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2004 AMC 12B Problems/Problem 20

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Problem

Each face of a cube is painted either red or blue, each with probability 1/2. The color of each face is determined independently. What is the probability that the painted cube can be placed on a horizontal surface so that the four vertical faces are all the same color?

\textbf{(A)}\ \frac14 \qquad \textbf{(B)}\ \frac {5}{16} \qquad \textbf{(C)}\ \frac38 \qquad \textbf{(D)}\ \frac {7}{16} \qqu...

Solution

There are 2^6 possible colorings of the cube. Consider the color that appears with greater frequency. The property obviously holds true if 5 or 6 of the faces are colored the same, which for each color can happen in 6 + 1 = 7 ways. If 4 of the faces are colored the same, there are 3 possible cubes (corresponding to the 3 possible ways to pick pairs of opposite faces for the other color). If 3 of the faces are colored the same, the property obviously cannot be satisfied. Thus, there are a total of 2(7 + 3) = 20 ways for this to occur, and the desired probability is \frac{20}{2^6} = \frac{5}{16}\ \mathbf{(B)}.

See also

2004 AMC 12B (ProblemsResources)
Preceded by
Problem 19
Followed by
Problem 21
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