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P&c: whose choice matters?

  1. Apr 21, 2015 #1
    1. The problem statement, all variables and given/known data
    The question says that :
    Twelve distinct balls are distributes among three distinct boxes. Find the probability that the first box contains 3 balls.

    2. Relevant equations
    None

    3. The attempt at a solution
    It can be done through:
    No. Of choices for each ball=3
    Total no. of choices=312
    We have to select 3 out of twelve and arrange the remaining nine balls in 29 ways so that the answer would be :
    110×210/312.
    But I was thinking that why a choice is given to the balls. Why not to boxes? Each box then will have 12 choices and total choices would be 123. Why this thinking is incorrect here?
     
  2. jcsd
  3. Apr 21, 2015 #2
    The first way is you are going to assign each ball to one of the 3 boxes until all the balls are used up. No balls are left over.

    The second way is you are attempting to assign each box to one of 12 balls until all the boxes are used up. Since the number of boxes is less than the number of balls then there will be 9 balls left over without any boxes assigned. You also calculated it wrong since you are assuming the balls would be re-used. The correct calculation should be 12⋅11⋅10 = 13,200.
     
  4. Apr 22, 2015 #3

    haruspex

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    But those choices are not independent. Two boxes cannot choose the same ball, but two balls can choose the same box.
    mooncrater's answer looks ok to me. 12C3(1/3)3(2/3)9.
    One thing bothers me... why are we told the balls are distinct?
    That's a bit vague. I'd rather it said "each ball is assigned independently and with equal probability to any of three boxes".
     
  5. Apr 22, 2015 #4
    Ok got it now. Thanks.
     
  6. Apr 22, 2015 #5

    Ray Vickson

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    When assigning balls to boxes, the box choices are independent from ball-to-ball, so you have a binomial distribution with n = 12 and the two categories ('box 1' or 'not box 1'). The probability that box 1 contains ##k## balls is
    $$P(k) = {12 \choose k} \, (1/3)^k \, (2/3)^{12-k}, \; k = 0,1,2, \ldots, 12.$$
    Substitute ##k = 3##
     
  7. Apr 22, 2015 #6
     
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