AMC 10 · 2004 · #23

Grade 7 geometry-3d
Archetype Casework by Position / Vertex Type └ Sub-archetype Count of Type
probability-basicspatial-visualization caseworksystematic-enumeration ↑ Prerequisites: probability-basic
📏 Medium solution 💡 3 insights
Problem
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.

Pick an answer.

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

AMC 10 2004 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Make a Systematic List

There are only 2⁶=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.

1STEP 1

Count all colorings and picture the goal

Six faces, two colors each, give 64 equally likely colorings; the four side faces form a ring, and a cube has only 3 rings.

total colorings=2⁶=64
2STEP 2

Case A: all six faces one color

If all six faces match, every ring matches too, so the cube works — that is 2 colorings, all red or all blue.

all red or all blue → 2 colorings
3STEP 3

Case B: five of one color, one of the other

Five matching faces plus one odd face: put the odd face on top and the ring works — 6 faces × 2 colors = 12 colorings.

6 (odd face) × 2 (majority color)=12
4STEP 4

Case C: four of one color, two of the other

In a four-two split a ring is one color only when the two odd faces are opposite: 3 pairs × 2 colors = 6; adjacent odd faces fail.

3 (opposite pair) × 2 (minority color)=6
5STEP 5

Add the cases and divide

The cases never overlap, so add them: 2+12+6=20 winners out of 64, and 20/64 reduces to 5/16 — choice (B).

2+12+6=20, 20/64=5/16 (B)
Answer
5/16
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 5/16≈ 0.31 is sensible. It lands exactly on choice (B), between (A) 1/4 and (C) 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.
💡Key takeaway

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 5/16.

  • Count all colorings and picture the goal
  • Case A: all six faces one color
  • Case B: five of one color, one of the other
  • Case C: four of one color, two of the other
  • Add the cases and divide

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