A FIRST COURSE IN PROBABILITY SOLUTION MANUAL SHELDON ROSS

A FIRST COURSE IN PROBABILITY SOLUTION MAN
UAL SHELDON ROSS

, Table of Contents



Chapter 1 .............................................................................. 1

Chapter 2 .............................................................................. 10

Chapter 3 .............................................................................. 20

Chapter 4 .............................................................................. 46

Chapter 5 .............................................................................. 64

Chapter 6 .............................................................................. 77

Chapter 7 .............................................................................. 98

Chapter 8 .............................................................................. 133

Chapter 9 .............................................................................. 139

Chapter 10 ............................................................................ 141

, Chapter 1

Problems
1. (a) By the generalized basic principle 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 sequence is a permutation of the
numbers 1, …, 20 and so there are 20! different possible assignments.

4. There are 4! possible arrangements. By assigning instruments to Jay, Jack, John and Jim, in
that order, we see by the generalized basic principle that there 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 where 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 principle 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
7!
(b) = 1260
2!2!
(c) 11! = 34,650
4!4!2!
7!
(d) = 1260
2!2!

(12)!
9. = 27,720
6!4!

10. (a) 8! = 40,320
(b) 2  7! = 10,080
(c) 5!4! = 2,880
(d) 4!24 = 384


Chapter 1 1

, 11. (a) 6!
(b) 3!2!3!
(c) 3!4!

12. (a) 305
(b) 30  29  28  27  26

 20 
13.  
2
 

52 
14.  
5
 

1012
15. There are  5  5 possible choices of the 5 men and 5 women. They can then be paired up
  
in 5! ways, since if we arbitrarily order the men then the first man can be paired with any of
1012
the 5 women, the next with any of the remaining 4, and so on. Hence, there are 5!  
5 5
  
possible results.
 6  7   4
16. (a)   = 42 possibilities.
     
2 2 2
     
(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 are
94 possible choices.

17. 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.

18.  5  6 4
    = 600
 2 2 3 
 8 4 8 2 4
19. (a) There are  = 896 possible committees.
      
38
 4    
3 3 1 2
 8  2  4
There are that do not contain either of the 2 men, and there are that
      
3 3  3  1  2
contain exactly 1 of them.
 6 6  2 6 6
(b) There are  = 1000 possible committees.
      
 3  3  1  2  3 




2 Chapter 7