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Math Help - Without replacement

  1. #1
    Senior Member DivideBy0's Avatar
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    Without replacement

    Just checking, thanks.

    7 balls are drawn from a box containing 6 blue balls and 3 red balls. Find the probability that only 1 of these is red.

    Is that:

    =\frac{{6 \choose 6}{3 \choose 1}}{{9 \choose 7}}=\frac{1}{12}


    There is a box containing 3 red balls, 5 green balls, and 9 blue balls. 10 Balls are drawn without replacement. Find the probability that of these, 2 are red and 3 green.

    Is that:

    =\frac{{9 \choose 5}{5 \choose 3}{3 \choose 2}}{{17 \choose 10}}=\frac{945}{4862}
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  2. #2
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    That looks right.
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