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M and M probability

  1. Feb 1, 2013 #1
    1. The problem statement, all variables and given/known data
    Lets say you have a mini bag of 20 M&Ms and find the first 19 are all blue. What is the probability that the 20th is also blue?

    According to the company the distribution is 24% blue, 14% brown, 16% green, 20% orange, 13% red, 14% yellow

    Thanks!



    2. Relevant equations
    I don't think this is relevant



    3. The attempt at a solution
    Would it just be the initial probability of getting a blue M&M (.24 according to the company), or does the fact that ALL of them in the bag are blue have an effect on the probability? I am assuming that the picking of each M&M are independent events, thus making the probability the same... but I am not sure if that's correct
     
  2. jcsd
  3. Feb 1, 2013 #2

    Dick

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    That would be correct.
     
  4. Feb 1, 2013 #3
    Thanks Dick. What's the reasoning behind it?
     
  5. Feb 1, 2013 #4

    Dick

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    They gave you the distribution of colors. And there's no reason not to assume the events are independent. The colors you've already seen don't affect the probability of the color of the next one. Of course, after pulling out 20 blues you might suspect the color distribution you've been given is wrong. But that really has nothing to do with this problem.
     
  6. Feb 1, 2013 #5

    Ray Vickson

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    You need to be careful. The point is that the contents of the bag were fixed at the M&M factory, so the question is about the number of blues that were put into the bag at the plant. The probability 0.24 governs what was put into the bag, not necessarily what will be taken out be a customer. In particular, you need to worry about whether the successive colors drawn out of the bag by a customer are truly independent.
     
  7. Feb 1, 2013 #6
    That's what I thought but I was involved in an argument here. Thanks.
    I am confused about the question below. It is a rider to the one above.I believe it is a combination since order is not spoken of, but I am not sure.

    If you are given 10 Brown M&Ms, 5 Yellow,16 Green, 7 Red, 11 orange and 6 blue, What would be FORMULA for the probability of obtaining ANY given combination of colors? Please define all variables used.
    Sorry for being such an Oliver Twist.
     
  8. Feb 1, 2013 #7

    Ray Vickson

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    I think the question wants P{X = 20|X >= 19} in a binomial distribution X~Bin(20,0.24). That conditional probability is definitely not equal to 0.24. Further, P{X = k+1|X >=k} depends on k, so the results are not independent!
     
    Last edited: Feb 1, 2013
  9. Feb 1, 2013 #8

    haruspex

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    Ray, I don't think that's relevant. That's where you're looking at the total number of blues in the bag, not the first so many drawn at random. The logic in the OP looks right to me. Pulling the Mth M&M from the Bth bag of N M&Ms is equivalent to pulling the (B-1)*N+Mth from the production line.
     
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