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發問:
圖片參考：http://imgcld.yimg.com/8/n/HA00167623/o/701202170143113873438050.jpg ANS ai)1/78 aII) 5/26 b) no c) no explain it with step plz
最佳解答:
91. (a) (i) No. of permutations when the 6 girls and 7 boys queue up randomly = P(13,13) = 13! No. of permutation when Yen and Mandy are at both end = P(2, 2) x P(11,11) = 2! x 11! P(Yen and Mandy are at both ends) = (2! x 11!)/13! = 2/(13 x 12) = 1/78 (a) (ii) Number of permutations when girls are at both ends = P(6, 2) x P(11, 11) = (6!/4!) x 11! = (6 x 5) x 11! = 30 x 11! P(Girls are at both ends) = (30 x 11!)/13! = 30/(13 x 12) = 5/(13 x 2) = 5/26 (b) No Girl is denoted as G, and boy is denoted by B. The number of B is one more than that of G. When any boys are not next to eachother, the only case is that : BGBGBGBGBGBGB However, the number of G is one less than that of B. When any girls are notnext to each other, there are a few cases : BGBGBGBGBGBGB, GBBGBGBGBGBGB, GBGBBGBGBGBGB ...... Obviously, P(Any girls are not next to each other) > P(Any boys are not nextto each other) (c) No "Boys are not next to each other" : BGBGBGBGBGBGB P(Boys are not next to each other) = P(7,7) x P(6,6)/13! = 7!6!/13! ≠ 1/2 P(Boys are next to each other) = 1 - P(Boys are not next to each other) ≠ 1/2 Hence, P(Boys are not next to each other) ≠ P(Boys are next to each other)
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probability發問:
圖片參考：http://imgcld.yimg.com/8/n/HA00167623/o/701202170143113873438050.jpg ANS ai)1/78 aII) 5/26 b) no c) no explain it with step plz
最佳解答:
91. (a) (i) No. of permutations when the 6 girls and 7 boys queue up randomly = P(13,13) = 13! No. of permutation when Yen and Mandy are at both end = P(2, 2) x P(11,11) = 2! x 11! P(Yen and Mandy are at both ends) = (2! x 11!)/13! = 2/(13 x 12) = 1/78 (a) (ii) Number of permutations when girls are at both ends = P(6, 2) x P(11, 11) = (6!/4!) x 11! = (6 x 5) x 11! = 30 x 11! P(Girls are at both ends) = (30 x 11!)/13! = 30/(13 x 12) = 5/(13 x 2) = 5/26 (b) No Girl is denoted as G, and boy is denoted by B. The number of B is one more than that of G. When any boys are not next to eachother, the only case is that : BGBGBGBGBGBGB However, the number of G is one less than that of B. When any girls are notnext to each other, there are a few cases : BGBGBGBGBGBGB, GBBGBGBGBGBGB, GBGBBGBGBGBGB ...... Obviously, P(Any girls are not next to each other) > P(Any boys are not nextto each other) (c) No "Boys are not next to each other" : BGBGBGBGBGBGB P(Boys are not next to each other) = P(7,7) x P(6,6)/13! = 7!6!/13! ≠ 1/2 P(Boys are next to each other) = 1 - P(Boys are not next to each other) ≠ 1/2 Hence, P(Boys are not next to each other) ≠ P(Boys are next to each other)
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