Stolyarov Finan P Answer Keys

Answer Keys for the 2008 Edition of Marcel B.
Finan’s “A Probability Course for the Actuaries”

Second Edition – G. Stolyarov II
Answer Keys for the 2008 Edition of

Marcel B. Finan's
A Probability Course for the Actuaries
Second Edition
G. Stolyarov II,
ASA, ACAS, MAAA, CPCU, ARe, ARC, API, AIS, AIE, AIAF
First Edition Published in July-August 2008

Second Edition Published in July 2014
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 has endeavored to make the answer keys to each section publicly
available. 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 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.
The answers in this document are current as of August 2008 and were previously published on
Associated Content / Yahoo! Voices. The contents of A Probability Course for the Actuaries
have been greatly expanded since that time, and some of the problem numbers have changed,
while new sections have been added. While updates to these answer keys are possible in the
future, Mr. Stolyarov’s priority in July 2014 has been to republish the existing answer keys
within this PDF document, so as to preserve them subsequent to the closure of Yahoo! Voices on
July 30, 2014.
1
Answer Keys for the 2008 Edition of Marcel B. Finan’s “A Probability Course for the Actuaries”
Section 1.
Answer 1.1.A={x:x shows a face with prime number}= {2, 3, 5}
Answer 1.2a. HHH, HHT, HTH, HTT, TTT, TTH, THT, THH
Answer 1.2b.
HTT, TTT, TTH, THT
Answer 1.2c. F in set-builder notation is
F={x│x is an element of S with more than one head} = {x|x is an element of S with at least
two heads}
Answer 1.3. E={HHH, HHT, HTH, HTT, TTH, THT, THH}. The outcomes with more than
one head are, HHH, HHT, HTH. Thus, F={HHH, HHT, HTH,THH}. Thus, F is a subset of E,
since F contains only elements that are also in E.
Answer 1.4. There are no elements of E and so E=Ø
Answer 1.9a. 55 tacos
Answer 1.9b. 40 tacos
Answer 1.9c. 10 tacos
Answer 1.10a. 20 students
Answer 1.10b. 5 students
Answer 1.10c. 11 students
Answer 1.10d. 42 students
Answer 1.10e. 46 students
Answer 1.10f. 46 students
Answer 1.11. n(S) = 24
Section 2.
Answer 2.3. (G U B U S)c.
Answer 2.5. There are 880 young, single, female policyholders.
2
Answer 2.6. 50% of the population owns an automobile or a house, but not both.
Answer 2.7. 5% of visits to a PCP's office result in both lab work and referral to a specialist.
Answer 2.8. n(A)=60
Answer 2.9. 53% of policyholders will renew at least one policy next year.
Answer 2.11. 50 players surveyed
Answer 2.12. 10
Answer 2.13a. 3 students
Answer 2.13b. 6 students
Answer 2.14. Mr. Brown owns 32 chickens.
Answer 2.15. 20% of the patients have a regular heartbeat and low blood pressure.
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
3
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
4
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
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}
5
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. Ec= {(1, 1), (1,2), (2, 1)}
Answer 6.4c. The probability of getting an outcome where the total score is at least six is 3/8; the
probability that the total score is less than 6 is 5/8.
Answer 6.4d. ¾=0.75
Answer 6.4e. 5/8= 0.625
Answer 6.5a. 12/25=0.48
Answer 6.5b. P(B)=0
Answer 6.5c. P(C)=1
Answer 6.5d. 9/25=0.36
Answer 6.5e. 1/25=0.04
Answer 6.6a. 3/8= 0.375
Answer 6.6b. ¼=0.25
Answer 6.6c. 1/2=0.5
Answer 6.6d.P(11)=0
Answer 6.6e. 3/8=0.375
Answer 6.6f. 1/8= 0.125
Answer 6.7a. 4/C(52,13)= about 6.29907809*10-12
Answer 6.7b. C(4,3)C(48,10)/C(52, 13) = about 0.041
Answer 6.8. ¼= 0.25
6
Answer 6.9. 1/8= 0.125
Answer 6.10a. 4/n2
Answer 6.10b. 5/n2
Answer 6.11. 1-200!/[100!*200100]
Answer 6.12a. 10 lawyers who are liars
Answer 6.12b. 2/5= 0.4
Answer 6.13. 1/10 = 0.1
Section 7.
Answer 7.1a. 0.78
Answer 7.1b. 0.57
Answer 7.1c. 0
Answer 7.2. 0.32
Answer 7.3. 4/13= about 0.30769
Answer 7.4. 5/9= about 0.55555556
Answer 7.6a. 2/11= about 0.1818181818
Answer 7.6b. 9/11= about 0.81818181818
Answer 7.6c. 6/11= about 0.54545454545
Answer 7.7. 8/9=about 0.888888888889
Answer 7.8. A and B cannot be mutually exclusive.
Answer 7.9. 0.52
Answer 7.10. 0.05
Answer 7.11. P(A)=0.60
Answer 7.12. P(T)= 0.48
7
Answer 7.13. 0.04= 1/25
Answer 7.14. 0.09= 9/100
Answer 7.15. 0.5 = ½
Section 8.
Answer 8.1. P(2 red balls)= 3/10 = 0.3
P(red ball, then blue ball)= 3/10 = 0.3
P(blue ball, then red ball)= 3/10 = 0.3
P(2 blue balls)= 1/10 = 0.1
Answer 8.2. P(2 red balls)= 9/25 = 0.36
P(red ball, then blue ball)= 6/25 = 0.24
P(blue ball, then red ball)= 6/25 = 0.24
P(2 blue balls)= 4/25 = 0.16
Answer 8.3. P(A)=3/5=0.6
P(B)=3/10=0.3
P(C)=1/10=0.1
Answer 8.4. 3/16=0.1875
Answer 8.5. 4/9= about 0.44444444
Answer 8.6. 1/6= about 0.16666666667
Answer 8.7. 3/5= 0.6
Answer 8.8. 12/25= 0.48
Answer 8.9. 8/57= about 0.1403508772
Answer 8.10. 36/65= about 0.5538461538
8
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
9
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
10
Section 11.
Answer 11.1a. Dependent
Answer 11.1b. Independent
Answer 11.2. 1/49= about 0.02041
Answer 11.3a. 36/169= about 0.21302
Answer 11.3b. 48/221= about 0.21719
Answer 11.4. 0.72
Answer 11.5. 4
Answer 11.6. about 0.32872
Answer 11.7. 1/10= 0.1
Answer 11.9. P(C)=0.56; P(D)= 0.38
Answer 11.10. 0.93
Answer 11.11. 0.43
Answer 11.13a. The sample space is the set of all possible outcomes. Let N denote the nickel, D
denote the dime, and Q denote the quarter. Let H denote heads and T denote tails. We will
express "the dime came up heads" as "DH."
S= {(NH, DH, QH), (NH, DH, QT), (NH, DT, QH), (NH, DT, QT), (NT, DH, QH),
(NT, DH, QT), (NT, DT, QH), (NT, DT, QT)}
A={(NH, DH, QH), (NH, DH, QT), (NH, DT, QH), (NT, DH, QH), (NT, DT, QH)}
B= {(NH, DH, QH), (NH, DT, QH), (NT, DH, QH), (NT, DT, QH)}
C= {(NH, DH, QH), (NH, DT, QH), (NT, DH, QT), (NT, DT, QT)}
Answer 11.13b. P(A)= 5/8, P(B)= ½, P(C)= ½.
Answer 11.13c. 0.8 = 4/5
Answer 11.13d. Since P(B and C)= P(B)*P(C)= ¼, B and C are independent.
11
Answer 11.14. 0.65= 13/20
Answer 11.15a. 0.70
Answer 11.15b. 0.06
Answer 11.15c. 0.24
Answer 11.15d. 0.72
Answer 11.15e. 0.4615
Section 12.
Answer 12.1. 15:1
Answer 12.2. 5/8= 0.625
Answer 12.3. 1:1
Answer 12.4. 4:6
Answer 12.5. 1/27=about 0.03704
Answer 12.6a. 1:5
Answer 12.6b. 1:1
Answer 12.6c. 1:0
Answer 12.6d. 0:1
Answer 12.7. 1:3
Answer 12.8a. 3/7= about 0.42857
Answer 12.8b. 3/10= 0.3
Answer 12.9. 2/3
Section 13.
Answer 13.1a. Continuous
Answer 13.1b. Discrete
12
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
13
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 Σ ((100-k)(99-k)(98-k)(97-k)/1,806,900,480) for x є integers, 1 ≤ x ≤ 96

