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Probability Question - Ball Urn Problem

  1. Mar 8, 2013 #1
    Using the binomial theorem you can solve ball urn problems. Like say for example a urn has 4 green balls and 3 white balls. You draw two balls. What's the probability you draw 2 white balls?

    I just made this problem up off the top of my head. But anyways using the binomial theorem

    (7C2)(3/7)^2*(4/7)^5

    However this is wrong correct? Because it's impossible to draw two balls simultaneously... I just realized that. So how do you solve this problem then?
     
  2. jcsd
  3. Mar 8, 2013 #2

    Ray Vickson

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    The issue is one of drawing *with* replacement (binomial applies) vs. drawing *without* replacement (binomial fails). For drawing without replacement you need to use the hypergeometric distribution.

    In your case, the probability that the first drawn ball is W is 3/7. Given that the first one is W that leaves 6 balls in the urn, of which 2 are white, so the probability the next one drawn is W is 2/6. The probability of drawing 2 whites is P{WW} = (3/7)*(2/6) = 1/7.

    For more on the hypergeometric distribution see
    http://en.wikipedia.org/wiki/Hypergeometric_distribution
    or
    http://mathworld.wolfram.com/HypergeometricDistribution.html
    or
    http://www.math.uah.edu/stat/urn/Hypergeometric.html .
     
  4. Mar 8, 2013 #3
    Well what if you don't know if the first ball is drawn is W or or G?

    1/7 if you draw W on the first draw
    but if you draw R on the first than the probability you draw two W is zero

    so how would you answer this question then since there are two answers I guess?
     
  5. Mar 8, 2013 #4

    Ray Vickson

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    The answer is the same, whether or not you *know* the color of the balls.
     
  6. Mar 8, 2013 #5
    Why is that? If you draw a Red ball on the first than there's no way to draw two white balls.
     
  7. Mar 8, 2013 #6

    Ray Vickson

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    You asked for the probability of two white balls.
     
  8. Mar 8, 2013 #7
    So given the question how do you know if there's replacement or not?
     
  9. Mar 8, 2013 #8

    Ray Vickson

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    You don't. Somebody needs to tell you, or you need more context. For example, in sampling industrial output by destructive testing, there cannot be any possibility of replacement.
     
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