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Difference between revisions of "2004 AMC 12B Problems/Problem 20"

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There are <math>2^6</math> possible colorings of the cube. Consider the color that appears with greater frequency. The property obviously holds true if <math>5</math> or <math>6</math> of the faces are colored the same, which for each color can happen in <math>6 + 1 = 7</math> ways. If <math>4</math> of the faces are colored the same, there are <math>3</math> possible cubes (corresponding to the <math>3</math> possible ways to pick pairs of opposite faces for the other color). If <math>3</math> of the faces are colored the same, the property obviously cannot be satisfied. Thus, there are a total of <math>2(7 + 3) = 20</math> ways for this to occur, and the desired probability is <math>\frac{20}{2^6} = \frac{5}{16}\ \mathbf{(B)}</math>.
 
There are <math>2^6</math> possible colorings of the cube. Consider the color that appears with greater frequency. The property obviously holds true if <math>5</math> or <math>6</math> of the faces are colored the same, which for each color can happen in <math>6 + 1 = 7</math> ways. If <math>4</math> of the faces are colored the same, there are <math>3</math> possible cubes (corresponding to the <math>3</math> possible ways to pick pairs of opposite faces for the other color). If <math>3</math> of the faces are colored the same, the property obviously cannot be satisfied. Thus, there are a total of <math>2(7 + 3) = 20</math> ways for this to occur, and the desired probability is <math>\frac{20}{2^6} = \frac{5}{16}\ \mathbf{(B)}</math>.
  
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==Solution 2==
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As with the previous solution, we will simply count the number of valid faces and divide by <math>2^6</math>. 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 <math>3</math> different rings (in each of the <math>3</math> 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 <math>2^2=4</math> ways. However, we have counted coloring that has each face the same color <math>3</math> times for both red and blue. This means that we must subtract the 4 overcounts to get <math>2*3*2^2-4=20</math>. Dividing by <math>2^6</math> total colorings, we get <math>20/2^6=\frac{5}{16}\mathbf{(B)}</math>.
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~bambithenambi
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2004|ab=B|num-b=19|num-a=21}}
 
{{AMC12 box|year=2004|ab=B|num-b=19|num-a=21}}
  
 
[[Category:Introductory Combinatorics Problems]]
 
[[Category:Introductory Combinatorics Problems]]
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{{MAA Notice}}

Latest revision as of 14:38, 10 September 2022

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

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 (Problems • Answer Key • Resources)
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

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