2009 AMC 10B Problem 25

P21-25~6 minPrint

Basic Probability Basic Counting Split into exhaustive cases and handle each Exploit symmetry to reduce work or pair up objects

Each 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?

Show solution without answering

Key insight

A loop must run through four faces forming a belt. There are 3 belts; each needs its 4 faces right, the other 2 faces are free, and belts never overlap.

Solution

Count all the paintings, then count the ones with a loop.

1. Each face has $2$ possible stripes. With $6$ faces there are $2^6 = 64$ equally likely paintings.
2. What does a loop look like? Hold the cube so one face is on top. A stripe that goes all the way around the cube must pass through the four side faces (front, right, back, left) and come back to where it started. It never touches the top or the bottom. Call these four faces a belt.
3. A cube has $3$ belts: the one around the top-bottom direction, the one around the front-back direction, and the one around the left-right direction.
4. Count paintings that complete one particular belt, say front, right, back, left:
- Each of those four faces must have its stripe running around the belt (the horizontal one). That is $1$ choice on each face.
- The top and bottom faces are not part of the loop, so each can be painted either way: $2 \times 2 = 4$ choices.
- So $1 \times 1 \times 1 \times 1 \times 4 = 4$ paintings complete this belt.
5. Can two belts be complete at the same time? Any two belts share two faces. For example, the front face is on the top-bottom belt and on the left-right belt. The first belt needs the front stripe horizontal, the second needs it vertical. A face has only one stripe, so this cannot happen.
6. So the three belts never overlap, and the number of good paintings is $3 \times 4 = 12$ .
7. Probability: $\frac{12}{64} = \frac{3}{16}$ .

The answer is $\boxed{\textbf{(B)}\ \frac{3}{16}}$ .

Why this works

First find the shapes that make the event happen (the three belts). Then count the ways to get each one, and check that no painting is counted twice before adding. The easy slip is forgetting the faces the loop does not use: they still have $2$ choices each, and those choices multiply the count.

Alternative approach

Work with chances instead of counts. For one belt, each of its four faces has a $\frac12$ chance of the right stripe, so the belt is complete with chance $\frac12 \times \frac12 \times \frac12 \times \frac12 = \frac{1}{16}$ . The top and bottom do not matter. The three belts can never happen together, so the total chance is $3 \times \frac{1}{16} = \frac{3}{16}$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Each face has $2$ possible stripes, so there are $2^6 = 64$ equally likely paintings.

A stripe on a face runs parallel to one pair of the cube's axes; think of each stripe as passing through the centers of two opposite edges of that face. A continuous loop around the cube must cross exactly four faces forming a belt: the four faces surrounding one axis. There are $3$ belts, one for each axis (top-bottom, left-right, front-back).

For the loop to exist on a given belt, each of its four faces must have its stripe running "along the belt," which is exactly one of that face's two options. That is $1$ choice on each of $4$ faces. The two faces not on the belt (the ends of the axis) can be painted either way: $2 \cdot 2 = 4$ choices. So each belt is completed in $4$ of the $64$ paintings.

Can two different belts be completed at once? Any two belts share two faces, and on a shared face the two belts need stripes in perpendicular directions. A face has only one stripe, so no painting completes two belts. The three cases are disjoint, and the number of good paintings is $3 \cdot 4 = 12$ .

The probability is $\frac{12}{64} = \frac{3}{16}$ .

The answer is $\boxed{\textbf{(B)}\ \frac{3}{16}}$ .

Review in the AoPS books

Introduction to Counting & Probability, Chapter 7: Introduction to Probability
Introduction to Counting & Probability, Chapter 1: Counting Is Arithmetic
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques (§2.2)
Introduction to Algebra, Chapter 22: Special Manipulations (§22.3)