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Probability/Combinatorial Question

  1. Feb 16, 2015 #1
    1. The problem statement, all variables and given/known data

    Now, this problem is not the typical stars and bars problem. I'm actually trying to solve for this:
    If I had n red marbles and m blue marble, I want the probability that I pick n red marbles and only 1 blue marble. The order is irrelevant, meaning, the only thing that is required is that after I have picked n+1 marbles, I had exactly n red marbles and 1 blue marble. How do I do this?


    2. Relevant equations

    (Number of possibilities for picking 1 marble and n red marbles)/(Total number of possiblities) = the probability

    3. The attempt at a solution

    So, I thought of it like this.

    I have n+1 open little crates (distinct boxes), and n indistinct objects (my red marbles). I am looking for the amount of ways to organize my n red marbles into n+1 crates, with the restriction that none of the crates can have more than one marble in it. By doing this, I can find the ways that I can pick exactly n marbles and leave one empty crate. This is basically the same thing as finding the ways that I can pick n marbles and 1 blue marble, although I am not really taking into account the amount of ways I can choose from the m blue marbles.

    I'm actually looking for a general way to find the amount of ways that I could organize n red marbles and k blue marbles in n+k slots, where each slot has exactly one item. Is there already a general formula for this? Thanks!
     
  2. jcsd
  3. Feb 16, 2015 #2

    Ray Vickson

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    Homework Helper

    Yes, there is a formula for P(exactly k blue marbles chosen); it is given by the so-called hypergeometric distribution; see, eg.,
    http://en.wikipedia.org/wiki/Hypergeometric_distribution
    http://stattrek.com/probability-distributions/hypergeometric.aspx
    In your case you have N = n+m objects, m of type I and n of type II. You draw n+1 items without replacement, and you want the probability of getting exactly k = 1 type I in your sample.

    Basically, you can have RRR...RB or RR...RBR or.... or BRR...RR, each of which (you can show) has the same probability. How many of these are there? What is the probability of any one of them, say P(RRR...RB)? Here is a hint to get you started: the first R has probability n/(n+m); that leaves n+m-1 balls, of which n-1 are red and m are blue. So, the probability of the second R is (n-1)/(n+m-1). Therefore, the probability that the first two are red is n*(n-1)/[(n+m)*(n+m-1)]. Keep going like that. Finally, what about that final B? Before picking the B there remain n+m-n = m balls, of which all are blue, so the probability of that final B is m/m = 1. To focus your understanding, look an another outcome, such as BRRR....R, and compute a formula for its probability; you will see that it gives the same final result as for RR...RB.
     
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