AMC 10 · 2004 · #23
Grade 7 geometry-3dEach 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?
Pick an answer.
AMC 10 2004 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Each of the six faces of a cube is painted red or blue, each color equally likely and each face chosen on its own. Find the probability that the cube can be set on a table so that its four side (vertical) faces are all one color.
Givens: A cube has $6$ faces.; Each face is painted red or blue with probability $\frac{1}{2}$ each, independently of the other faces.; We may turn the cube any way we like before setting it down; only one placement needs to work.; Answer choices: (A) $\frac{1}{4}$, (B) $\frac{5}{16}$, (C) $\frac{3}{8}$, (D) $\frac{7}{16}$, (E) $\frac{1}{2}$.
Unknowns: The probability that some placement makes all four vertical faces the same color.
Understand
Restated: Each of the six faces of a cube is painted red or blue, each color equally likely and each face chosen on its own. Find the probability that the cube can be set on a table so that its four side (vertical) faces are all one color.
Givens: A cube has $6$ faces.; Each face is painted red or blue with probability $\frac{1}{2}$ each, independently of the other faces.; We may turn the cube any way we like before setting it down; only one placement needs to work.; Answer choices: (A) $\frac{1}{4}$, (B) $\frac{5}{16}$, (C) $\frac{3}{8}$, (D) $\frac{7}{16}$, (E) $\frac{1}{2}$.
Plan
Primary tool: #2 Make a Systematic List
Secondary: #17 Visualize Spatial Relationships, #7 Identify Subproblems
There are only $2^6=64$ ways to paint the cube, so we can count the winning ones exactly. Tool #17 (Visualize Spatial Relationships) turns the words 'four vertical faces' into a picture: a ring of four faces around the cube, with the leftover opposite pair as top and bottom. Tool #2 (Make a Systematic List) then sorts the winning colorings into a few tidy cases by how the two colors are split ($6$ of one color, $5$ and $1$, or $4$ and $2$). Tool #7 (Identify Subproblems) lets us count each case on its own and add the totals, being careful that the cases never overlap so nothing is counted twice.
Execute — Answer: B
7.SP.C.8 Step 1 Count all colorings and picture the goal
- Each of the $6$ faces is painted in one of $2$ colors, independently, so there are $2^6=64$ equally likely colorings.
- A placement's four vertical faces form a ring that wraps around the cube; the top and bottom faces are the one opposite pair left out of that ring.
- A cube has exactly $3$ rings (choose any of the $3$ opposite pairs to be top and bottom).
- So the cube 'works' exactly when at least one of these $3$ rings is painted all one color.
- We count the colorings that have such a monochrome ring.
💡 Four vertical faces are just a ring around the cube, and the cube has only three such rings to check.
7.SP.C.8 Step 2 Case A: all six faces one color
- If every face is the same color, then of course every ring is that color, so the cube works.
- There are exactly $2$ such colorings: all red, or all blue.
💡 A cube painted a single color trivially has a matching ring.
7.SP.C.8 Step 3 Case B: five of one color, one of the other
- Suppose five faces share a color and one face is the odd one out.
- Put that odd face on top: then the four vertical faces and the bottom are all the majority color, so the cube works.
- Count these colorings by choosing which face is the odd one ($6$ choices) and which color the other five share ($2$ choices): $6\times 2=12$ colorings.
💡 Stand the single stray face on top, and the ring plus bottom are all the same color.
7.SP.C.8 Step 4 Case C: four of one color, two of the other
- Now split the colors four and two.
- A ring can be monochrome only if the four matching faces themselves form a ring, which happens exactly when the two minority faces are an opposite pair (so they become the top and bottom).
- Count these: choose which of the $3$ opposite pairs is the minority pair ($3$ ways) and which color that pair is ($2$ ways), giving $3\times 2=6$ colorings.
- If instead the two minority faces are adjacent, no ring is all one color, so those do not work.
- And any more balanced split (three and three) can never fill a ring, since a ring needs four faces of one color.
- So Case C contributes $6$ colorings.
💡 The only way four matching faces form a ring is when the two odd faces sit opposite each other as top and bottom.
4.NF.A.1 Step 5 Add the cases and divide
- The three cases split colorings by how many faces are the majority color ($6$, $5$, or $4$), so they never overlap.
- Adding gives $2+12+6=20$ winning colorings out of $64$.
- The probability is $\frac{20}{64}=\frac{5}{16}$, so the answer is (B).
💡 Add the disjoint cases, then reduce the fraction by dividing top and bottom by $4$.
7.SP.C.8 Each of the $6$ faces is painted in one of $2$ colors, independently, so there a 7.SP.C.8 If every face is the same color, then of course every ring is that color, so the 7.SP.C.8 Suppose five faces share a color and one face is the odd one out. Put that odd f 7.SP.C.8 Now split the colors four and two. A ring can be monochrome only if the four mat 4.NF.A.1 The three cases split colorings by how many faces are the majority color ($6$, $ Review
Reasonableness: The count $20$ is comfortably between the extremes: it is more than the $2+12=14$ colorings from the all-same and five-one cases alone, and far below $64$, so a probability near $\frac{5}{16}\approx 0.31$ is sensible. It lands exactly on choice (B), between (A) $\frac{1}{4}$ and (C) $\frac{3}{8}$. Splitting by majority-color count ($6$, $5$, $4$) guarantees no coloring is counted twice, which is the main danger in a problem like this.
Alternative: Use inclusion-exclusion on the $3$ rings instead of casework. For one fixed ring, being all one color means $2$ color choices for the ring and $2^2=4$ free choices for the top and bottom, so $2\times 4=8$ colorings; across $3$ rings that is $3\times 8=24$. Two different rings share a face, so if both are monochrome all six faces must match, giving $2$ colorings for each of the $3$ pairs of rings, i.e. $6$ to subtract. All three rings monochrome again means all six faces match, $2$ colorings to add back. Inclusion-exclusion gives $24-6+2=20$, matching $\frac{20}{64}=\frac{5}{16}$.
CCSS standards used (min grade 7)
7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Setting up the $64$ equally likely colorings and counting the winning ones by organized cases to get a probability.)4.NF.A.1Explain why a fraction is equivalent to another fraction (Reducing the probability $\frac{20}{64}$ to $\frac{5}{16}$ by dividing numerator and denominator by $4$.)
⭐ Picture the four side faces as a ring around the cube; count the colorings with an all-one-color ring by cases (all six match, five-and-one, or four-and-two with the odd pair on top and bottom): $2+12+6=20$ out of $64$, which is $\frac{5}{16}$.
⭐ Picture the four side faces as a ring around the cube; count the colorings with an all-one-color ring by cases (all six match, five-and-one, or four-and-two with the odd pair on top and bottom): $2+12+6=20$ out of $64$, which is $\frac{5}{16}$.
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