2004 AMC 12B Problems/Problem 20

Revision as of 14:38, 10 September 2022 by Bambithenambi (talk | contribs) (Added second soln)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)


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?

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


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

Solution 2

As with the previous solution, we will simply count the number of valid faces and divide by $2^6$. 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 $3$ different rings (in each of the $3$ 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 $2^2=4$ ways. However, we have counted coloring that has each face the same color $3$ times for both red and blue. This means that we must subtract the 4 overcounts to get $2*3*2^2-4=20$. Dividing by $2^6$ total colorings, we get $20/2^6=\frac{5}{16}\mathbf{(B)}$. ~bambithenambi

See also

2004 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png