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Solution Manual For First Course in Probability 10th Edition by Sheldon Ross
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Solution Manual For First Course in Probability 10th Edition by Sheldon Ross
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INSTRUCTOR ’S SOLUTIONS MANUAL A FIRST COURSE IN PROBABILITY TENTH EDITION Sheldon Ross University of Southern California The author and publisher of this book have used their best effo rts in preparing this book. These efforts include the development, research, and testing of the theories and prog rams to determine their effectiveness. The author and publisher make no warranty of any kind, expressed or implie d , w i t h r e g a r d t o t h e s e p r o g r a m s o r t h e documentation contained in this book. The author and publisher shall not be liable in any event for incidental or consequential damages in connection with, or arising out of, th e furnishing, performance, or use of these programs. Reproduced by Pearson from electronic files supplied by the aut hor. Copyright © 2019, 2014, 2010 Pearson Education, Inc. Publishing as Pearson. All rights reserved. No part of this publication may be reprodu ced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopyi ng, recording, or otherw ise, without the prior written permission of the publisher. Printed in the United Stat es of America. 1 18 ISBN-13: 978-0-13-475371-3 ISBN-10: 0-13-475371-2 1 Copyright © 2018 Pear son Education, Inc. Chapter 1 Problems 1. (a) By the generalized basic pr inciple of counting there are 26 ⋅ 26 ⋅ 10 ⋅ 10 ⋅ 10 ⋅ 10 ⋅ 10 = 67,600,000 (b) 26 ⋅ 25 ⋅ 10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6 = 19,656,000 2. 64 = 1296 3. An assignment is a sequence i1, …, i20 where ij is the job to which person j is assigned. Since only one person can be assigned to a job, it follows that the s equence is a permutation of the numbers 1, …, 20 and so there a re 20! different possible assign ments. 4. There are 4! possible arra ngements. By assigning instrument s to Jay, Jack, John and Jim, in that order, we see by the genera lized basic principle that ther e are 2 ⋅ 1 ⋅ 2 ⋅ 1 = 4 possibilities. 5. There were 8 ⋅ 2 ⋅ 9 = 144 possible codes. There were 1 ⋅ 2 ⋅ 9 = 18 that started with a 4. 6. Each kitten can be identified by a code number i, j, k, l wh ere each of i, j, k, l is any of the numbers from 1 to 7. The number i represents which wife is carrying the kitten, j then represents which of that wife’s 7 sacks contain the kitten; k represents which of the 7 cats in sack j of wife i is the mother of the kitten; and l represents the number of the kitten of cat k in sack j of wife i. By the generalized princ iple there are thus 7 ⋅ 7 ⋅ 7 ⋅ 7 = 2401 kittens 7. (a) 6! = 720 ( b ) 2 ⋅ 3! ⋅ 3! = 72 (c) 4!3! = 144 ( d ) 6 ⋅ 3 ⋅ 2 ⋅ 2 ⋅ 1 ⋅ 1 = 72 8. (a) 5! = 120 ( b ) 7!
2!2! = 1260 ( c ) 11!
4!4!2! = 34,650 ( d ) 7!
2!2! = 1260 9. (12)!
6!4! = 27,720 2 Chapter 1 Copyright © 2018 Pears on Education, Inc. 10. (a) 8! = 40,320 ( b ) 2 ⋅ 7! = 10,080 (c) 5!4! = 2,880 ( d ) 4 ! 24 = 384 11. (a) 6! (b) 3!2!3! ( c ) 3 ! 4 ! 12. 10
3 − 10 ⋅ 9 ⋅ 8 = 280 numbers have at least 2 equal values. 280 − 10 = 270 have exactly 2 equal values. 13. With ni equal to the number of length i, n1 = 3, n2 = 8, n3 = 12, n4 = 30, n5 = 30, giving the answer of 83. 14. (a) 30
5 (b) 30 ⋅ 29 ⋅ 28 ⋅ 27 ⋅ 26 15. 20
2
16. 52
5
15. There are 10 12
55
possible choices of the 5 men and 5 women. They can then be p aired up in 5! ways, since if we arbitrarily order the men then the firs t man can be paired with any of the 5 women, the next with any of the remaining 4, and so on. Hence, there are 10 125!55
possible results. 18. (a) 674
222++ = 42 possibilities. (b) There are 6 ⋅ 7 choices of a math and a science book, 6 ⋅ 4 choices of a math and an economics book, and 7 ⋅ 4 choices of a science and an economics book. Hence, there ar e 94 possible choices. 19. The first gift can go to any of the 10 children, the second to any of the remaining 9 children, and so on. Hence, there are 10 ⋅ 9 ⋅ 8 ⋅ ⋅ ⋅ 5 ⋅ 4 = 604,800 possibilities.
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