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Probability combination Question

  1. Sep 17, 2007 #1
    1. The problem statement, all variables and given/known data

    We want to put 4 Maroon, 3 Green, and 2 Yellow plates in a row on a table. They are arranged randomly. What is the probability that the 4 Maroon plates are together?

    2. Relevant equations

    3. The attempt at a solution

    I couldn't figure out how to set this problem up properly.

    All I could think to do was to break it into the six outcomes we are interested in.

    Let X denote a non-maroon plate and M denote a maroon plate.

    So the outcomes are:
    1- M M M M X X X X X
    2- X M M M M X X X X
    3- X X M M M M X X X
    4- X X X M M M M X X
    5- X X X X M M M M X
    6- X X X X X M M M M

    P(1) = (4/9)(3/7)(2/6)(1/5) = 1/126
    P(2) = (5/9)(4/8)(3/7)(2/6)(1/5) = 1/126
    P(3) = (5/9)(4/8)(4/7)(3/6)(2/5)(1/5) = 1/126
    P(4) = ....... = 1/126
    P(5) = ....... = 1/126
    P(6) = ....... = 1/126

    So for a final answer I get 1/21. I'm not very confident in that answer, plus there has to be some way to set this up much simpler. Using the (n choose k) deal..

    Any advice would be awesome.
    Last edited: Sep 17, 2007
  2. jcsd
  3. Sep 17, 2007 #2


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    Homework Helper

    Once you've picked a place to put the first maroon plate, all six patterns are equally likely (as you've probably noticed) (4/9)*(3/8)*(2/7)*(1/6)=1/126 (chance first maroon lies on selected place, time chance second maroon lies on second selected place etc). And you also correctly decided there are 6 equally probable ways to place them. I think you've done quite well. But you didn't need to write down a different product for each combination. Just IGNORE the non-maroon plates. They will sort themselves out.
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