AMC 10 · 2009 · #25
Grade 7 geometry-3dEach face of a cube is given a single narrow stripe painted from the center of one edge to the center of the opposite edge. The choice of the edge pairing is made at random and independently for each face. What is the probability that there is a continuous stripe encircling the cube?
Pick an answer.
AMC 10 2009 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 a cube's $6$ faces gets one stripe joining the midpoint of an edge to the midpoint of the opposite edge. A square face has two pairs of opposite edges, so each face's stripe has $2$ equally likely orientations, chosen independently. Find the probability that some stripes join up into a single loop that runs all the way around the cube.
Givens: A cube has $6$ faces; each face gets exactly one stripe from an edge-midpoint to the opposite edge-midpoint.; Each face's stripe has $2$ possible orientations, chosen at random and independently, giving $2^6 = 64$ equally likely outcomes.; An 'encircling stripe' is a closed loop formed by the stripes on $4$ faces that make a band around the cube.; Answer choices: (A) $\frac 18$, (B) $\frac{3}{16}$, (C) $\frac 14$, (D) $\frac 38$, (E) $\frac 12$.
Unknowns: The probability that at least one continuous encircling stripe is formed.
Understand
Restated: Each of a cube's $6$ faces gets one stripe joining the midpoint of an edge to the midpoint of the opposite edge. A square face has two pairs of opposite edges, so each face's stripe has $2$ equally likely orientations, chosen independently. Find the probability that some stripes join up into a single loop that runs all the way around the cube.
Givens: A cube has $6$ faces; each face gets exactly one stripe from an edge-midpoint to the opposite edge-midpoint.; Each face's stripe has $2$ possible orientations, chosen at random and independently, giving $2^6 = 64$ equally likely outcomes.; An 'encircling stripe' is a closed loop formed by the stripes on $4$ faces that make a band around the cube.; Answer choices: (A) $\frac 18$, (B) $\frac{3}{16}$, (C) $\frac 14$, (D) $\frac 38$, (E) $\frac 12$.
Plan
Primary tool: #17 Visualize Spatial Relationships
Secondary: #1 Draw a Diagram, #2 Make a Systematic List, #7 Identify Subproblems
The whole problem turns on seeing, in 3D, what an encircling stripe actually is: a belt of $4$ faces wrapping around the cube while skipping one opposite pair. Tool #17 (Visualize Spatial Relationships) is the unlock for that mental picture and for spotting why two belts cannot both close at once. Tool #1 (Draw a Diagram) keeps a labeled cube sketch so 'this face's stripe must run this way' stays concrete. Tool #7 (Identify Subproblems) splits the job into: (i) the chance one chosen belt closes, then (ii) combine the three belts. Tool #2 (Make a Systematic List) both fixes the size of the sample space ($2^6 = 64$) and gives a clean favorable-count cross-check ($12$ of $64$).
Execute — Answer: B
7.SP.C.8 Step 1 Count all equally likely outcomes
- Each of the $6$ faces independently takes one of $2$ stripe orientations.
- By the multiplication principle the number of equally likely stripe patterns on the whole cube is $2^6 = 64$.
- Every pattern is equally likely because each face is a fair, independent $\tfrac12$-$\tfrac12$ choice.
💡 Six independent two-way choices multiply into one big sample space of $64$ equally likely cubes.
6.G.A.4 Step 2 Picture an encircling stripe as a band
- Hold the cube and trace a loop that goes all the way around.
- Such a loop crosses exactly $4$ faces that form a belt, and it never touches the remaining $2$ faces, which sit on opposite sides like a top and a bottom.
- A cube has $3$ pairs of opposite faces, so there are exactly $3$ possible belts an encircling stripe could follow.
💡 A belt around a cube always skips one opposite pair, and there are only three such pairs.
7.SP.C.8 Step 3 Probability one chosen band closes
- Fix one of the three belts.
- Its $4$ faces must each take the single orientation that runs along the belt so the stripes meet at the shared edge midpoints; the other orientation would cut across and break the loop.
- Each of the $4$ faces has probability $\tfrac12$ of being aligned, and the choices are independent, so the belt closes with probability $\left(\tfrac12\right)^4 = \tfrac1{16}$.
- The $2$ skipped faces may be oriented either way.
💡 Four faces must each hit their one 'aligned' side, and $\tfrac12$ four times over is $\tfrac1{16}$.
