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Closed form expression for this?

  1. Apr 10, 2012 #1
    Say we have NA blue balls and NB red balls mixed together in an urn. When we pick a ball it leaves the urn. I want to find the probability of picking n red balls in a row.
    The probability of picking a red ball is:
    NA/(NA+NB)
    But each time a red ball leaves the urn the probability of picking the next will be:
    (NA-1)/(NA+NB-1)
    So in general for n successive red ball picks the probability must be:

    p(n) = ∏i=0n(NA-1)/(NA+NB-i)

    First of all I want to ask if this is correctly deduced. Secondly, I want to know if you can evaluate a product like the above in a closed form.
     
  2. jcsd
  3. Apr 10, 2012 #2

    Ray Vickson

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    I'll change to a nicer notation: we start with N balls altogether, R red and B blue, with R+B=N. If you draw n <= R balls, what is the probability all n are red? Note: this is very different from asking that if we pick all balls, one at a time, we ever get n red balls in a row; that would be a much harder problem.

    As you said, the answer is
    [tex] p(n) = \frac{R}{N} \frac{R-1}{N-1} \cdots \frac{R-n+1}{N-n+1}.[/tex]
    This can be written in terms of factorials and binomial coefficients (if you think that is nicer and fits your definition of "closed form"):
    [tex] p(n) = \frac{R!}{(R-n)!} \frac{(N-n)!}{N!}= \frac{C(R,n)}{C(N,n)},[/tex]
    where [tex] C(a,b) = \frac{a!}{b!(a-b)!}[/tex] is the binomial coefficient.

    RGV
     
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