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Thread: 8 marbles. Probablity of third one being black?

  1. #1
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    8 marbles. Probablity of third one being black?

    There are 8 marbels in a bag, 2 black and 6 red. What's the probability of the third marble you take out is black?
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  2. #2
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    Re: 8 marbles. Probablity of third one being black?

    There are 3 ways to accomplish this

    rrb
    rbb
    brb

    $P[rrb]=\dfrac{6}{8}\dfrac{5}{7}\dfrac{2}{6} = \dfrac{5}{28}$

    $P[rbb]=\dfrac{6}{8}\dfrac{2}{7}\dfrac{1}{6} = \dfrac{1}{28}$

    $P[brb] = \dfrac{2}{8}\dfrac{6}{7}\dfrac{1}{6} = \dfrac{1}{28}$

    Their sum is $\dfrac{7}{28}=\dfrac{1}{4}$
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  3. #3
    Member Cervesa's Avatar
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    Re: 8 marbles. Probablity of third one being black?

    P(3rd black) = 1 - P(3rd red)

    $P[rrr] = \dfrac{6}{8} \cdot \dfrac{5}{7} \cdot \dfrac{4}{6} = \dfrac{5}{14}$

    $P[brr] = \dfrac{2}{8} \cdot \dfrac{6}{7} \cdot \dfrac{5}{6} = \dfrac{5}{28}$

    $P[rbr] = \dfrac{6}{8} \cdot \dfrac{2}{7} \cdot \dfrac{5}{6} = \dfrac{5}{28}$

    $P[bb] = \dfrac{2}{8} \cdot \dfrac{1}{7} = \dfrac{1}{28}$

    $1 - \left(\dfrac{5}{14}+\dfrac{5}{28}+\dfrac{5}{28}+ \dfrac{1}{28} \right) = 1 - \dfrac{3}{4} = \dfrac{1}{4}$
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