Results 1 to 3 of 3
Like Tree2Thanks
  • 1 Post By mfb
  • 1 Post By Soroban

Math Help - Probability of arranging students into groups

  1. #1
    Senior Member
    Joined
    Apr 2009
    Posts
    306

    Probability of arranging students into groups




    I am not quite sure of how the book did this question. I did it another way using counting.

    Consider students 1, 2, 3, 4 are all in different groups, then there are 12!/(3!)^4 different ways that the other 12 students can be put into groups.

    There are 16!/((4!)^4*4!) to put 16 students into 4 groups of 4, the *4! accounting for the overcounting of the order of the groups.

    The final probability follows.

    However with the book's working, how is P(A_3) = P(A_1 n A_2 n A_3)?

    Isn't A_1 n A_2 = A_1?

    Since (student 1,2 in different groups) n (student 1, 2, 3 in different groups) = (student 1,2 in different groups) ?

    Also how did they get P(A_1) = 12/15?

    Like I just don't understand the intuition of how they arrived at the probability.

    Thanks.
    Last edited by usagi_killer; June 25th 2012 at 08:46 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    mfb
    mfb is offline
    Junior Member
    Joined
    Jun 2012
    From
    Germany
    Posts
    72
    Thanks
    5

    Re: Probability of arranging students into groups

    If A_1 and A_2, then (1&2 are in different groups) and (1&2&3 are in different groups), which is equivalent to the latter statement alone. Therefore, P(A_3) = P(A_1 n A_2 n A_3).

    Also how did they get P(A_1) = 12/15?
    Put student 1 in a random group. Now there are 15 open slots, 3 of them are in the same group, 12 in others. Therefore, student 2 has a 12/15 probability to join a different group: P(A_1)=12/15.

    In the same way, assume that student 1 and 2 are in different groups (A_1). Now there are 14 slots left, 8 in different groups. Student 3 has a 8/14 probability to join a new group: P(A_2|A_1)=8/14
    Assume that students 1&2&3 are in different groups. The probability that student 4 will join the 4th group is 4/13: P(A_3|A_2)=P(A_3|A_2 and A_1)=4/13

    Total probability: P=12/15*8/14*4/13=64/455.
    I get the same number with your approach.
    Thanks from usagi_killer
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,659
    Thanks
    598

    Re: Probability of arranging students into groups

    Hello, usagi_killer!

    A class consisting of 4 graduate and 12 undergraduate students
    is randomly divided into four groups of four students.
    What is the probability that each group includes a graduate student?

    Partition the 16 students into 4 ordered groups of 4 students each.

    There are: . {16\choose4,4,4,4} \:=\:\frac{16!}{4!\,4!\,4!\,4!} \:=\:63,\!063,\!000 ways.


    Now place a graduate student in each of the four groups:
    . . . . |\,g\,\_\,\_\,\_\,|\,g\,\_\,\_\,\_\,|\,g\,\_\,\_\,  \_\,|\,g\,\_\,\_\,\_\,|

    There are 4! ways to place the graduate students.

    There are {12\choose3,3,3,3} ways to place the 12 undergrads.

    Hence, there are: . 4!\!\cdot\!\frac{12!}{3!\,3!\,3!\,3!} \:=\: 8,\!870,\!400 desirable partitions.


    The proability is: . \frac{8,\!870,\!400}{63,\!063,\!000} \;=\;\frac{64}{455}
    Thanks from usagi_killer
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: June 26th 2011, 04:26 PM
  2. Probability of Students being left handed..
    Posted in the Statistics Forum
    Replies: 7
    Last Post: December 26th 2010, 06:04 PM
  3. N is divided into 6 groups...find the probability
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: May 7th 2010, 06:59 PM
  4. Probability of students passing med school
    Posted in the Statistics Forum
    Replies: 2
    Last Post: October 14th 2009, 12:38 PM
  5. Probability Question: 2 students and 1 professor
    Posted in the Statistics Forum
    Replies: 1
    Last Post: October 22nd 2007, 09:44 AM

Search Tags


/mathhelpforum @mathhelpforum