2009 AMC 10B Problem 25

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

Identify the structures that satisfy the condition (belts), count how many paintings realize each, and confirm they are mutually exclusive before adding. The "free faces" factor is the classic omission: faces the condition does not mention still contribute a multiplier to the count.

Probability version: fix a belt; its four faces are each correct with probability $\frac12$, so the belt forms with probability $\frac{1}{16}$. The three belt events are disjoint, so the total is $3 \cdot \frac{1}{16} = \frac{3}{16}$.

• Forgetting the 2 x 2 = 4 free choices on the two faces outside the belt, which gives 3/64, or double-counting configurations with two belts, which cannot happen.
• Multiplying $\frac{1}{16}$ by $6$ (one per face) instead of by $3$ (one per axis), giving $\frac38$.