2004 AMC 10B Problems/Problem 23
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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?
Solution
Label the six sides of the cube by numbers 
to 
as on a classic dice. Then the "four vertical faces" can be: 
, 
, or 
.
Let 
be the set of colorings where 
are all of the same color, similarly let 
and 
be the sets of good colorings for the other two sets of faces.
There are 
possible colorings, and there are 
good colorings. Thus the result is 
. We need to compute .
Using the Principle of Inclusion-Exclusion we can write
Clearly 
, as we have two possibilities for the common color of the four vertical faces, and two possibilities for each of the horizontal faces.
What is 
? The faces must have the same color, and at the same time faces 
must have the same color. It turns out that 
the set containing just the two cubes where all six faces have the same color.
Therefore
, and the result is 
.
See also
| 2004 AMC 10B (Problems • Resources) | ||
| Preceded by Problem 22 | Followed by Problem 24 | |
| 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 | ||
