AMC12 2004 B

AMC12 2004 B · Q20

AMC12 2004 B · Q20. It mainly tests Basic counting (rules of product/sum), Probability (basic).

Each face of a cube is painted either red or blue, each with probability $1/2$. The color of each face is determined independently. What is the probability that the painted cube can be placed on a horizontal surface so that the four vertical faces are all the same color?

一个立方体的每个面以概率 $1/2$ 被涂成红色或蓝色，各面颜色独立决定。求涂色后的立方体能放在水平面上，使得四个竖直侧面全为同一种颜色的概率。

(A) \frac{1}{4} \frac{1}{4}

(B) \frac{5}{16} \frac{5}{16}

(D) \frac{7}{16} \frac{7}{16}

(E) \frac{1}{2} \frac{1}{2}

Answer

Correct choice: (B)

正确答案：（B）

Solution

There are $2^6$ possible colorings of the cube. Consider the color that appears with greater frequency. The property obviously holds true if $5$ or $6$ of the faces are colored the same, which for each color can happen in $6 + 1 = 7$ ways. If $4$ of the faces are colored the same, there are $3$ possible cubes (corresponding to the $3$ possible ways to pick pairs of opposite faces for the other color). If $3$ of the faces are colored the same, the property obviously cannot be satisfied. Thus, there are a total of $2(7 + 3) = 20$ ways for this to occur, and the desired probability is $\frac{20}{2^6} = \frac{5}{16}\ \mathbf{(B)}$.

共有 $2^6$ 种可能的涂色方式。考虑出现次数更多的那种颜色。若有 $5$ 个或 $6$ 个面被涂成同一种颜色，则该性质显然成立；对每种颜色，这分别有 $6 + 1 = 7$ 种方式。若有 $4$ 个面被涂成同一种颜色，则有 $3$ 种可能的立方体（对应于为另一种颜色选择哪一对相对面，共有 $3$ 种选择）。若有 $3$ 个面被涂成同一种颜色，则该性质显然无法满足。因此，共有 $2(7 + 3) = 20$ 种方式使其成立，所求概率为 $\frac{20}{2^6} = \frac{5}{16}\ \mathbf{(B)}$。

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