x
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
14
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 Σ C(95,10-k)*C(5,k)/C(100,k) for y є integers, 0 ≤ y ≤ 5

y
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,
15
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
Section 15.
Answer 15.1. 7
Answer 15.2. 50/3= about $16.67
Answer 15.3. Break even; E(X) = 0
Answer 15.4. E(X)= -1= -$1.00
Answer 15.5. E(X)= 1/8= about 0.125
Since E(X)>0, the owner can expect to make money over an extended period of time using
this spinner.
Answer 15.6. The company should charge each customer a premium of $26 to cover the payouts
and make a $20 profit per person.
Answer 15.7. $110
Answer 15.8. E(X)= -6/11= about -$0.545
16
Thus, we can expect to lose about 54.5 cents per game on average.
Answer 15.9. About $896.91
Answer 15.10a. 209/783= about 0.2669220945
Answer 15.10b. 352/783=about 0.4495530113
Answer 15.10c. 16/15= about 1.066666667
Answer 15.11a. q= 0.3
Answer 15.11b. P(X2 > 2)= 0.7
Answer 15.11c. 0
Answer 15.11d. p(x)=
0.2 for x=-2,
0.3 for x=0,
0.1 for x=2.2,
0.3 for x=3,
0.1 for x=4,
0 otherwise
Answer 15.11e. E(X)= 1.12
Answer 15.12a. 2400
Answer 15.12b. No.
Answer 15.13a. p(1) = 7/210; p(2) = 63/210; p(3) = 105/210; p(4) = 35/210
Answer 15.13b. F(x) = 0 for x < 1;
7/210 for 1 ≤ x < 2;
70/210 for 2 ≤ x < 3;
175/210 for 3 ≤ x < 4;
17
1 for 4 ≤ x.
Answer 15.13c. E(X) = 2.8; Var(X) = 0.56
Section 16.
Answer 16.1a. 1/30= about 0.0333333333
Answer 16.1b. 10/3= about 3.33333333333333
Answer 16.1c. 127/15= about 8.466666666666667
Answer 16.2a. c=1/9
Answer 16.2b. p(-1)= 2/9, p(1)=1/3, and p(2)= 4/9.
Answer 16.2c. E(X)=1; E(X2) = 7/3= about 2.33333333333333
Answer 16.3a. p(x)=x/21 for x = 1, 2, 3, 4, 5, 6, and p(x) = 0 otherwise.
Answer 16.3b. 4/7= about 0.5714285714
Answer 16.3c. 13/3= about 4.33333333333
Answer 16.5. E(F(X))=0.62
Answer 16.6. The expected payment for hospitalization is $220 under this policy.
Answer 16.7. 43/1370= about 0.0313868613
Answer 16.8. 0.24
Answer 16.9a. p(x)=
4/7, for x= -1,
3/7 for x=2,
0 otherwise.
Answer 16.9b. E(2X)= 2
Answer 16.10a. P= 3C+ 8A+ 5S- 300
Answer 16.10b. $1,101
18
Answer 16.12. For n = 1, E(S) = 108. For n = 2, E(S) = 115.2. For n = 3, E(S) = 121.6.
Section 17.
Answer 17.1. 0.45= 45%.
Answer 17.2. 374.4
Answer 17.3a. c= 1/55
Answer 17.3b. E(X) = -12/11= about -1.090909090909;Var(X)= 644/605= about 1.06446281
Answer 17.4a. p(x)=
1/30 for x=1,
3/10 for x=2,
½ for x=3,
1/6 for x=4,
0 otherwise
Answer 17.4b. E(X)= 2.8; Var(X)= 0.56
Answer 17.5a. c=1/30
Answer 17.5b. E(X)= 10/3; E(X(X-1))= 127/15
Answer 17.5c. Var(X)= 31/45 = about 0.6888889
Answer 17.6. E(Y)=21; Var(Y)= 144
Answer 17.7b. 4/7= about 0.5714285714
Answer 17.7c. p(X)=
0.24 for x=7,
0.16 for x=8,
0.18 for x=12,
0.42 for x=15
19
Answer 17.7d. E(X)= 11.42; Var(X)= 12.0036
Answer 17.8a. p(x)=
4/7, for x= -1,
3/7 for x=2
Answer 17.8b. E(X)= 2/7= about 0.2857142857; E(X2) = 16/7= about 2.2857142857
Answer 17.8c. Var(X)= 108/49= about 2.204081633
Answer 17.9. Var(X)= 9.86; SD(X)=about 3.136877428
Answer 17.10. E(X) = p; Var(X) = p(1-p)
Section 18.
Answer 18.1. 0.38263752
Answer 18.2. 27/128= 0.2109375
Answer 18.3. p(x)=
1/8, for x=0,
3/8, for x=1,
3/8, for x=2,
1/8 for x=3,
0 otherwise
Answer 18.4. about 0.0956576009
Answer 18.5. $60
Answer 18.6. 61/1250= 0.0488
Answer 18.7. about 0.9245163262
Answer 18.8. about 0.1441136931
Answer 18.9a. 3/16= 0.1875
20
Answer 18.9b. 0.5= ½
Answer 18.10. about 0.1198380223
Answer 18.11. about 0.6241903616
Answer 18.12. about 1.82284263*10-5
Answer 18.13. SD(X)= about 0.6324523697
Answer 18.14. E(X2)= 154
Answer 18.15. 0.135
Answer 18.16. 0.784
Answer 18.17c. about 0.1238
Section 19.
Answer 19.1. e- 4=about 0.0183156389= probability of no car accident in one day. 8e-8= about
0.002683701= probability of 1 car accident in two days.
Answer 19.2. About 0.1251100357= probability of receiving 10 calls in 5 minutes.
Answer 19.3. e-15 = about 3.059023205*10-7= probability of having no errors on a page.
Answer 19.4a. about 0.5768099189
Answer 19.4b. e-3 = about 0.0497870684
Answer 19.5a. about 0.9473469827
Answer 19.5b. about0.7618966944
Answer 19.5c. about 0.160623141
Answer 19.7a. 1- e-5/6=about 0.5654017915
21
Answer 19.7b. about 0.4963317258
Answer 19.8. about 0.4231900811
Answer 19.9. λ= variance= 2
Answer 19.10. about$7231.30
Answer 19.11.about$698.90
Answer 19.12. about 0.1550277818
Answer 19.13. about 0.758563549
Answer 19.14. λ = 4
Section 20.
Answer 20.1a. 1/10= 0.1
Answer 20.1b. 9/100= 0.09
Answer 20.1c. 9k-1/10 k
Answer 20.2. about 0.387420489
Answer 20.3. about 0.9158381456
Answer 20.4a. 0.001999
Answer 20.4b. 1000 times
Answer 20.5a. 3/64= 0.046875
Answer 20.5b. 1/256= 0.00390625
Answer 20.6. about 0.3595464678
Answer 20.7a. P(X=3) = 4/27= about 0.148148148148; P(X=50) = about 7.84164273*10-10
22
Answer 20.7b. E(X) = 3
Answer 20.8a. P(X = n) = (0.15)n(0.85) for all whole-number values of n.
Answer 20.8b. P(Y = n) =(0.15)n-1(0.85) for all natural-number values of n.
Answer 20.9a. about 0.1198380223
Answer 20.9b. about 0.3998629849
Answer 20.11. about 0.0527265219
Answer 20.12a. 10
Answer 20.12b. 0.81
Answer 20.13a. X is a geometric distribution with pmf p(x) = 0.4(0.6)x-1; x = 1, 2, etc.
Answer 20.13b. X is a binomial random variable with pmf p(x) = C(20, x)(0.60)x(0.40)20-x,
where x = 0, 1, ... , 20.
Answer 20.14. 108/625= 0.1728
Answer 20.15. 96/3125= 0.03072
Answer 20.16a. C(n-1, 2)(1/13)3(12/13)n-3
Answer 20.16b. about 0.017934507
Answer 20.17. E(X)= 24; Var(X)= 120
Answer 20.18. 7/64= 0.109375
Answer 20.19. 3/16= 0.1875
Answer 20.20. 0.289792
Answer 20.21. 0.0221184
Answer 20.22a. about 0.1198380223
Answer 20.22b. about 0.0253527238
Answer 20.23. 20
23
Answer 20.24. about 0.0644972544
Answer 20.25. C(t-1, 2)(1/6)3(5/6)t-3
Answer 20.26. 1625/4998= about 0.325130052
Answer 20.27. 1911/9614= about 0.1987726233
Answer 20.28. P(X=k)= C(6, k)*C(47, 6-k)/C(53, 6), for k= 0, 1, 2, 3, 4, 5, or 6
Answer 20.29. 80/323= about 0.2476780186
Answer 20.30. about 0.0727151097
Answer 20.31a. 3/14= 0.2142857143
Answer 20.31b. mean=3 batteries; Var(X)= 3/7= about 0.4285714286
Answer 20.32. 6149/7752= about 0.7932146543
Answer 20.33. C(2477,3)*C(121373, 97)/C(123850, 100)
Answer 20.34. 1/30= about 0.0333333333333333
Answer 20.36. x = 4
Answer 20.37. 3/8 = 0.375
Section 21.
Answer 21.1a. p(x) =

