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數學 概率問題

Permutation, Combination and Probability

1.) 7 boys and 5 girls are randomly divided into four groups of equal size. Find the probability that

(a) one of the groups consists of all girls;

(b) there is at least one girl in each group.

2.) 9 seats are arranged in a row for 8 students from 3 different classes. There are 2 students from 5A, 3 from 5B and 3 from 5C.

(a)(i)Find the number of ways in which these 8 students can take the seats.

(ii)Find the number of ways in which these 8 students can take the seats if the 2 students from 5A must sit next to each other. (Two students sitting next to each other means there’s no occupied/unoccupied seat between them.)

(b)The 9 seats are now rearranged into 3 separate rows of 3 seats each. Find the number of ways in which these 8 students can take the seats, if the 2 students from 5A must sit next to each other.

概率一直是我的弱項,可否詳細解釋?感激不盡!

1 個解答

評分
  • 5 年前
    最愛解答

    1a

    there are 3 people in a group.

    Probability of having one of the groups consists of all girls

    =[5C3 (way to choose a group consists of all girls) *9C3*6C3*3C3/3! (there are repeated grouping for the remaining 3 groups)] / [12C3*9C3*6C3*3C3/4! (there are repeated grouping for all groups)]

    =2800/15400

    =2/11

    # for 12 numbers, 1-12, to form groups of 3,

    1 2 3, 4 5 6, 7 8 9, 10 11 12

    4 5 6, 7 8 9, 10 11 12, 1 2 3

    7 8 9, 10 11 12, 1 2 3, 4 5 6

    10 11 12, 1 2 3, 4 5 6, 7 8 9 are the same, but they are counted as 4 different combinations using (12C3*9C3*6C3*3C3), so we divide it by 4!, similar to that when dividing 3!, but 5C3 is specific for group containing all girls, so only divide by 3!

    i.e.

    if 1-5 are girls, 6-12 are boys,

    123, 456, 789, 101112

    123, 456,101112, 789

    123, 789, 456, 101112

    123, 789, 101112, 456

    123, 101112, 456, 789

    123, 101112, 789, 456

    when 123 is the group of all girls, there are 6(3!) repeated counting for the remaining 3 groups. the female group is specific and doesnt need to be take into account.~

    tiring~

    2015-05-16 19:42:46 補充:

    1b

    probability

    =5P3*(8C2*6C2*4C2*2C2/4!)/15400

    =9/22

    2015-05-16 19:43:13 補充:

    1b

    probability

    =5P3*(8C2*6C2*4C2*2C2/4!)/15400

    =9/22

    2015-05-16 19:47:15 補充:

    2ai

    the number of ways

    =9P8

    =362880

    2015-05-16 21:40:52 補充:

    2aii

    take the two students as a unit, so we can assume that there are only 7 students, with 8 seats

    probability

    =8P7*2!

    =80640

    2! comes from the possible positions of the two students as a unit, i.e. AB or BA

    2015-05-16 21:41:12 補充:

    2b

    there are 6 possible seats for 5A students: 1&2 or 2&3 or 4&5 or 5&6 or 7&8 or 8&9

    so probability

    =7P6*2!*6

    =60480

    If I am wrong, feel free to inform me:)

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