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  1. #1
    Jon
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    Last edited by Jon; March 12th 2006 at 04:22 PM. Reason: Delete
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  2. #2
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    Quote Originally Posted by Jon
    A bowl contains 6 red, 4 blue and 2 green marbles. A sample of 3 marbles is drawn at random w/o replacement. Find the probability that the sample contains the following:

    All Blue marbles
    All Green marbles
    No Red marbles
    Exactly 2 red Marbles
    At most 2 red marbles.
    All blue:
    \frac{4}{12}\times\frac{3}{11}\times\frac{2}{10}= \frac{1}{55}
    All green:
    \frac{2}{12}\times\frac{1}{11}\times \frac{0}{10}=0
    No red:
    \frac{6}{12}\times\frac{5}{11}\times\frac{5}{10}= \frac{5}{44}
    2 Red:
    \frac{_6C_2\cdot _6C_1}{_{12}C_3}=\frac{9}{22}
    At most 2: (None, 1, or 2):
    \frac{5}{44}+\frac{18}{44}+\frac{18}{44}=\frac{41}  {44}
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