1/10 for x=2,
2/5 for x=6,
1/10 for x=10,
1/5 for x=11,
1/5 for x=15,
0 otherwise
24
Answer 21.1b. F(x)=
0, for x < 2,
1/10, for 2 ≤ x < 6,
½, for 6 ≤ x < 10,
3/5, for 10 ≤ x < 11,
4/5, for 11 ≤ x < 15
1, for 15 ≤ x
Answer 21.1c.`E(X) = 8.8 cents
Answer 21.2. The cdf of X can be expressed via the following table:
Interval for x--(-∞,0)---[0,1)-----[1,2)------[2, 3)------[3, ∞)
F(x)--------------0--------57/115---209/230--229/230---1
Answer 21.3a. P(X=1)= ¼; P(X=2)=1/6; P(X=3)= 1/12
Answer 21.3b. P(1/2 < X < 3/2)= ½
Answer 21.4. p(x) =
½ for x=0,
1/10 for x=1,
1/5 for x=2,
1/10 for x=3,
1/10 for x=3.5,
0 otherwise
Answer 21.5a. p(x)=
0.1, for x= -2,
0.2, for x=1.1,
25
0.3, for x=2,
0.4, for x=3,
0 otherwise
Answer 21.5b. P(2 < X < 3)=0
Answer 21.5c. P(X ≥ 3)= P(X=3)= 0.4
Answer 21.5d. 4/9= 0.444444444444444
Answer 21.6a. p=0.1
Answer 21.6b. F(X)=
0, for x < -1.9
0.1, for -1.9 ≤ x < -0.1
0.2, for -0.1 ≤ x < 2
0.5, for 2 ≤ x < 3
0.6, for 3 ≤ x < 4
1, for 4 ≤ x
Answer 21.6c. F(0)= 0.2; F(2)=0.5; F(F(3.1))= 0.2
Answer 21.6d. ½=0.5
Answer 21.6e. E(F(X))= 0.64
Answer 21.7a. p(x)=
0.3, for x= -4,
0.4 for x= 1,
0.3 for x=4,
0 otherwise
Answer 21.7b. Var(X)= 9.84; SD(X)= about 3.136877428
26
Answer 21.8a. p(x)=
1/12 for x=1,
1/6 for x=2,
1/12 for x=3,
1/6 for x=4,
1/12 for x=5,
1/6 for x=6,
1/12 for x=8,
1/12 for x=10,
1/12 for x= 12,
0 otherwise
Answer 21.8b. F(x)=
0, for x < 1
1/12, for 1 ≤ x < 2
1/4, for 2 ≤ x < 3
1/3, for 3 ≤ x < 4
1/2, for 4 ≤ x < 5
7/12, for 5 ≤ x < 6
3/4, for 6 ≤ x < 8
5/6, for 8 ≤ x < 10
11/12, for 10 ≤ x < 12
1, for 12 ≤ x
Answer 21.8c. P(X1/3. P(Xnot the same as F(4).
27
Answer 21.9a. P(X=0)= 0; P(X=1)= ¼; P(X=2)= ¼
Answer 21.9b. 7/16= 0.4375
Answer 21.9c. 3/16= 0.1875
Answer 21.9d. 3/8= 0.375
Answer 21.10a. 1/8= 0.125
Answer 21.10b. about 0.5839466094
Answer 21.10c. 1/2= 0.5
Answer 21.10d. ¼= 0.25
Answer 21.11a. e−1/2= P(A)= about 0.6065306597
Answer 21.11b. e−0.4- e−0.8= P(B) = about 0.2209910819
Answer 21.11c. (e−0.5- e−0.8)/(e−0.4- e−0.8)= P(A|B)=about 0.7113485948
Section 22.
Answer 22.1. c=2
Answer 22.2a. e−2 = about 0.1353352832
Answer 22.2b. e−1 - e−2 = about 0.2325441579
Answer 22.2c. F(x)= 1-e−x/5if x ≥ 0, 0 otherwise
Answer 22.3. k= 3/1000= 0.003; 0.027= 27/1000
Answer 22.4. 15/16= 0.9375
Answer 22.5a. Yes, F is the cdf of a continuous random variable.
Answer 22.5b. f(x)=
1/2, 0 ≤ x < 1,
1/6, 1 ≤ x < 4,
0 otherwise
28
Answer 22.6a. k=1
Answer 22.6b. 2e-1= about0.7357588823
Answer 22.7a. f(x)= ex/(ex + 1)2
Answer 22.7b. e/(e + 1) - ½=about 0.2310585786
Answer 22.8. About 369.0426555 gallons
Answer 22.9. 1/9= about 0.111111111111111
Answer 22.10. 15/32= 0.46875
Answer 22.11. 32/243= about 0.1316872428
Answer 22.12. 37/64= about 0.578125
Answer 22.13. C=0.3
Answer 22.14. 511/512= about 0.998046875
Answer 22.15. θ= 2
Answer 22.16.
F(x)/F(a) for x ≤ a;
1 for x > a.
Section 23.
Answer 23.1a. c=1.2
Answer 23.1b. F(x)=
0, for x ≤ -1,
0.2x+0.2, for −1 < x ≤ 0,
0.2+ 0.2x+ 0.6x2, for 0 < x ≤ 1,
1, for x > 1
Answer 23.1c. 0.25= ¼
29
Answer 23.1d. E(X)= 0.4 = 2/5
Answer 23.2a. a=0.6=3/5 and b=1.2=6/5
Answer 23.2b. F(x)=
0, for x
0.6x+0.4x3, for 0 ≤ x ≤ 1,
1, for x > 1
Answer 23.3a. E(X)= 4
Answer 23.3b. E(X)=0
