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Probability homework problem

  1. Apr 22, 2012 #1
    1. The problem statement, all variables and given/known data
    Bobo, Coco, and 8 other students are arranged to sit in 2 rows of 5 students. If these 10 students take their seat randomly, find the probability that Bobo and Coco are sitting in the same row but not next to each other.



    3. The attempt at a solution
    Bobo and Coco together have 12 ways to sit, they 2 can exchange, other 8 students sit randomly.
    Probability = [itex]\frac{12×2×8!}{10!}[/itex] = [itex]\frac{4}{15}[/itex]

    Is this correct? Thank you!!
     
  2. jcsd
  3. Apr 22, 2012 #2

    HallsofIvy

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    Re: Probability

    Correct answer to what question? Where did you include the "sitting in the same row"? Or "not next to each other?
     
  4. Apr 23, 2012 #3
    Re: Probability

    I mean, say the seats are arranged like this:
    1 2 3 4 5
    6 7 8 9 10

    Bobo and Coco sitting in the same row but not next to each other, they can take:
    1,3
    2,4
    3,5
    1,4
    2,5
    1,5
    6,8
    7,9
    8,10
    6,9
    7,10
    6,10
    totally 12 ways.

    The above arrangement can be for Bobo|Coco or Coco|Bobo, so times 2.

    Is my answer correct? Thanks again.
     
  5. Apr 23, 2012 #4

    HallsofIvy

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    Re: Probability

    Yes, that is correct- here is how I would have done it- a different way to get the same answer:
    Coco can sit anywhere in a given row. Then Bobo must sit in the same row but not next to Coco.
    There is a slight complication here- If Coco is sitting in an end seat there are 3 seats where Bobo can sit. But if Coco is sitting in any other seat, there are only 2 seats where Coco can sit. That is the number possible ways for both Bobo and Coco to sit are 2*3+ 3*2= 12. Since there are 2 rows there are 2*12= 24 ways Bobo and Coco can sit in the two rows.
     
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