標題:
[數學] 唔識2條大約中5-6既數!!(20點)
發問:
four English, two Chemistry and four Physics books are to be arranged on a shelf. In how many ways can this be done if: a) there are no restrictions. b) the chemistry books must remain together. c) the books must stay together by subject. 教下我!thx! 更新: 另一條: the probability that an aniaml will still be alive in 12years is 0.55 and the probability that its mate will be alive in 12 years is 0.6 Find the probability: 更新 2: a)both will still be alive in 12 years. b)only the mate will still be alive in 12 years c) at least one of them still be alive in 12 years.
最佳解答:
1.Assume that books in each subjects are different, i.e. there are ten different books. a) The number of ways the books to be arranged = 10! = 10x9x8x...x3x2x1 = 3628800 Note: there are 10 choices for the 1st book, then 9 (i.e. 10-1) choices for the 2nd, 8 (i.e. 10-2) choices for the 3nd, ... b) The number of ways the books to be arranged with the chemistry books remaining together = 9! * 2! = 725760 Note: there are 9 groups of books to be arranged - 4 English, 4 Physics and 1 for the two Chemistry. However, the 2 chemistry books can be arranged in [C1][C2] or [C2][C1], therefore, the number of permutations for the Chemistry books are 2!. c) The number of arrangements that books with same subjects staying together = 3! * 4! * 2! * 4! = 6912 Similar to b), there are 3 groups (subjects) of books to be arranged - English, Chemistry and Physics. The number of permutations for each groups depends on the number of books in each subject, i.e. 4!, 2! and 4! respectively. 2. a) P(both alive in 12 years) = P(the animal alive 12 years) * P(the mate alive in 12 years) = 0.55 * 0.6 = 0.33 b) P(only the mate alive in 12 years) = P(the animal dies 12 years) * P(the mate alive in 12 years) = (1 - 0.55) * 0.6 = 0.27 c) P(at least one alive in 12 years) = 1 - P(both die in 12 years) = 1 - (1 - 0.55)(1 - 0.6) = 1 - 0.45 * 0.4 = 1 - 0.18 = 0.82 OR P(at least one alive in 12 years) = P(both alive in 12 years) + P(only the mate alive in 12 years) + P(only the animal alive in 12 years) = 0.33 + 0.27 + 0.55 * (1 - 0.6) ......... by a) & b) = 0.82
1(a) No. of ways=10!/4!4!2!=3150 1(b) No. of ways=9!/4!4!=630 1(c) No. of ways=3!=6 2(a) P(both in 12 years)=0.55x0.6=0.33 2(b) P(only the mate alive in 12 years)=0.6x(1-0.55)=0.27 2(c) P(at least one of them alive in 12 years)=0.33+0.27+0.45x(1-0.6)=0.78
[數學] 唔識2條大約中5-6既數!!(20點)
發問:
four English, two Chemistry and four Physics books are to be arranged on a shelf. In how many ways can this be done if: a) there are no restrictions. b) the chemistry books must remain together. c) the books must stay together by subject. 教下我!thx! 更新: 另一條: the probability that an aniaml will still be alive in 12years is 0.55 and the probability that its mate will be alive in 12 years is 0.6 Find the probability: 更新 2: a)both will still be alive in 12 years. b)only the mate will still be alive in 12 years c) at least one of them still be alive in 12 years.
最佳解答:
1.Assume that books in each subjects are different, i.e. there are ten different books. a) The number of ways the books to be arranged = 10! = 10x9x8x...x3x2x1 = 3628800 Note: there are 10 choices for the 1st book, then 9 (i.e. 10-1) choices for the 2nd, 8 (i.e. 10-2) choices for the 3nd, ... b) The number of ways the books to be arranged with the chemistry books remaining together = 9! * 2! = 725760 Note: there are 9 groups of books to be arranged - 4 English, 4 Physics and 1 for the two Chemistry. However, the 2 chemistry books can be arranged in [C1][C2] or [C2][C1], therefore, the number of permutations for the Chemistry books are 2!. c) The number of arrangements that books with same subjects staying together = 3! * 4! * 2! * 4! = 6912 Similar to b), there are 3 groups (subjects) of books to be arranged - English, Chemistry and Physics. The number of permutations for each groups depends on the number of books in each subject, i.e. 4!, 2! and 4! respectively. 2. a) P(both alive in 12 years) = P(the animal alive 12 years) * P(the mate alive in 12 years) = 0.55 * 0.6 = 0.33 b) P(only the mate alive in 12 years) = P(the animal dies 12 years) * P(the mate alive in 12 years) = (1 - 0.55) * 0.6 = 0.27 c) P(at least one alive in 12 years) = 1 - P(both die in 12 years) = 1 - (1 - 0.55)(1 - 0.6) = 1 - 0.45 * 0.4 = 1 - 0.18 = 0.82 OR P(at least one alive in 12 years) = P(both alive in 12 years) + P(only the mate alive in 12 years) + P(only the animal alive in 12 years) = 0.33 + 0.27 + 0.55 * (1 - 0.6) ......... by a) & b) = 0.82
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其他解答:1(a) No. of ways=10!/4!4!2!=3150 1(b) No. of ways=9!/4!4!=630 1(c) No. of ways=3!=6 2(a) P(both in 12 years)=0.55x0.6=0.33 2(b) P(only the mate alive in 12 years)=0.6x(1-0.55)=0.27 2(c) P(at least one of them alive in 12 years)=0.33+0.27+0.45x(1-0.6)=0.78
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