Answer 23.3c. E(X)= ∞
Answer 23.4. E(X)= 2/3= about 0.6666666666 Var(X)= 2/9= about 0.22222222222
Answer 23.5a. E(X)= 1/3; SD(X)= √(2)/3= about 0.4714045208
Answer 23.5b. 15/16
Answer 23.6. E(lnX)= ½
Answer 23.7. Var(X)= 3/50= 0.06
Answer 23.8. E(Y)=7/3= about 2.333333333; Var(Y)= 34/45= about 0.75555555555556
Answer 23.9. 28/15= about 1.8666666667
Answer 23.10. About 328.1973799
Answer 23.11. E(X)= ½= 0.5
Answer 23.12. About 123.8291962
Answer 23.13. Var(X)= 5/36= about 0.138888888889
Answer 23.14. E(Y)= 3/2 5/2∫(πy/4)(cos πy)(1 + sin πy)3dy
When evaluated, E(Y)= about 2.2265625.
Answer 23.15. 2694.4
30
Answer 23.16. √(ln 16) + a
Answer 23.17. 3ln(2) = about 2.079441542
Answer 23.18. The mode of X is 2.
Answer 23.19. 21/20 = 1.05
Answer 23.20. k=6
Answer 23.21. m = about 0.8409
Answer 23.22a. f X (x) = 4x3/106
Answer 23.22b. 80
Section 24.
Answer 24.1a. f(x)= 1/4, for 3 ≤ x ≤ 7, 0 otherwise
Answer 24.1b. P(X
Answer 24.1c. 1/2
Answer 24.2a. f(x)=1/10, 5 ≤ x ≤ 15, 0 otherwise
Answer 24.2b. 3/10= 0.3
Answer 24.2c. E(X)=10; Var(X)= 25/3= about 8.333333333333333
Answer 24.3a. F(x)=
0, for x ≤ 0
x, for 0 < x < 1
1, for x ≥1
Answer 24.4. 1/(n+1)
Answer 24.5. ln(2)= about 0.6931471806
Answer 24.6. 2/3= about 0.6666666666667
Answer 24.7. 500
31
Answer 24.8. Var(Y)= 128/75= about 1.706666666666667
Answer 24.9. SD(X-250)= about 403.4357652
Answer 24.10a. E(e−X) = e/2 - 1/2e = about 1.175201194
Answer 24.10b. Var(e−X) =about 0.4323323575
Answer 24.11. a = 3
Answer 24.12. 7/10 = 0.7
Answer 24.13. P(X > X2) = min{1, 1/a}
Section 25
Answer 25.1d. about 88.12
Answer 25.2a. 0.7157
Answer 25.3a. 0.5
Answer 25.5. about 0.0228= 57/2500
32
Answer 25.9. The standard deviation of X is 75.
Answer 25.10a. P(Y= 1060)=e-2/[30√(2π)]
Answer 25.10b. x= 1098
Answer 25.11a. P(X > 5.5)= about 0.4721
Answer 25.11b. P(4 < X < 6.5) = about 0.1389
Answer 25.11c. P(X < 8) = about 0.6664
Answer 25.11d. P(|X − 7| ≥ 4) = about 0.5808
Answer 25.12. σ =about 0.01287
Answer 25.14b. E(X) = about 1070.4
Answer 25.15. 0.1151
Answer 25.16a. 0.1056
Answer 25.16b. 362.84
Answer 25.17. 1:18 p.m.
Answer 25.19c. about 26.81 milliamperes
Section 26
Answer 26.1. 1-e-0.9= about 0.5934303403
33
Answer 26.3. 1- e−0.5= about 0.3934693403
Answer 26.4. 1-e-1/8= about 0.1175030974
Answer 26.5. e-3/5= about 0.5488116361
Answer 26.6a. e-5/3= about 0.1888756028
Answer 26.6b. e-2/3- e-4/3= about 0.2498199809
Answer 26.7a. e-2= about 0.1353352832
Answer 26.7b. e-4/3- e-7/3= about 0.1666251703
Answer 26.7c. e-3= about 0.0497870684
Answer 26.8. about 0.1339745962
Answer 26.9a. e-1.68= about 0.186373976
Answer 26.9b. e-0.015- e-0.03= about 0.0146664061
Answer 26.10. about 0.4348589372
Answer 26.11. about 10255.89899
Answer 26.12. E(X)= 2+ 3e-2/3
E(X)= about 3.540251357
Answer 26.13. x= about 5644.226968
Answer 26.14. m= about 173.2867951
Answer 26.15. about 0.4204482076
Answer 26.16. about 8.325475924
Answer 26.17. about 0.0181327887
Section 27
Answer 27.1. about 0.1238883826
Answer 27.2. E(X)=3/2=1.5; Var(X)= ¾= 0.75
34
Answer 27.3. about 0.0947578682
Answer 27.4. about 0.4191864763
Answer 27.5. about 0.014387678
Answer 27.6a. f(x)= (1/16)x2e-0.5x, x ≥ 0, 0 otherwise;E(X)= 6 million kilowatt hours.

