Answers to Section 32 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 32

Answer 32.1. p_X|Y(x|y) where Y = 1 is

0.25, when x=0,

0.75, when x=1,

0 otherwise.

Answer 32.2a. p_XY(x,y)= 1/(5x) for x є {1, 2, 3, 4, 5} and 0 otherwise.

Answer 32.2b. p_X|Y(x|1)= 60/137 for x=1, 30/137 for x=2, 20/137 for x=3, 15/137 for x=4, 12/137 for x=5, 0 otherwise.

p_X|Y(x|2)= 30/77 for x=2, 20/77 for x=3, 15/77 for x=4, 12/77 for x=5, 0 otherwise.

p_X|Y(x|3)= 20/47 for x=3, 15/47 for x=4, 12/47 for x=5, 0 otherwise.

p_X|Y(x|4)= 5/9 for x=4, 4/9 for x=5, 0 otherwise.

p_X|Y(x|5)= 1 for x=5, 0 otherwise.

Answer 32.2c. X and Y are not independent.

Answer 32.3a. P(X=3|Y = 4)= 2/7

Answer 32.3b. P(X=4|Y = 4)= 3/7

Answer 32.3c. P(X=5|Y = 4)= 2/7

Answer 32.4. p_X|Y(x|1)=

1/6 for x=0,

½ for x=1,

1/3 for x=2,

0 otherwise.

Answer 32.5. p_Y|X(y|1)= 1 for y=1, 0 otherwise.

p_Y|X(y|2)= 2/3 for y=1, 1/3 for y=2, 0 otherwise.

p_Y|X(y|3)= 2/5 for each y є {1, 2}, 1/5 for y=3, 0 otherwise.

p_Y|X(y|4)= 2/7 for each y є {1, 2, 3}, 1/7 for y=4, 0 otherwise.

p_Y|X(y|5)= 2/9 for each y є {1, 2, 3, 4}, 1/9 for y=5, 0 otherwise.

p_Y|X(y|6)= 2/11 for each y є {1, 2, 3, 4, 5}, 1/11 for y=6, 0 otherwise.

X and Y are not independent.

Answer 32.6a. p_Y(y) = C(n, y) p^y(1−p)^n−y, y = 0, 1, · · · , n, 0 otherwise.

Answer 32.6b. p_X|Y(x|y)= y^xe^{− y}/x! , y = 0, 1, · · · , n ; x = 0, 1, · · ·, 0 otherwise. X and Y are not independent.

Answer 32.7. p_X|Y(x|0)= 1/11 for x=0, 4/11 for x=1, 6/11 for x=2, 0 otherwise.

p_X|Y(x|1)= 3/7 for x=0, 3/7 for x=1, 1/7 for x=2, 0 otherwise.

p_Y|X(y|0)= ¼ for y=0, ¾ for y=1, 0 otherwise.

p_Y|X(y|1)= 4/7 for y=0, 3/7 for y=1, 0 otherwise.

p_Y|X(y|2)= 6/7 for y=0, 1/7 for y=1, 0 otherwise.

Answer 32.8a. c= 1/(2^N-1)

Answer 32.8b. p_X(x)= 2^x/(2^N-1) for x = 0, 1, · · · , N − 1, 0 otherwise.

Answer 32.8c. p_Y|X(y|x)= (1 − 2^−x)^y/2^x for x = 0, 1, · · · , N − 1, and y= 0, 1, 2, · · ·, 0 otherwise.

Answer 32.9. P(X = k | X + Y = n)= C(n, k)/2ⁿ

Answer 32.10a. The joint probability distribution of X and Y is

P(X = 0, Y = 0) = 188/221;

P(X = 1, Y = 0) = 16/221;

P(X = 1, Y = 1) = 16/221;

P(X = 2, Y = 1) = 1/221;

and 0 otherwise.

Answer 32.10b. The marginal distribution of Y is

P(Y = 0) = 12/13;

P(Y = 1) = 1/13;

and 0 otherwise.

Answer 32.10. The conditional distribution of X given Y = 1 is

p_X│Y(x│1)= 16/17 for x=1,

1/17 for x=2,

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