AMC 10 · 2004 · #20

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 a cube's six faces is painted red or blue, each colour equally likely and each face chosen on its own. Find the probability that the cube can be set down with its four vertical faces all one colour.

Pick an answer.

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

There are 64 colourings, and a cube has only three rings.

total colorings=2⁶=64
2STEP 2

Case A: all six faces one color

All one colour always works, giving 2 colourings.

all red or all blue → 2 colorings
3STEP 3

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

Five and one works by putting the odd face on top, giving 12.

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

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

Four and two works only when the minority pair is opposite, giving 6.

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

Add the cases and divide

The cases never overlap, so the probability is 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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