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Expected number of red balls drawn before drawing a white ball

  1. Jun 23, 2010 #1

    txy

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    1. The problem statement, all variables and given/known data

    There are 50 balls in a bag, all are the same except for the color. 4 of them are white while the rest (46) of them are red. Now randomly draw balls from the bag, one at a time, without replacement. What is the expected number of red balls that you will draw before you first draw a white ball?

    2. Relevant equations

    If a discrete random variable X takes only positive values, then its expectation
    [tex]E(X) = \sum_{k=1}^{+\infty}P(X \geq k)[/tex]

    3. The attempt at a solution
    Let X be the number of red balls drawn before first drawing a white ball. I can't seem to find [tex]P(X \geq k)[/tex]. Can somebody give me a hint?
     
  2. jcsd
  3. Jun 23, 2010 #2
    Say it out loud :)
    What is the probability, that you draw 1 or more red balls? It is of course the probability of the first ball being red.
    What is the probability, that you draw 2 or more red balls?
    Etc.
    You should get a nice sum, which you will be able to write in a closed form.
     
  4. Jun 24, 2010 #3

    txy

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    The probability of the first k balls is red is [tex]\frac{C_{46}^{k}}{C_{50}^{k}}[/tex]

    This is so difficult to sum!
     
  5. Jun 24, 2010 #4
    Is it OK for the answer to just be a sum? It might be the case that this doesn't have a better looking form than a sum.
     
  6. Jun 25, 2010 #5

    txy

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    But the answer is 46/5 ...
     
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