answer keys for the 2008 edition of marcel b. finan’s “a...

Answer Keys for the 2008 Edition of Marcel B. Finan’s “A Probability Course for the Actuaries” Second Edition – G. Stolyarov II

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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.

http://faculty.atu.edu/mfinan/actuarieshall/Pbook.pdf




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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.


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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



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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




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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}


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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


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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


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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


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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


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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


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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.


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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


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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


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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]*I2,3,4,5,6,7,8,9,10,11,12(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)*I1,2,3,4,5,6(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]*I0,1,2(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=1xΣ ((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


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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=0yΣ 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,


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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


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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;


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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


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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


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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


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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


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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


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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


24

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


25

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,


26

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


27

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).


28

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


29

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


30

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)= 3/25/2∫(πy/4)(cos πy)(1 + sin πy)3dy

When evaluated, E(Y)= about 2.2265625.

Answer 23.15. 2694.4


31

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. fX(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


32

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





33



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


34

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


35

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


36

Section 28

Answer 28.1. fY(y)= [1/{√(2π)a}]exp{−([y-b]/a−μ)2/2}

Answer 28.2. fY(y)= 2(y+1)/9, -1 ≤ y ≤ 2, 0 otherwise.

Answer 28.3. fY(y)=(1/6)y-1/3, 0 ≤ y ≤ 8, 0 otherwise.

Answer 28.4. fY(y) = λy−λ-1, y ≥ 1, 0 otherwise.

Answer 28.5. fY(y)= √(2)cy1/2 m-3/2e−2βy/m, y ≥ 0

Answer 28.6. fY(y)= e-y for 0 < y < ∞, 0 otherwise

Answer 28.7. fY(y)= 1/[π√(1-y2)] for − 1 ≤ y ≤ 1, 0 otherwise.

Answer 28.8a. fY(y)=(1/α)y(-α+1)/α, for 0 < y < 1, 0 otherwise; E(Y)= 1/[α+1]

Answer 28.8b. fC(c)= ec for -∞< c < 0, 0 otherwise; E(C)=-1 Answer 28.8c. fR(r)= 1/r for 1< r < e, 0 otherwise; E(R)= e-1 Answer 28.8d. fN(n)= 2/[π√(1-n2)] for 0 < n < 1, 0 otherwise; E(N)=2/π Answer 28.9. about 998.7196821

Answer 28.10. fY(y)=4y-2 for y > 4, 0 otherwise.

Answer 28.11. FV(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. fY(y)= 0.125(0.1y)0.25exp{-(0.1y)1.25}, for y > 0, 0 otherwise.

Answer 28.13. fR(r)= 5/(2r2) for 5/4< r < 5/6, 0 otherwise.

Answer 28.14. fB(b)= fA(b/2)/2


Answer 28.15b. fY(y)= [1/{2y√(2π)}]exp{−( ln(y)−1)2/8}

Answer 28.16. fY(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).


37

Answer 28.17a. fY(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. fY(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. pX(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


38

Answer 29.5. 0.0512

Answer 29.6a. FXY(x, y)= [1- exp{(−x2)/2}][1- exp{(−y2)/2}], for 0 < x, y, 0 otherwise.

Answer 29.6b. fX(x) = x*exp{(−x2)/2} for x>0, 0 otherwise;

fY(y) = y*exp{(−y2)/2} for y>0, 0 otherwise.

Answer 29.7. In order to make fXY(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. fY(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. θ1= 1/4; θ2 = 0

Answer 29.21. 3/8 = 0.375

Answer 29.22a. pXY(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


39

Answer 29.22b. FXY(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. fX(x)= 3x2-6x+3, for 0 ≤ x < 1, 0 otherwise

fY(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. fX(x)= 2x, 0 ≤ x ≤ 1, 0 otherwise

fY(y)= 2y, 0 ≤ y ≤ 1, 0 otherwise

Answer 30.4c. X and Y are independent.

Answer 30.5a. k=6

Answer 30.5b. fX(x)=2x, 0 ≤ x ≤ 1, 0 otherwise

fY(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


40

Answer 30.6b. X and Y are independent.

Answer 30.6c. 16/21= about 0.7619047619

Answer 30.7a. fX(x)= x2/4+1/6, 0 ≤ x ≤ 2, 0 otherwise

fY(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. fX(x)= 4/3-8x/9, 0 ≤ x ≤ 3/2, 0 otherwise.

fY(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. fZ(z)= e−z/2-e−z, z > 0

Answer 30.16. fX(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.


