2009 AMC 12B Problem 17

Attempt Problem 17 of the 2009 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2009 AMC 12B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

17.

Each face of a cube is given a single narrow stripe painted from the center of one edge to the center of its 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?
18\dfrac{1}{8}
316\dfrac{3}{16}
14\dfrac{1}{4}
38\dfrac{3}{8}
12\dfrac{1}{2}
Answer: B
Concepts:basic probabilitycube geometrycasework
Small hint:
Each face’s stripe has 22 orientations, giving 26=642^6 = 64 equally likely configurations
Big hint:
An encircling stripe runs around one of 33 pairs of opposite faces; count the configurations forming a loop
Solution:
Each of the 66 faces has 22 equally likely stripe orientations, for 26=642^6 = 64 configurations.
An encircling stripe runs around one of the 33 pairs of opposite faces. Fixing such a band, the four faces it passes through must be aligned, with probability (12)4=116,\left(\dfrac{1}{2}\right)^4 = \dfrac{1}{16}, while the two remaining faces are free. The 33 possible bands are disjoint events, so the probability is 3⋅116=316.3 \cdot \dfrac{1}{16} = \dfrac{3}{16}.
Thus, the correct answer is B.
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