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Math Help - Probability

  1. #1
    Junior Member Fnus's Avatar
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    Probability

    In a box there are 22 cubes, 12 red and 10 blue.
    7 cubes are chosen at random.
    Find the probability for the following events:

    a. Among the seven cubes there are 4 blue cubes and 3 red cubes

    There are others, but I've tried to do them, and for a I get something like..
    0.00780 or something like that.
    Which I think might be wrong..
    So if anyone could just show me what to do, I think I can do the rest myself, thanks
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  2. #2
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    Hi Fnus,

    The probability of getting one blue cube is \frac{10}{22} . Due to the fact that the cubes aren't being replaced the probability of getting another blue cube is \frac{9}{21} and so on and so fourth.

    This gives an answer of \frac{10}{1539} \approx 0.0065
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  3. #3
    Junior Member Fnus's Avatar
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    Hmm, I think that was what I did..

    And I'd multiply them, right?

    Maybe I just pressed something wrong.
    I'll try again, thanks.
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  4. #4
    MHF Contributor

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    Here is the correct answer:
    \frac {{10 \choose 4}{12 \choose 3}}{{22 \choose 7}} =0.270897832817337
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  5. #5
    Moo
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  6. #6
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Moo View Post
    I thought it was that...

    Oohh...I'm gonna have to deal with that guy and all these other weird distributions this fall in my hardcore stats class...once I survive probability

    --Chris
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  7. #7
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    Quote Originally Posted by Plato View Post
    Here is the correct answer:
    \frac {{10 \choose 4}{12 \choose 3}}{{22 \choose 7}} =0.270897832817337
    I need to sleep! Thank you for correcting Plato.

    Quote Originally Posted by Chris L T521 View Post
    I thought it was that...

    Oohh...I'm gonna have to deal with that guy and all these other weird distributions this fall in my hardcore stats class...once I survive probability

    --Chris
    Good luck. I'm sure you will enjoy
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