7.SP.C.7 Step 4 Bands cannot overlap
- Check whether two belts could close at the same time.
- Any two of the three belts share a side face.
- To complete one belt that face's stripe must run one way; to complete the other belt the same face's stripe must run the other way.
- A single stripe cannot do both, so no two belts can close together.
- The three band-events are mutually exclusive.
💡 A shared face can point only one way, so at most one belt can ever close on a given cube.
6.RP.A.3 Step 5 Add the three disjoint bands
- Because the three belts are mutually exclusive, the probability that some belt closes is the sum of their separate probabilities: $3 \times \tfrac1{16} = \tfrac{3}{16}$.
- Cross-check by counting: each belt gives $2 \times 2 = 4$ patterns (the two skipped faces are free), so there are $3 \times 4 = 12$ favorable patterns out of $64$, and $\tfrac{12}{64} = \tfrac{3}{16}$.
- The probability of a continuous encircling stripe is $\tfrac{3}{16}$, choice (B).
💡 Disjoint events just add, and the tidy $\tfrac{12}{64}$ count confirms the $\tfrac{3}{16}$ answer.
7.SP.C.8 Each of the $6$ faces independently takes one of $2$ stripe orientations. By the 6.G.A.4 Hold the cube and trace a loop that goes all the way around. Such a loop crosses 7.SP.C.8 Fix one of the three belts. Its $4$ faces must each take the single orientation 7.SP.C.7 Check whether two belts could close at the same time. Any two of the three belts 6.RP.A.3 Because the three belts are mutually exclusive, the probability that some belt c Review
Reasonableness: The answer $\tfrac{3}{16} = \tfrac{12}{64} \approx 0.19$ is a small probability, which fits: getting four specific faces to line up is demanding. Two independent counts agree — the disjoint-sum $3 \times \tfrac1{16}$ and the direct favorable count $\tfrac{12}{64}$. The distractors flag the classic slips: $\tfrac1{16}$ forgets there are three belts, $\tfrac38 = 6 \times \tfrac1{16}$ double-counts as if belts could overlap, and $\tfrac14$ or $\tfrac12$ overcount the aligned faces. Since the belts are genuinely mutually exclusive (Step 4), no double-counting occurs and $\tfrac{3}{16}$ stands. Answer (B).
Alternative: Tool #2 (Make a Systematic List), pure counting: fix the top stripe's orientation, say front-to-back. An encircling loop through the top must continue down the front, under the bottom, and up the back — one forced pattern on those four faces — while the two side faces are free ($2 \times 2 = 4$ patterns), giving $4$ favorable of the $2^5 = 32$ patterns once the top is fixed, i.e. probability $\tfrac{4}{32} = \tfrac18$ for a loop that includes the top. But that is really two of the three belts. Summing correctly over all three belts with no double-count returns $\tfrac{3}{16}$, matching the disjoint-sum result.
CCSS standards used (min grade 7)
7.SP.C.8Find probabilities of compound events using organized lists, tables, tree diagrams, and simulation (Sizing the sample space at $2^6 = 64$ and multiplying the four independent $\tfrac12$ face-choices into the per-band probability $\tfrac1{16}$.)6.G.A.4Represent three-dimensional figures using nets made up of rectangles and triangles (Seeing an encircling stripe as a belt of four faces around the cube that skips one opposite pair, with three such belts available.)7.SP.C.7Develop a probability model and use it to find probabilities of events (Establishing that the three belt-events are mutually exclusive, so their probabilities may simply be added.)6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Combining the three equal $\tfrac1{16}$ chances and reducing the favorable-count ratio $\tfrac{12}{64}$ to $\tfrac{3}{16}$.)
⭐ A stripe going all the way around the cube is just a belt of $4$ faces that skips a top-and-bottom pair; there are $3$ such belts, each closes with chance $\left(\tfrac12\right)^4 = \tfrac1{16}$, and since two belts can never close at once you add them: $3 \times \tfrac1{16} = \tfrac{3}{16}$, choice (B).
⭐ A stripe going all the way around the cube is just a belt of $4$ faces that skips a top-and-bottom pair; there are $3$ such belts, each closes with chance $\left(\tfrac12\right)^4 = \tfrac1{16}$, and since two belts can never close at once you add them: $3 \times \tfrac1{16} = \tfrac{3}{16}$, choice (B).
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