41

Section 31

Answer 31.1. pZ(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. pZ(z)= C(30, z)0.2z0.830-z, for z є {integers between 0 and 30, inclusive}, 0 otherwise.

Answer 31.3. pX+Y(n)= (n-1)p2(1-p)n-2, n = 2, 3, ..., 0 otherwise

Answer 31.4. pZ(z)=

1/12 , for z=3,

1/3 , for z=4,

1/3 , for z=5,

¼ , for z=6.

0 otherwise

FZ(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


42

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. pX+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. pX+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 pX+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. fX+Y(a)= 2λe-λa-2λe-2λa, for a > 0, 0 otherwise.

Answer 31.14. fX+Y(a)= 1-e-λa for 0 ≤ a ≤ 1,


43

e-λa(eλ-1) for a>1,

0 otherwise

Answer 31.15. fX+2Y(a)= -∞∞∫[fX(a-2y)fY(y)]dy

Answer 31.16. fX+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. fX+Y(a) = 2a3/3 if a є (0, 1],

-2a3/3 + 4a - 8/3 if a є (1,2),

0 otherwise.

Answer 31.18. fX+Y(a) = [(αβ)(e-αa-e-βa)]/(β-α)

Answer 31.19. f2T1+ T2(a)= e-a/2-e-a, a > 0, 0 otherwise.

Answer 31.20. fX+Y(a) = a/4- ½ if 2 ≤ a ≤ 4,

3/2 - a/4 if 4 ≤ a ≤ 6,

0 otherwise.

Answer 31.21. fX+Y(a) = a2/8 if 0 < a ≤ 2,

- a2/8 +a/2 if 2 < a < 4,

0 otherwise.

Answer 31.22. fX+Y(a) = [1/√(2π[σ12+ σ2

2])] exp{−(a−[μ1+ μ2])2/(2[σ12+ σ2

2])}

Answer 31.23. 0.125= 1/8

Answer 31.24. fZ(z) = ze-z, z ≥ 0, 0 otherwise.

Answer 31.25. 1-2e-1 = about 0.2642411177


44

Section 32

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

0.25, when x=0,

0.75, when x=1,

0 otherwise.

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

Answer 32.2b. pX|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.

pX|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.

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

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

pX|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. pX|Y(x|1)=

1/6 for x=0,

½ for x=1,

1/3 for x=2,

0 otherwise.

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

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

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


45

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

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

pY|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. pY(y) = C(n, y) py(1−p)n−y, y = 0, 1, · · · , n, 0 otherwise.

Answer 32.6b. pX|Y(x|y)= yxe− y/x! , y = 0, 1, · · · , n ; x = 0, 1, · · ·, 0 otherwise. X and Y are not independent.

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

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

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



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

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

Answer 32.8c. pY|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


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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

pX│Y(x│1)= 16/17 for x=1,

1/17 for x=2,

0 otherwise.

Section 33

Answer 33.1. fX|Y(x|y)= 3x2/(2 -2y3); fX|Y(x|0.5)= 12x2/7 for −1 ≤ x ≤ -0.5 or 0.5 ≤ x ≤ 1, 0 otherwise.

Answer 33.2. fX|Y(x|y)= 2xy-2, 0 ≤ x < y ≤ 1, 0 otherwise.

Answer 33.3. fY|X(y|x) = 3y2/x3, 0 ≤ y < x ≤ 1, 0 otherwise.

Answer 33.4. fX|Y(x|y)= (y+1)2xe−x(y+1), x ≥ 0, y ≥ 0, 0 otherwise.

fY|X(y|x) = xe−xy, x ≥ 0, y ≥ 0, 0 otherwise.

Answer 33.5a. fXY(x, y)= [y2e−x/2]

Answer 33.5b. fX|Y(x|y)= e−(x-y)

Answer 33.6a. fX|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. fX|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


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Answer 33.12. about 0.1222526851

Answer 33.13. fX(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. fZW(z, w)=│1/(ad-bc)│fXY((zd-wb)/(ad-cb), (zc-wa)/(bc-da))

Answer 34.2. fy1y2(y1, y2)= λ2e-λy1/y2, y1 ≥ ln(2), y2 > 1, 0 otherwise.

Answer 34.3. fRФ(r, φ)= (r)fXY[rcos(φ), rsin(φ)], r > 0, −π < Ф ≤ π.

