1. ## 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.

2. 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:

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

There are three ways to get two orange faces:

. . $\displaystyle \begin{array}{|c||c|c|c|}\hline \text{Case}& A & B & C \\ \hline 1 & \text{orange} & \text{orange} & \text{blue} \\ 2 & \text{orange} & \text{blue} & \text{orange} \\ 3 & \text{blue} & \text{orange} & \text{orange} \\ \hline \end{array}$

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

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