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

m (Solution 2)
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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>
  
==Solution==
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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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Therefore
 
Therefore
 
<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==
 
==Solution 3==
  

Revision as of 22:52, 17 December 2015

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 2

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 3

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 $3*2=6$

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

III. All faces same There are 2 colors, 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. The are $2^6$ ways to color the cube, so the answer is $\frac{20}{64}=\boxed{\frac{5}{16}}$

See also

2004 AMC 10B (ProblemsAnswer KeyResources)
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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