First Course In Probability Compress 7th Edition By Sheldon Ross Test Bank - Questions & Answers (Graded A+) | Latest 2023

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 (b) (c) 7! 2!2! 11! = 1260 = 34,650 4!4!2! (d) 7! 2!2! = 1260 9. (12)! = 27,720 6!4! 10. (a) 8! = 40,320 (b) 2  7! = 10,080 (c) 5!4! = 2,880 (d) 4!24 = 384