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Math Help - Hypergeometric

  1. #1
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    Hypergeometric

    From a box containing 5 white balls and 4 reds, two are randomly selected w/o replacement. Find

    (a) Exactly 1 is white.
    X number of white selected.
    P(X=1), \ N=9, \ k=5, \ n=2

    \displaystyle p(1)=\frac{\binom{5}{1}\binom{4}{1}}{\binom{9}{2}}  =\frac{5}{9}

    The book says the answer is 0.14. I don't know what I did wrong since what I have done seems correct.
    Last edited by dwsmith; March 22nd 2011 at 12:26 PM.
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  2. #2
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    I also have a question in regards to part (c).

    Two white balls are selected, given at least one white is selected.
    \displaystyle P(2W|\text{at least} \ 1W)=\frac{P(2W\cap \text{at least} \ 1W)}{\sum_{x=1}^2\frac{\binom{5}{x}\binom{4}{2-x}}{\binom{9}{2}}=\frac{5}{6}}

    I don't know how to determine that intersection.
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  3. #3
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    Question: what does "w/replacement" mean?
    Is it with replacement? Or without replacement?
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  4. #4
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    Quote Originally Posted by Plato View Post
    Question: what does "w/replacement" mean?
    Is it with replacement? Or without replacement?
    I forgot the o. It shold be w/o=without
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  5. #5
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    P(\text{2W}\cap\text{at least one W})=P(\text{2W})
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