∞
Answer 27.6b. P(X>12)= 12 ∫(1/16)x2e-0.5xdx
Answer 27.7. 480
Answer 27.8a. k=60
Answer 27.8b. E(X)= 4/7= about 0.5714285714; Var(X)= 3/98= about 0.0306122449
Answer 27.9a. F(X)=
0 for x
3x2-2x3 for 0 ≤ x ≤ 1,
1 for x > 1
Answer 27.9b. 99/250= 0.396
Answer 27.10a. f(x)= 60x3(1-x)2 for 0 < x < 1, 0 otherwise.
Answer 27.10b. F(X)=
0 for x ≤ 0,
60x3−120x4+ 60x5, for 0 < x < 1,
1 for x ≥ 1.
Answer 27.11. 0.47178
Answer 27.12a. f(x)= 43.6486021x1.0952(1-x)4.6648, for 0 < x
Answer 27.12b. about 0.0829070132
Answer 27.13a. k=12
Answer 27.13b. 0.6875= 11/16
35
Section 28
Answer 28.1. f Y (y)= [1/{√(2π)a}]exp{−([y-b]/a−μ)2/2}
Answer 28.2. f Y (y)= 2(y+1)/9, -1 ≤ y ≤ 2, 0 otherwise.
Answer 28.3. f Y (y)=(1/6)y-1/3, 0 ≤ y ≤ 8, 0 otherwise.
Answer 28.4. f Y (y) = λy−λ-1, y ≥ 1, 0 otherwise.
Answer 28.5. f Y (y)= √(2)cy1/2 m-3/2e−2βy/m, y ≥ 0
Answer 28.6. f Y (y)= e-y for 0 < y < ∞, 0 otherwise
Answer 28.7. f Y (y)= 1/[π√(1-y2)] for − 1 ≤ y ≤ 1, 0 otherwise.
Answer 28.8a. f Y (y)=(1/α)y(-α+1)/α, for 0 < y < 1, 0 otherwise; E(Y)= 1/[α+1]
Answer 28.8b. f C (c)= ec for -∞< c < 0, 0 otherwise; E(C)=-1
Answer 28.8c. f R (r)= 1/r for 1< r < e, 0 otherwise; E(R)= e-1
Answer 28.8d. f N (n)= 2/[π√(1-n2)] for 0 < n < 1, 0 otherwise; E(N)=2/π
Answer 28.9. about 998.7196821
Answer 28.10. f Y (y)=4y-2 for y > 4, 0 otherwise.
Answer 28.11. F V (v) =25ln(v/10000)-1 for v є (10000e0.04, 10000e0.08),
0 for v ≤ 10000e0.04,
1 for v ≥ 10000e0.08
Answer 28.12. f Y (y)= 0.125(0.1y)0.25exp{-(0.1y)1.25}, for y > 0, 0 otherwise.
Answer 28.13. f R (r)= 5/(2r2) for 5/4< r < 5/6, 0 otherwise.
Answer 28.14. f B (b)= f A (b/2)/2
Answer 28.15b. f Y (y)= [1/{2y√(2π)}]exp{−( ln(y)−1)2/8}
Answer 28.16. f Y (y) =[1/B(1/2, 1)] (y)1/2-1(1-y)1-1, which is the pdf of a beta random variable
with parameters (1/2, 1).
36
Answer 28.17a. f Y (y) = (1/18)y2 − 1/3, -3 ≤ y ≤ 3, 0 otherwise
Answer 28.17b. (3/2)z2- 9z + 25/2, 2 ≤ z ≤ 4, 0 otherwise
Answer 28.18. f Y (y) = y-1/2-1, 0 < y < 1, 0 otherwise
Answer 28.19. e2yexp{−(e2y)/2}, y є {Real Numbers}
Section 29
Answer 29.1b. ¼=0.25
Answer 29.1c. 11/20= 0.55
Answer 29.1d. p X (x)=
0.3 for x=0,
0.5 for x=1,
0.125 for x=2,
0.075 for x=3,
0 otherwise
Answer 29.2. 0.25= ¼
Answer 29.3a. 0.08
Answer 29.3b. 0.36
Answer 29.3c. 0.86
Answer 29.3d. 0.35
Answer 29.3e. 0.6
Answer 29.3f. 0.65
Answer 29.3g. 0.4
Answer 29.4a. 0.4
Answer 29.4b. 0.8
37
Answer 29.5. 0.0512
Answer 29.6a. F XY (x, y)= [1- exp{(−x2)/2}][1- exp{(−y2)/2}], for 0 < x, y, 0 otherwise.
Answer 29.6b. f X (x) = x*exp{(−x2)/2} for x>0, 0 otherwise;
f Y (y) = y*exp{(−y2)/2} for y>0, 0 otherwise.
Answer 29.7. In order to make f XY (x, y) a joint pdf, we need to multiply it by a factor of (2-
a)(a+1).
Answer 29.8. 5/8=0.625
Answer 29.9. About 0.708333333333333= 17/24
Answer 29.10. 83/144= about 0.5763888888889
Answer 29.11. 0.488
Answer 29.12. f Y (y) = 15y3/2-15y2, 0 ≤ y ≤ 1, 0 otherwise
Answer 29.13. 0.171875= 11/64
Answer 29.14. 52/9= about 5.7777777777778
Answer 29.15. 0.8333333333333= 5/6
Answer 29.16. 1/125= 0.008
Answer 29.17. 0.35= 7/20
Answer 29.18. 1-2e−1 = about 0.2642411177
Answer 29.19. 12/25 = 0.48
Answer 29.20. θ 1 = 1/4; θ 2 = 0
Answer 29.21. 3/8 = 0.375
Answer 29.22a. p XY (x, y) =

