AMC10 2009 B

AMC10 2009 B · Q25

AMC10 2009 B · Q25. It mainly tests Basic counting (rules of product/sum), Symmetry.

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?

立方体每个面上从一条边的中心画一条窄条纹到对边中心的中心。每面边对选择随机独立。存在一条环绕立方体的连续条纹的概率是多少？

(A) 1/8 1/8

(B) 3/16 3/16

(D) 3/8 3/8

(E) 1/2 1/2

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): The stripe on each face of the cube will be oriented in one of two possible directions, so there are $2^6=64$ possible stripe combinations on the cube. There are 3 pairs of parallel faces so, if there is an encircling stripe, then the pair of faces that do not contribute uniquely determine the stripe orientation for the remaining faces. In addition, the stripe on each face that does not contribute may be oriented in 2 different ways. Thus a total of $3\cdot2\cdot2=12$ stripe combinations on the cube result in a continuous stripe around the cube, and the requested probability is $\frac{12}{64}=\frac{3}{16}$.

答案（B）：立方体每个面的条纹方向有两种可能，因此立方体上一共有 $2^6=64$ 种条纹组合。立方体有 3 对互相平行的面，所以如果存在一条环绕的条纹，那么那一对“不起唯一贡献”的面会唯一确定其余各面的条纹方向。另外，这两个“不起贡献”的面上的条纹各自仍可有 2 种取向。因此，使立方体出现连续环绕条纹的组合总数为 $3\cdot2\cdot2=12$，所求概率为 $\frac{12}{64}=\frac{3}{16}$。

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