2004 AMC 12B Problems/Problem 20
Contents
Problem
Each face of a cube is painted either red or blue, each with probability . 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?
Solution
There are possible colorings of the cube. Consider the color that appears with greater frequency. The property obviously holds true if or of the faces are colored the same, which for each color can happen in ways. If of the faces are colored the same, there are possible cubes (corresponding to the possible ways to pick pairs of opposite faces for the other color). If of the faces are colored the same, the property obviously cannot be satisfied. Thus, there are a total of ways for this to occur, and the desired probability is .
Solution 2
As with the previous solution, we will simply count the number of valid faces and divide by . We say that a cube has a ring of color when four vertical faces are all the same color. If we fix the cube in space, there can be different rings (in each of the axis), each of which can take on 2 different colors. This leaves us to choose the remaining two faces (which are opposite one another) in ways. However, we have counted coloring that has each face the same color times for both red and blue. This means that we must subtract the 4 overcounts to get . Dividing by total colorings, we get . ~bambithenambi
See also
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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