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Math Help - Confused about basic probability

  1. #1
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    Confused about basic probability

    A box contains 5 white balls, 3 red balls and 2 green balls.

    The balls are taken from box. With replacement and taking a total of 3 balls,

    P(all balls have different colour)=(5/10)x(3/10)x(2/10) x 3! =0.18

    This answer is the same as at the back of the book. My explanation is that a white, red or green ball must be taken and there are 3! ways of arranging the 3 scenarios.

    Is this correct????

    Then secondly,
    Without replacement and taking a total of 5 balls

    My wrong method:
    P(exactly 2 balls are white) = (5/10)x(4/9)x(5/8)x(4/7)x(3/6) x 5!

    This probability cannot be correct since it is above 1. What is the loophole that I did not see???

    I dont need the answer to this question (I can solve using counting total number of cases) but I want to know the loophole in my thinking.

    Thanks,
    Last edited by qazxsw11111; November 12th 2009 at 01:50 AM.
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  2. #2
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    Lexington, MA (USA)
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    Hello, qazxsw11111!

    A box contains 5 white balls, 3 red balls and 2 green balls.

    Three balls are taken from box with replacement.


    My work:

    P(\text{different colors}) \:=\:\frac{5}{10}\times \frac{3}{10} \times \frac{2}{10} \times 3! \:=\:0.18

    My explanation is that a white, red or green ball must be taken
    and there are 3! ways of arranging the 3 scenarios.

    Correct!



    Five balls are taken without replacement.

    My wrong method:

    P(\text{exactly 2 White} \:=\: \frac{5}{10} \times\frac{4}{9} \times\frac{5}{8}\times\frac{4}{7} \times \frac{3}{6} \times {\color{red}5!}

    This probability cannot be correct since it is above 1.
    What is the loophole that I did not see?

    We want 2 Whites and 3 Others.

    There are: . _5C_2 \:=\:\frac{5!}{2!\,3!} \:=\: 10 scenarios ... not 120.

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  3. #3
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    Mar 2009
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    Thank you for your answer. But can you help to clarify when to add factorial sign and when not to?

    Thanks.
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