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  1. #1
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    Question Probability Question

    Suppose we have box containing 50 balls of which 6 are black and the rest red.
    Without seeing the balls we take 23 balls out of the box. My question is what is the probability that 6 of these 23 balls are black.
    Thank you.
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  2. #2
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    Quote Originally Posted by ciccio View Post
    Suppose we have box containing 50 balls of which 6 are black and the rest red.
    Without seeing the balls we take 23 balls out of the box. My question is what is the probability that 6 of these 23 balls are black.
    Thank you.
    Read this: Hypergeometric distribution - Wikipedia, the free encyclopedia
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  3. #3
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    thanks. I almost got it, need to read it again ,
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  4. #4
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    Hello, ciccio!

    We have a box containing 50 balls of which 6 are black and the rest red.
    Without seeing the balls we take 23 balls out of the box.
    What is the probability that 6 of these 23 balls are black?

    There are 50 balls. We take 23 of them.
    . . There are: . _{50}C_{23} possible outcomes.


    There are 6 black balls. We want 6 of them.
    . . There is: . _6C_6\:=\:1 way.
    There are 44 red balls, and we want 17 of them.
    . . There are: . _{44}C_{17} ways.

    Hence, there are: . (1)(_{44}C_{17}) \:=\:_{44}C<br />
_{17} ways to draw 6 black balls.


    Therefore: . P(\text{6 black balls}) \;=\;\frac{_{44}C_{17}}{_{50}C_{23}} \;=\;\frac{44!}{17!\,27!}\cdot\frac{23!\,27!}{50!}

    I think it comes out to: . \frac{4807}{756,\!700}

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  5. #5
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    thanks ,very clear answer
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