2009 AMC 10B Problem 25

Below is the professionally curated solution for Problem 25 of the 2009 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2009 AMC 10B solutions, or check the answer key.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:basic probabilityindependent eventssymmetry

Difficulty rating: 2090

25.

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\dfrac18

316\dfrac{3}{16}

14\dfrac14

38\dfrac38

12\dfrac12

Solution:

Each face's stripe has 22 orientations, giving 26=642^6=64 equally likely configurations.

An encircling stripe runs around one of the 33 pairs of opposite faces. For a given band, the 44 faces it crosses must each be oriented to continue it, a probability of (12)4=116.\left(\dfrac12\right)^4=\dfrac{1}{16}.

The three bands are mutually exclusive, so the probability is 3116=316.3\cdot\dfrac{1}{16}=\dfrac{3}{16}.

Thus, the correct answer is B.

Problem 25 in Other Years