hw2 solution - 29. a. b. c. d. (26)(26) = 676; (36)(36) =...

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29. a. (26)(26) = 676; (36)(36) = 1296 b. (26) 3 = 17,576; (36) 3 = 46,656 c. (26) 3 = 456,976; (36) 4 = 1,679,616 d. 1 – 97,786/(36) 4 = .942 30. a. Because order is important, we’ll use P 8,3 = 8(7)(6) = 336. b. Order doesn’t matter here, so we use C 30,6 = 593,775. c. From each group we choose 2: 160 , 83 2 12 2 10 2 8 d. The numerator comes from part c and the denominator from part b: 14 . 775 , 593 160 , 83 e. We use the same denominator as in part d. We can have all zinfandel, all merlot, or all cabernet, so P(all same) = P(all z) + P(all m) + P(all c) = 002 . 775 , 593 1162 6 30 6 12 6 10 6 8 35. a. . 760 , 38 6 20 P(all from day shift) = 0048 . 060 , 145 , 8 760 , 38 6 45 0 25 6 20 b. P(all from same shift) =
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This note was uploaded on 04/23/2011 for the course STAT 503 taught by Professor Staff during the Spring '08 term at Purdue University-West Lafayette.

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hw2 solution - 29. a. b. c. d. (26)(26) = 676; (36)(36) =...

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