0.2 for (x, y) = (1, 1),
0.5 for (x, y) = (1, 2),
0.2 for (x, y) = (2, 1),
0.1 for (x, y) = (2, 2),
0 otherwise
38
Answer 29.22b. F XY (x, y), the joint cdf of X and Y is expressed by the following table:
X/Y---1----2
1-----0.2---0.7
2-----0.4---1
Section 30
Answer 30.1a. X and Y are independent.
Answer 30.1b. ½= 0.5
Answer 30.1c. 1-e−a
Answer 30.2c. π/4= about 0.7853981634
Answer 30.3a. 0.484375= 31/64
Answer 30.3b. f X (x)= 3x2-6x+3, for 0 ≤ x < 1, 0 otherwise
f Y (y)=6y(1-y) for 0 < y ≤ 1, 0 otherwise
Answer 30.3c. X and Y are not independent.
Answer 30.4a. k=4
Answer 30.4b. f X (x)= 2x, 0 ≤ x ≤ 1, 0 otherwise
f Y (y)= 2y, 0 ≤ y ≤ 1, 0 otherwise
Answer 30.4c. X and Y are independent.
Answer 30.5a. k=6
Answer 30.5b. f X (x)=2x, 0 ≤ x ≤ 1, 0 otherwise
f Y (y)= 3y2, 0 ≤ y ≤ 1, 0 otherwise
Answer 30.5c. 0.15= 3/20
Answer 30.5d. 0.875= 7/8
Answer 30.5e. X and Y are independent.
Answer 30.6a. k=8/7
39
Answer 30.6b. X and Y are independent.
Answer 30.6c. 16/21= about 0.7619047619
Answer 30.7a. f X (x)= x2/4+1/6, 0 ≤ x ≤ 2, 0 otherwise
f Y (y)= 1/3+y/6, 0 ≤ y ≤ 2, 0 otherwise
Answer 30.7b. X and Y are not independent.
Answer 30.7c. 31/72= about 0.43055555555555555556
Answer 30.8a. f X (x)= 4/3-8x/9, 0 ≤ x ≤ 3/2, 0 otherwise.
f Y (y)= 4y/9 for 0 ≤ y ≤ 1.5,
4/3-4y/9 for 1.5 ≤ y ≤ 3,
0 otherwise.
Answer 30.8b. 2/3
Answer 30.9. about 0.4691535082
Answer 30.10. about 0.1914554452
Answer 30.11. 2/5= 0.4
Answer 30.12. 0.19= 19/100
Answer 30.13. 0.295
Answer 30.14. about 0.4137812139
Answer 30.15. f Z (z)= e−z/2-e−z, z > 0
Answer 30.16. f X (x)= 2/(1+2x)2
Answer 30.17. P(X2 ≥ Y3)= 3/5 = 0.6
Answer 30.18. It is impossible that X and Y are independent, as .4 = P(X = 1|Y = 1) ≠ P(X =
1|Y = 0) = 0.7.
40
Section 31
Answer 31.1. p Z (z)=
0.025, for z=0,
0.09 , for z=1,
0.19 , for z=2,
0.29 , for z=3,
0.265, for z=4,
0.14 , for z=5,
0 otherwise
Answer 31.2. p Z (z)= C(30, z)0.2z0.830-z, for z є {integers between 0 and 30, inclusive}, 0
otherwise.
Answer 31.3. p X+Y (n)= (n-1)p2(1-p)n-2, n = 2, 3, ..., 0 otherwise
Answer 31.4. p Z (z)=
1/12 , for z=3,
1/3 , for z=4,
1/3 , for z=5,
¼ , for z=6.
0 otherwise
F Z (z)=
0, for z < 3,
1/12, for 3≤ z < 4,
5/12, for 4≤ z < 5,
¾ , for 5≤ z < 6,
1 , for 6≤ z
41
Answer 31.5. 1/64= 0.015625
Answer 31.7a. P(X + Y = 2)= e-λp(1-p+λ)
Answer 31.7b. P(Y > X) = e-pλ
Answer 31.8. p X+Y (a) =
¼ if a= 0,
¼ if a = 1,
5/16 if a=2,
1/8 if a=3,
1/16 if a=4,
0 otherwise
Answer 31.9. p X+Y (a) =
1/6 if a = 1,
5/18 if a=2,
1/3 if a=3,
1/6 if a=4,
1/18 if a=5,
0 otherwise
Answer 31.10. Hint: Show that p X+Y (n)= C(n-1, 2-1)(p2)(1-p)n-2.
Answer 31.11. 9e-8 = about 0.0030191637
Answer 31.12. e-10λ(10λ)10/10! =1562500e-10λλ10/567
Answer 31.13. f X+Y (a)= 2λe-λa-2λe-2λa, for a > 0, 0 otherwise.
Answer 31.14. f X+Y (a)= 1-e-λa for 0 ≤ a ≤ 1,
42
e-λa(eλ-1) for a>1,
0 otherwise
∞
Answer 31.15. f X+2Y (a)= -∞ ∫[f X (a-2y)f Y (y)]dy
Answer 31.16. f X+Y (a)= a3/6- 3a2/2+ 2a for 0 ≤ a ≤ 1,
7/6-a/2 for 1 ≤ a ≤ 2,
9/2-9a/2+3a2/2- a3/6 for 2 ≤ a ≤ 3, 0 otherwise.
Answer 31.17. f X+Y (a) = 2a3/3 if a є (0, 1],
-2a3/3 + 4a - 8/3 if a є (1,2),
0 otherwise.
Answer 31.18. f X+Y (a) = [(αβ)(e-αa-e-βa)]/(β-α)
Answer 31.19. f 2T1+ T2 (a)= e-a/2-e-a, a > 0, 0 otherwise.
Answer 31.20. f X+Y (a) = a/4- ½ if 2 ≤ a ≤ 4,
3/2 - a/4 if 4 ≤ a ≤ 6,
0 otherwise.
Answer 31.21. f X+Y (a) = a2/8 if 0 < a ≤ 2,
- a2/8 +a/2 if 2 < a < 4,
0 otherwise.
Answer 31.22. f X+Y (a) = [1/√(2π[σ 1 2+ σ 2 2])] exp{−(a−[μ 1 + μ 2 ])2/(2[σ 1 2+ σ 2 2])}
Answer 31.23. 0.125= 1/8
Answer 31.24. f Z (z) = ze-z, z ≥ 0, 0 otherwise.
Answer 31.25. 1-2e-1 = about 0.2642411177
43
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 32.3a. 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.
44
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) py(1−p)n−y, y = 0, 1, · · · , n, 0 otherwise.
Answer 32.6b. p X|Y (x|y)= yxe− 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.
Answer 32.8a. c= 1/(2N-1)
Answer 32.8b. p X (x)= 2x/(2N-1) for x = 0, 1, · · · , N − 1, 0 otherwise.
Answer 32.8c. p Y|X (y|x)= (1 − 2−x)y/2x 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
45
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.
Section 33
Answer 33.1. f X|Y (x|y)= 3x2/(2 -2y3); f X|Y (x|0.5)= 12x2/7 for −1 ≤ x ≤ -0.5 or 0.5 ≤ x ≤ 1, 0
otherwise.
Answer 33.2. f X|Y (x|y)= 2xy-2, 0 ≤ x < y ≤ 1, 0 otherwise.
Answer 33.3. f Y|X (y|x) = 3y2/x3, 0 ≤ y < x ≤ 1, 0 otherwise.
Answer 33.4. f X|Y (x|y)= (y+1)2xe−x(y+1), x ≥ 0, y ≥ 0, 0 otherwise.
f Y|X (y|x) = xe−xy, x ≥ 0, y ≥ 0, 0 otherwise.
Answer 33.5a. f XY (x, y)= [y2e−x/2]
Answer 33.5b. f X|Y (x|y)= e−(x-y)
Answer 33.6a. f X|Y (x|y)=6x(1 − x) , for 0 < x < 1, 0 otherwise. X and Y are independent.
Answer 33.6b. ¼= 0.25
Answer 33.7a. f X|Y (x|y)= [(1/3)x − y + 1]/(3/2-y) , 1 ≤ x ≤ 2, 0 ≤ y ≤ 1, 0 otherwise.
Answer 33.7b. 11/24= about 0.4583333333
Answer 33.8. ¼= 0.25
Answer 33.9. 8/9= about 0.8888888888888888
Answer 33.10. 5/12= 0.4166666666667
Answer 33.11. 0.875= 7/8
46
Answer 33.12. about 0.1222526851
Answer 33.13. f X (x) = 2(1 - x), x є (0, 1), 0 otherwise.
Answer 33.14. E(Y)=1/3; Var(Y)= 1/18
Section 34
Answer 34.1. f ZW (z, w)=│1/(ad-bc)│f XY ((zd-wb)/(ad-cb), (zc-wa)/(bc-da))
Answer 34.2. fy 1 y 2 (y 1 , y 2 )= λ2e-λy1/y 2 , y 1 ≥ ln(2), y 2 > 1, 0 otherwise.
Answer 34.3. f RФ (r, φ)= (r)f XY [rcos(φ), rsin(φ)], r > 0, −π < Ф ≤ π.
Answer 34.4. f ZW (z,w)= [z/(1+ w2)]f XY [z*cos(arctan(w)), z*sin(arctan(w))]
Answer 34.5. f UV (u, v)= λα+βe-λuuαvα-1(u-uv)β-1/[Γ(α)Γ(β)], for u ≥ 0, v ≥ 0, 0 otherwise.
Answer 34.6. fy 1 y 2 (y 1 , y 2 )= y 1 e− y1, y 1 ≥ 0, 0 < y 2
Answer 34.7. fy 1 y 2 (y 1 , y 2 )= (1/28π)exp{-(3y 1 + y 2 )2/98}exp{-(y 1 -2y 2 )2/392}
Answer 34.8. Hint: Show that f U (u) = [exp{-u2/2}/√(2π)], the pdf of a standard normal random
variable, and f V (v) = [exp{-v2/2}/√(2π)], the pdf of a standard normal random variable.
Answer 34.9. f U (u)= -∞ ∞∫f XY (u-v, v)dv= -∞ ∞∫f XY (u-y, y)dy;
This is irrespective of whether X and Y are independent or dependent.
In the case where X and Y are independent,

