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Math Help - Probability

  1. #1
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    Probability

    Die A has orange on one face and blue on five faces, Die B has orange on two faces and blue on four faces, Die C has orange on three faces and blue on three faces. If the 3 dice are rolled, find the probability that exactly 2 of the 3 dice come up orange.
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  2. #2
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    Hello, antman!

    Die A has orange on one face and blue on five faces.
    Die B has orange on two faces and blue on four faces.
    Die C has orange on three faces and blue on three faces.
    If the 3 dice are rolled, find the probability that exactly 2 faces come up orange.
    Here are the probabilities:

    . . \begin{array}{|c||c|c|} \hline<br />
\text{Die} & \text{orange} & \text{blue} \\ \hline \hline \\[-4mm]<br />
A & \frac{1}{6} & \frac{5}{6} \\ \\[-4mm]<br />
B & \frac{2}{6} & \frac{4}{6} \\ \\[-4mm]<br />
C & \frac{3}{6} & \frac{3}{6} \\ \\[-4mm] \hline \end{array}


    There are three ways to get two orange faces:

    . . \begin{array}{|c||c|c|c|}\hline<br />
\text{Case}& A &  B & C \\ \hline<br />
1 & \text{orange} & \text{orange} & \text{blue} \\<br />
2 & \text{orange} & \text{blue} & \text{orange} \\<br />
3 & \text{blue} & \text{orange} & \text{orange} \\ \hline \end{array}


    Then: . \begin{array}{ccccc}P(\text{Case 1}) &=& \tfrac{1}{6}\cdot\tfrac{2}{6}\cdot\tfrac{3}{6} &=& \frac{1}{36} \\<br />
P(\text{Case 2}) &=&  \frac{1}{6}\cdot\frac{4}{6}\cdot\frac{3}{6} &=& \frac{2}{36} \\<br />
P(\text{Case 3}) &=& \frac{5}{6}\cdot\frac{2}{6}\cdot\frac{3}{6} &=&\frac{5}{36}\end{array}


    Therefore: . P(\text{2 orange}) \:=\:\frac{1}{36} + \frac{2}{36} + \frac{5}{36} \:=\:\frac{8}{36} \:=\:\boxed{\frac{2}{9}}

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