Answer 34.4. fZW(z,w)= [z/(1+ w2)]fXY[z*cos(arctan(w)), z*sin(arctan(w))]

Answer 34.5. fUV(u, v)= λα+βe-λuuαvα-1(u-uv)β-1/[Γ(α)Γ(β)], for u ≥ 0, v ≥ 0, 0 otherwise.

Answer 34.6. fy1y2(y1, y2)= y1e− y1, y1 ≥ 0, 0 < y2

Answer 34.7. fy1y2(y1, y2)= (1/28π)exp{-(3y1+ y2)2/98}exp{-(y1-2y2)2/392}

Answer 34.8. Hint: Show that fU(u) = [exp{-u2/2}/√(2π)], the pdf of a standard normal random variable, and fV(v) = [exp{-v2/2}/√(2π)], the pdf of a standard normal random variable.

Answer 34.9. fU(u)= -∞∞∫fXY(u-v, v)dv= -∞

∞∫fXY(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,

fU(u)= -∞∞∫fX(u-v)fV(v)dv= -∞

∞∫fX(u-y)fY(y)dy

Also, by convolution, we get

fU(u)= -∞∞∫fX(u-y)fY(y)dy

Answer 34.10. fU(u)= -∞∞∫fXY(v-u, v)dv= -∞

∞∫fXY(y-u, y)dy

Answer 34.11. fU(u)= -∞∞∫│1/v│*fXY(v, u/v)dv= -∞

∞∫│1/x│*fXY(x, u/x)dx

Answer 34.12. fU(u)= 1/(u+1)2 for u > 0 and 0 otherwise.

Section 35

Answer 35.1. E(|X − Y |)= (m+1)(m-1)/(3m)


48

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, fXY(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


49

Answer 36.6b. ρ(X1 + X2, X3 + X4)= 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. fXY(x, y)= ½ for 0 < x < 1, 0 < y < 2, 0 otherwise.

Answer 36.19b. fZ(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. fZ(z)= z/2 for z є (0, 2), 0 otherwise.

Answer 36.20b. fX(x)= 1-x/2 for 0 < x < 2, 0 otherwise. fY(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


50

Answer 36.21. Cov(X, Y)= -0.15

Answer 36.22. Var(X)= 1/6

Answer 36.23. Var(X1 + 2X2 − X3) = 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------pX(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 pY(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


51

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. MX(t)= 1 when t=0 and ∞ otherwise.

Answer 38.7. MY(t)= e-2t[λ/(λ-3t)], 3t < λ

Answer 38.8. X is a binomial random variable such that

pX(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

pY(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(t1, t2)= exp{(t12+ t2

2)}


52

Answer 38.11. SD(X)= 5000

Answer 38.12. E(X3)= 10,560

Answer 38.13. MY(t)= (8/27)et + (19/27)

Answer 38.14. 0.8384

Answer 38.15a. MXi(t)= pet/[1- et(1 − p)]

Answer 38.15b. {pet/[1-(1-p)et]}n

Answer 38.16. P(M1 > V2)= 0.6915

Answer 38.17. Var(X)= 2

Answer 38.18. MX(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(t1, t2)= [(et1x-1)(et2x-1)]/(t1t2)

Answer 38.24. Var(X)=2/9

Answer 38.25. MX+2Y(t)= exp{13t2+4t}

Answer 38.26. E(2X1 − X2)= 0.4

Answer 38.27. Cov(X, Y)= -0.9375

Answer 38.28. MV(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


53

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(X1 + X2 + · · · + X20 > 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. FY_n(x) = 0 for x ≤ 0;

FY_n(x) = 1 - (1 - x)n for 0 < x < 1;

FY_n(x) = 1 for x ≥ 1.

Section 40





Answer 40.5. about 0.1357 or about 0.1367, depending on the rounding used.





54

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


55

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. 01∫x

x+1∫f(x, y)dydx

Answer 43.2. 01∫0

x/2∫f(x, y)dydx or 01/2∫2y

1∫f(x, y)dxdy

Answer 43.3. 1/21∫1/2

1∫f(x, y)dydx

Answer 43.4. 01/2∫1/2

1∫f(x, y)dydx + 1/21∫0

1∫f(x, y)dydx

Answer 43.5. 2030∫20

50-x∫f(x, y)dydx

Answer 43.6. 01∫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. 0.51∫0.5

1∫f(x, y)dydx

Answer 43.11. 7/8

Answer 43.12. 3/8

answer keys for the 2008 edition of marcel b. finan’s “a...

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