∞
f U (u)= -∞ ∫f X (u-v)f V (v)dv= -∞ ∞∫f X (u-y)f Y (y)dy
Also, by convolution, we get

∞
f U (u)= -∞ ∫f X (u-y)f Y (y)dy
Answer 34.10. f U (u)= -∞ ∞∫f XY (v-u, v)dv= -∞ ∞∫f XY (y-u, y)dy
Answer 34.11. f U (u)= -∞ ∞∫│1/v│*f XY (v, u/v)dv= -∞

∞
∫│1/x│*f XY (x, u/x)dx
Answer 34.12. f U (u)= 1/(u+1)2 for u > 0 and 0 otherwise.
Section 35
Answer 35.1. E(|X − Y |)= (m+1)(m-1)/(3m)
47
Answer 35.2. 7/12= about 0.583333333333
Answer 35.3. E(|X − Y|)= 1/3= about 0.3333333333
Answer 35.4. E(X2Y) = 7/36= 0.194444444444; E(X2 + Y2)= 5/6= 0.8333333333
Answer 35.5. E(3X + 4Y − 7)=0
Answer 35.6. E[(3X − 4)(2Y + 7)]= 33
Answer 35.7. E(│X-Y│)= L/3
Answer 35.8. We can expect 1 person to select his own hat.
Answer 35.9. 30/19= about 1.578947368
Answer 35.10a. 9/10= 0.9
Answer 35.10b. 49/10= 4.9
Answer 35.10c. 21/5 = 4.2
Answer 35.11a.E[(2 + X)2]=14
Answer 35.11b. Var(4 + 3X)= 45
Answer 35.12. E[T1+T2]= 292/51= about 5.725490196
Answer 35.13. E[T12+T22]= 2L2/3
Answer 35.14. E[(X + 1)2(Y − 1)2]= 27
Section 36
Answer 36.1. E[(X − Y )2] = 2σ2
Answer 36.2. Cov(X,Y)= -25/204= about -0.1225490196; ρ(X,Y)= -1/3= about -

0.3333333333333
Answer 36.3. Cov(X,Y)= 1/12= about 0.083333333; ρ(X,Y)= √(2)/2= about 0.7071067812
Answer 36.4a. Thus, f XY (x, y)=5 for -1 < x < 1, x2 < y < 0.1+x2, 0 otherwise
Answer 36.4b. Cov(X,Y)= 0; ρ(X,Y)= 0
Answer 36.6a. 0.5 = ½
48
Answer 36.6b. ρ(X 1 + X 2 , X 3 + X 4 )= 0
Answer 36.7. Cov(X, Y)= -n2/36
Answer 36.9. Cov(X,Y)= 3/160= 0.01875; ρ(X,Y)= about 0.3973597071
Answer 36.10. Cov(2X − 5, 4Y + 2)=24
Answer 36.11. $19,300
Answer 36.12. Var(Z)= 11
Answer 36.13. 200
Answer 36.14. Cov(X,Y)=0
Answer 36.15. 28/675= about 0.0414814815
Answer 36.16. Cov(X, Y)= 6
Answer 36.17. Cov(C1,C2)= 8.8
Answer 36.18. 0.2743
Answer 36.19a. f XY (x, y)= ½ for 0 < x < 1, 0 < y < 2, 0 otherwise.
Answer 36.19b. f Z (a)= a/2, for 0 < a ≤ 1,
½, for 1 < a ≤ 2,
3/2-a/2, for 2 < a ≤ 3,
0 otherwise
Answer 36.19c. E(X)= ½; Var(X)= 1/12; E(Y)= 1; Var(Y)= 1/3
Answer 36.19d. E(Z)= 3/2; Var(Z)= 5/12
Answer 36.20a. f Z (z)= z/2 for z є (0, 2), 0 otherwise.
Answer 36.20b. f X (x)= 1-x/2 for 0 < x < 2, 0 otherwise.

