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Difference between revisions of "2004 AMC 10B Problems/Problem 23"

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<math> \mathrm{(A) \ } \frac{1}{4} \qquad \mathrm{(B) \ } \frac{5}{16} \qquad \mathrm{(C) \ } \frac{3}{8} \qquad \mathrm{(D) \ } \frac{7}{16} \qquad \mathrm{(E) \ } \frac{1}{2} </math>
 
<math> \mathrm{(A) \ } \frac{1}{4} \qquad \mathrm{(B) \ } \frac{5}{16} \qquad \mathrm{(C) \ } \frac{3}{8} \qquad \mathrm{(D) \ } \frac{7}{16} \qquad \mathrm{(E) \ } \frac{1}{2} </math>
  
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==Solution 1 (Alcumus)==
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==Solution 1==
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If the orientation of the cube is fixed, there are <math>2^6=64</math> possible arrangements of colors on the faces. There are
 
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<cmath>2\dbinom{6}{6}=2</cmath>
 
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arrangements in which all six faces are the same color and
 
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<cmath>2\dbinom{6}{5}=12</cmath>
 
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arrangements in which exactly five faces have the same color. In each of these cases the cube can be placed so that the four vertical faces have the same color. The only other suitable arrangements have four faces of one color, with the other color on a pair of opposing faces. Since there are three pairs of opposing faces, there are <math>2(3)=6</math> such arrangements. The total number of suitable arrangements is therefore <math>2+12+6=20</math>, and the probability is <math>\dfrac{20}{64}=\boxed{\dfrac{5}{16}}.</math>
 
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==Solution 2==
 
  
 
Label the six sides of the cube by numbers <math>1</math> to <math>6</math> as on a classic dice. Then the "four vertical faces" can be: <math>\{1,2,5,6\}</math>, <math>\{1,3,4,6\}</math>, or <math>\{2,3,4,5\}</math>.
 
Label the six sides of the cube by numbers <math>1</math> to <math>6</math> as on a classic dice. Then the "four vertical faces" can be: <math>\{1,2,5,6\}</math>, <math>\{1,3,4,6\}</math>, or <math>\{2,3,4,5\}</math>.
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<math>|A\cup B\cup C| = 8+8+8-2-2-2+2 = 20</math>, and the result is <math>\frac{20}{64}=\boxed{\frac{5}{16}}</math>.
 
<math>|A\cup B\cup C| = 8+8+8-2-2-2+2 = 20</math>, and the result is <math>\frac{20}{64}=\boxed{\frac{5}{16}}</math>.
  
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==Solution 3 (Similar to solution 1)==
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==Solution 2 ==
  
 
Suppose we break the situation into cases that contain four vertical faces of the same color:
 
Suppose we break the situation into cases that contain four vertical faces of the same color:

Latest revision as of 18:14, 11 June 2024

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?

$\mathrm{(A) \ } \frac{1}{4} \qquad \mathrm{(B) \ } \frac{5}{16} \qquad \mathrm{(C) \ } \frac{3}{8} \qquad \mathrm{(D) \ } \frac{7}{16} \qquad \mathrm{(E) \ } \frac{1}{2}$

Solution 1

Label the six sides of the cube by numbers $1$ to $6$ as on a classic dice. Then the "four vertical faces" can be: $\{1,2,5,6\}$, $\{1,3,4,6\}$, or $\{2,3,4,5\}$.

Let $A$ be the set of colorings where $1,2,5,6$ are all of the same color, similarly let $B$ and $C$ be the sets of good colorings for the other two sets of faces.

There are $2^6=64$ possible colorings, and there are $|A\cup B\cup C|$ good colorings. Thus the result is $\frac{|A\cup B\cup C|}{64}$. We need to compute $|A\cup B\cup C|$.

Using the Principle of Inclusion-Exclusion we can write \[|A\cup B\cup C| = |A|+|B|+|C| - |A\cap B| - |A\cap C| - |B\cap C| + |A\cap B\cap C|\]

Clearly $|A|=|B|=|C|=2^3=8$, 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 $A\cap B$? The faces $1,2,5,6$ must have the same color, and at the same time faces $1,3,4,6$ must have the same color. It turns out that $A\cap B=A\cap C=B\cap C= A\cap B\cap C =$ the set containing just the two cubes where all six faces have the same color.

Therefore $|A\cup B\cup C| = 8+8+8-2-2-2+2 = 20$, and the result is $\frac{20}{64}=\boxed{\frac{5}{16}}$.

Solution 2

Suppose we break the situation into cases that contain four vertical faces of the same color:

I. Two opposite sides of same color: There are 3 ways to choose the two sides, and then two colors possible, so $3 \cdot 2=6$.

II. One face different from all the others: There are 6 ways to choose this face, and 2 colors, so $6 \cdot 2=12$.

III. All faces are the same: There are 2 colors, and so two ways for all faces to be the same.

Adding them up, we have a total of $20$ ways to have four vertical faces the same color. There are $2^6$ ways to color the cube, so the answer is $\frac{20}{64}=\boxed{\frac{5}{16}}$.

See also

2004 AMC 10B (Problems • Answer Key • 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
All AMC 10 Problems and Solutions

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