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Math Help - simple question about w/o replacement

  1. #1
    Newbie
    Joined
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    simple question about w/o replacement

    a box contains 15 balls (4 yellow, 5 purple, 6 blue) 3 balls are drawn at random...

    without replacement ..calculate P( all 3 balls are of the same color)

    The solution says it should be (4/15)^3 + (5/15)^3 +(6/15)^3

    I think it should be (4*3*2)/(15*14*13)...Anyone can explain it to me?

    Thanks a lot!
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  2. #2
    Newbie Mirado's Avatar
    Joined
    Apr 2009
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    New York
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    solved

    Ok, so you are close to the answer. Basically, remember that the probability of getting 3 yellows are: (4 \times 3 \times 2)/(15 \times 14 \times 13)

    3 purples are: (5 \times 4 \times 3)/(15 \times 14 \times 13)

    3 blue are: (6 \times 4 \times 3)/(15 \times 14 \times 13)

    Hence, when you combine the terms, you get ((4 \times 3 \times 2) + (5 \times 4 \times 3) + (6 \times 4 \times 3))/(15 \times 14 \times 13)

    hope that clears it up, have fun.
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