2004 AMC 12B Problems/Problem 20

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?

$\textbf{(A)}\ \frac14 \qquad \textbf{(B)}\ \frac {5}{16} \qquad \textbf{(C)}\ \frac38 \qquad \textbf{(D)}\ \frac {7}{16} \qquad \textbf{(E)}\ \frac12$

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 $\frac{20}{2^6} = \frac{5}{16}\ \mathbf{(B)}$ .

2004 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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All AMC 12 Problems and Solutions

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2004 AMC 12B Problems/Problem 20

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