Complete Solution Manual A First Course in Probability 10th Edition Questions & Answers with rationales

A First Course in Probability 10th Edition Solution Manual Problems Chapter 1 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 thu s 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! (c) 11! = 1260 = 34,650 4!4!2! (d) 7! 2!2! = 1260 9. (12)! = 27,720 6!4! 1 2 5  5  5 5 5  2     2 2 Chapter 1 12. 103 − 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) 305 (b) 30  29  28  27  26 15. 16.  20       52      15. There are 10 12  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 the 5 women, the next with any of the remaining 4, and so on. Hence, there are possible results. 10 12  5!      18. (a)  6  +  7  +  4  = 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 are 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. 2 10. (a) 8! = 40,320 (b) (c) (d) 2  7! = 10,080 5!4! = 2,880 4!24 = 384 11. (a) 6! (b) 3!2!3! (c) 3!4!  2  2  3   3 3     3 1 2  3  3     1 2 3 3 3 3 1 2  3  3      2 3 3 2  3  3  2  3 3 2  5        1 4 5 3     Chapter 1 3 20.  5  6  4  = 600     21. (a) There are  8 4  +  8 2  4         = 896 possible committees. There are  8 4  that do not contain either of the 2 men, and there are  8 2  4  that       contain exactly 1 of them. (b) There are  6  6  +  2  6  6  = 1000 possible committees.                (c) There are  7  5 +  7  5 +  7  5           = 910 possible committees. There are  7  5 in    which neither feuding party serves;  7  5 in which the feuding women serves; and     7  5        in which the feuding man serves. 22.  6  +  2  6 ,  6  +  6           23. 7! 3!4! = 35. Each path is a linear arrangement of 4 r’s and 3 u’s (r for right and u for up). For instance the arrangement r, r, u, u, r, r, u specifies the path whose first 2 steps are to the right, next 2 steps are up, next 2 are to the right, and final step is up. 24. There are 4! 2!2! paths from A to the circled point; and 3! 2!1! paths from the circled point to B. Thus, by the basic principle, there are 18 different paths from A to B that go through the circled point. 25. 3!23 26. (a) n  n  2k = (2 + 1)n k =0  k  (b) n  n  xk = ( x + 1)n k =0  k           3   3   5  5 5 4 Chapter 1 28.  52  13, 13, 13, 13 30.  12  = 12!  3, 4, 5 3!4!5! 31. Assuming teachers are distinct. (a) 48 (b)  8  = 8! = 2520.  2, 2, 2, 2  (2)4 32. (a) (10)!/3!4!2! (b) 3 3  7!  2  4!2! 33. 2  9! − 228! since 2  9! is the number in which the French and English are next to each other and 228! the number in which the French and English are next to each other and the U.S. and Russian are next to each other. 34. (a) number of nonnegative integer solutions of x1 + x2 + x3 + x4 = 8. Hence, answer is 11   = 165 (b) here it is the number of positive solutions —hence answer is  7  = 35   35. (a) number of nonnegative solutions of x1 + … + x6 = 8 answer = 13   (b) (number of solutions of x1 + … + x6 = 5)  (number of solutions of x1 + … + x6 = 3) = 10  8       