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Probability, Bag of Marbles - quantity grabbed > quantity needed

  1. Sep 16, 2012 #1
    1. The problem statement, all variables and given/known data
    Bob has a marble sack with 20 marbles. 6 - purple, 8 - blue, 4 - white, 2 green.
    If Bob randomly grabs 5 marbles what is the probability that at least 2 are purple and at least 2 are blue?


    2. Relevant equations
    ?


    3. The attempt at a solution
    I guess the part that is throwing me is taking one more than required. So lets say Bob only grabs 4 marbles. Would the probability be (6/20)*(5/19)*(8/18)*(7/17) = 14/969? If this is incorrect for picking 4 please explain as well. I would also appreciate it if you used the probability notation like p(b) as I need practice, but it is not necessary.This isn't for any math class, but a different subject. I've modified the problem, but the principle is the same. I don't have much background in probability (it has been 3 years since I've done some). What are some good resources for learning/relearning probability? I've been watching some videos on khanacademy.org.
     
  2. jcsd
  3. Sep 17, 2012 #2

    HallsofIvy

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    6/20 is the probability the first marble you pick is purple. 5/19 is the probability the second marble you pick is also purple. 8/18 is the probability the third marble you pick is blue. 7/17 is the probability the fourth marble you pick is blue. So what you have calcuated is the probability of getting two purple and two blue marbles in that order.

    You can extend that easily to 5 marbles by noting that if the fifth marble is NOT purple or blue it must be green or white and there are 6 such marbles in the 16 marbles that are left. The probability of getting two purple, two bue, and one other marble, in that order, is (6/20)(5/19)(8/18)(7/17)(6/16).

    If you were to do that same analysis for "one purple, two blue, one "other", one purple marble" in that order you will, of course, stll have 20, 19, 18, 17, and 16 in the denominator and the same numbers in the numerator, just rearranged. So the probability of "two blue, two purple, one other" in any specific order is the same. To find the probability of that in any order, you just have to multiply by the number of possible orders.

    Now, how many different "orders" are there for "PPBBO"? Once you have multiplied by that you will have the probability of getting "two purple, two blue, and one other marble" in any order.

    But the problem says "at least 2 purple marbles" and "at least 2 blue marbles" so you also have to consider "three purple and two blue marbles" and "two purple and three blue marbles" in any order.

     
    Last edited: Sep 17, 2012
  4. Sep 17, 2012 #3

    Ray Vickson

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    Let the outcomes be labeled as (P,B,O), where P = number purple, B =number blue, O = number of "others" (neither blue nor purple). The event you want is E = {(3,2,0),(2,3,0),(2,2,1)}. Each outcome's probability is easy to calculate using a multi-class version of the hypergeometric distribution.

    RGV
     
    Last edited: Sep 17, 2012
  5. Sep 18, 2012 #4
    Thanks for the replies everyone. This is what I did, I'm not sure if the computation is correct.

    P(3,2,0) = 0.036119711
    P(2,3,0) = 0.0541795666
    P(2,2,1) = 0.1625386997
    So then do I add the events together to get 0.2528379773?



    Edit: I think I understand it conceptually now. I couldn't get past the three different events. So if all I have to do is add them together I'm set. Once again thanks a lot!
     
    Last edited: Sep 18, 2012
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