f Y (y)= 1-y/2 for 0 < y < 2, 0 otherwise.
Answer 36.20c. E(X)= 2/3; Var(X)= 2/9
Answer 36.20d. Cov(X, Y) = -1/9
49
Answer 36.21. Cov(X, Y)= -0.15
Answer 36.22. Var(X)= 1/6
Answer 36.23. Var(X 1 + 2X 2 − X 3 ) = 5/12
Answer 36.24. Var(Y-X)= 1.04
Answer 36.25. c=√(π)/2
Answer 36.26. E(W)=4; Var(W)= 67
Answer 36.27. ρ(X, Y) = -1/5
Answer 36.28. a= 2
Answer 36.29. ρ(U, V) = (n-2)/(n+2)
Answer 36.30a. The following table gives the marginal distributions:
XY--------0------1--------2------p X (x)
0----------0.25----0.08---0.05---0.38
1----------0.12----0.20---0.10---0.42
2----------0.03----0.07---0.10---0.20
p Y (y)-----0.40----0.35---0.25-----1
Answer 36.30b. E(X) = 0.82; E(Y) = 0.85
Answer 36.30c. E(XY) = 0.243
Answer 36.30d. E(C) = $147.75
Answer 36.31. σ= √(865) = about 29.41088234
Section 37
Answer 37.1. E(X|Y)= 2y/3; (2/3)(1+x+x2)/(1+x)
Answer 37.2. E(X)= ½; Var(X)= 1/12; E(Y|X)= 3x/4; Var(Y|X)= 3x2/80; Var(Y)= 19/320;
E[Var(Y|X)]= 1/80; Var[E(Y |X)]= 3/64
Answer 37.3. P(X > Y) = λ/(λ+μ)
Answer 37.4. E(X|X > 0.5)= ¾= 0.75
Answer 37.5. E(Y |X = 2)= 2.3
50
Answer 37.6. (2/3)(1-x6)/(1-x4)= E(Y|X)
Answer 37.7. E(Y)= 7/9
Answer 37.8. E(X)= 15; E(Y)= 5
Answer 37.9. E(Y)= -1.4
Answer 37.10. E(Y)=15 people
Answer 37.11. E(X|Y = 1)= 2; E(X|Y = 2)= 5/3= about 1.6666666667; E(X|Y = 3)= 12/5= 2.4.
X and Y are not independent.
Answer 37.12. 10/49= about 0.2040816327
Answer 37.13. Var(Y|X=x)=1/12
Answer 37.14. Var(X|Y=0)= 0.9856
Answer 37.15. Var(Y)= 13
Answer 37.16. Var(Y |X = x)= (1/12)x2 - (1/6)x+ (1/12)
Answer 37.17. β2(λ + λ2) + α2λ
Section 38
Answer 38.1. E(X)= (n+1)/2; Var(X)= (1/12)n2-1/12
Answer 38.2. E(X)= 1/p; Var(X)= (1-p)/p2
Answer 38.4. E(X)= α/λ; Var(X)= a/λ2
Answer 38.6. M X (t)= 1 when t=0 and ∞ otherwise.
Answer 38.7. M Y (t)= e-2t[λ/(λ-3t)], 3t < λ
Answer 38.8. X is a binomial random variable such that
p X (x)= C(15, x)(3/4)x(1/4)15-x, x є {0, 1, 2, ... 15}, 0 otherwise.
Answer 38.9. Thus, Y is a random variable such that
p Y (y)= (1/3)*C[15, (y+2)/3]*(3/4)(y+2)/3*(1/4)15-(y+2)/3, y є {-2, 1, 4, ... 43}, 0 otherwise.
Answer 38.10. M(t 1 , t 2 )= exp{(t 1 2+ t 2 2)}
51
Answer 38.11. SD(X)= 5000
Answer 38.12. E(X3)= 10,560
Answer 38.13. M Y (t)= (8/27)et + (19/27)
Answer 38.14. 0.8384
Answer 38.15a. M Xi (t)= pet/[1- et(1 − p)]
Answer 38.15b. {pet/[1-(1-p)et]}n
Answer 38.16. P(M 1 > V 2 )= 0.6915
Answer 38.17. Var(X)= 2
Answer 38.18. M X (t)= (3t2et-6tet+6et- 6)/t3
Answer 38.19. E[(X-2)3]= -38
Answer 38.20. P(X = 2)= e−1/2= about 1.359140914
Answer 38.21. exp{k2/2}
Answer 38.22. b=4
Answer 38.23. M(t 1 , t 2 )= [(et1x-1)(et2x-1)]/(t 1 t 2 )
Answer 38.24. Var(X)=2/9
Answer 38.25. M X+2Y (t)= exp{13t2+4t}
Answer 38.26. E(2X 1 − X 2 )= 0.4
Answer 38.27. Cov(X, Y)= -0.9375
Answer 38.28. M V (t) = (0.7+0.3et)9
Section 39
Answer 39.1. Hint: Show that for all possible values of X, P(│X│≥ ε) = 1 and Chebyshev's
Inequality is in this case an equality.
Answer 39.2. n = 100
52
Answer 39.3. 4/9= about 0.4444444444
Answer 39.4. P(X ≥ 104) ≤ 10−7.
Answer 39.5. P (0 < X < 40) ≥ 19/20 or P (0 < X < 40) ≥ 0.95
Answer 39.6. P(X 1 + X 2 + · · · + X 20 > 15) ≤ 1
Answer 39.7. P(65 < X < 85) ≥ ¾
Answer 39.9. The probability of not having enough stock left by the end of the week is at
most 25/38 = about 0.6578947368.
Answer 39.10. P(0.475 ≤ X ≤ 0.525) ≥ 0.99
Answer 39.12. P(X > 120) ≤ 100/121 = about 0.826446281; P(70 < X
Answer 39.13a. P(100 ≤ X ≤ 140) ≥ 0.79
Answer 39.13b. P(100 ≤ X ≤ 140) = about 0.9709
Answer 39.14. X- converges to E(X) = 2/3.
Answer 39.15a. F Y_n (x) = 0 for x ≤ 0;
F Y_n (x) = 1 - (1 - x)n for 0 < x < 1;
F Y_n (x) = 1 for x ≥ 1.
Section 40
Answer 40.5. about 0.1357 or about 0.1367, depending on the rounding used.
53
Answer 40.9. There must be at least 23 components in stock.
Answer 40.11. 6,342,525
Answer 40.13. 16 bulbs
Answer 40.18a. X- is approximated by a normal distribution with mean 100 and variance 4.
Section 41
Answer 41.1a. 1/6
Answer 41.1b. P(X ≥ 6) ≤ 7/12
Answer 41.1c. P(X ≥ 6) ≤ 7/15= about 0.466666666666667
Answer 41.1d. P(X ≥ 6) ≤ 7/22= about 0.31818181818
Answer 41.2. The Chernoff bounds for this random variable are
P(X ≥ a) ≤ e-ta(pet + 1 - p)n, t > 0 and
P(X ≤ a) ≤ e-ta(pet + 1 - p)n, t < 0
Answer 41.3. P(X ≥ 5) ≤ 9/13 = about 0.6923076923
Answer 41.4. P(X ≥ 1500) ≤ 1/16 = 0.0625
Answer 41.5. The Chernoff bounds for this random variable are
P(X ≥ a) ≤ e-taexp[λ(t-1)], t > 0 and
54
P(X ≤ a) ≤ e-taexp[λ(t-1)], t < 0
Answer 41.6a. p ≤ 10/13 = about 0.7692307692
Answer 41.6b. P(X ≥ 26) ≤ e-26texp[20(t-1)], t > 0.
Answer 41.6c. P(X ≥ 26) ≤ 5/14 = about 0.3571428571
Answer 41.6d. p = about 0.1093.
Answer 41.8a. E(X) = a/(a-1)
Answer 41.8b. E(1/X) = a/(a+1)
Answer 41.8d. Hint: Show that E(1/X) ≥ 1/E(X).
Section 43
Answer 43.1. 0 1∫ x x+1∫f(x, y)dydx
Answer 43.2. 0 1∫0 x/2∫f(x, y)dydx or 0 1/2∫ 2y 1∫f(x, y)dxdy
Answer 43.3. 1/2

1
∫1/2 1∫f(x, y)dydx
Answer 43.4. 0
1/2
∫1/2 1∫f(x, y)dydx + 1/2 1∫0 1∫f(x, y)dydx
Answer 43.5. 20 30∫20 50-x∫f(x, y)dydx
Answer 43.6. 0 1∫0 x+1∫f(x, y)dydx + 1 ∞∫ x-1 x+1∫f(x, y)dydx
Answer 43.7. 1 - 2e-1
Answer 43.8. 1/6
Answer 43.9. L4/3
Answer 43.10. 0.5 1∫0.5 1∫f(x, y)dydx
Answer 43.11. 7/8
Answer 43.12. 3/8
55

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Answer Keys for the 2008 Edition of Marcel B.

Finan’s “A Probability Course for the Actuaries”

Answer Keys for the 2008 Edition of

First Edition Published in July-August 2008

Answer 1.1.A={x:x shows a face with prime number}= {2, 3, 5}

HTT, TTT, TTH, THT

Answer 1.2c. F in set-builder notation is

Answer 1.4. There are no elements of E and so E=Ø

Answer 1.9a. 55 tacos

Answer 1.9b. 40 tacos

Answer 1.9c. 10 tacos

Answer 1.10a. 20 students

Answer 1.10b. 5 students

Answer 1.10c. 11 students

Answer 1.10d. 42 students

Answer 1.10e. 46 students

Answer 1.10f. 46 students

Answer 1.11. n(S) = 24

Answer 2.3. (G U B U S)c.

Answer 2.5. There are 880 young, single, female policyholders.

Answer 2.8. n(A)=60

Answer 2.11. 50 players surveyed

Answer 2.13a. 3 students

Answer 2.13b. 6 students

Answer 2.14. Mr. Brown owns 32 chickens.

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

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 4.6d. 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