation: P(x)= [(2-│1-x│)/4]*I_0,1,2(x)

Answer 14.6. F(n)= 1-(2/3)ⁿ⁺¹

Answer 14.7a. Pr(X=1)= 5/100= 1/20=0.05;

Pr(X=k) = _k=1^xΣ ((100-k)(99-k)(98-k)(97-k)/1,806,900,480) for x є integers, 1 ≤ x ≤ 96

Answer 14.7b. We can define F(y) as follows:

F(y)= 0 for y

0.5837523669 for 0≤y

0.9321424769 for 1≤y

0.9933612861 for 2≤y

0.9997448142 for 3≤y

0.9999958518 for 4≤y

1 for y≥5

We can also express F(y) as the following sum:

F(y)= Pr(Y≤y)= _k=0^yΣ C(95,10-k)*C(5,k)/C(100,k) for y є integers, 0 ≤ y ≤ 5

Answer 14.8. P(x)= 3/10 for x=-4,

2/5 for x=1,

3/10 for x=4

Answer 14.9. P(x)=

4/7 for x= -1,

3/7 for x=2

Answer 14.10a. P(x)=

8/27 for x=0,

4/9 for x=1,

2/9 for x=2,

1/27 for x=3

Answer 14.10b. F(x)= 0 for x

8/27 for 0≤ x

20/27 for 1≤ x

20/27 for 1≤ x

26/27 for 2≤ x

1 for x≥3

Answer 14.11. P(x) =

1/36, for x=2,

1/18, for x=3,

1/12, for x=4,

1/9, for x= 5,

5/36, for x= 6,

1/6, for x=7,

5/36 for x=8,

1/9 for x=9,

1/12 for x= 10,

1/18 for x = 11,

1/36 for x= 12,

0 otherwise.

Answer 14.12. p(0) = 44/91; p(1) = 198/455; p(2) = 36/455; p(3) = 1/455

See a list of all the answer keys to Dr. Finan's study guide for Exam 1/P here.