AMC8 2008

AMC8 2008 · Q19

AMC8 2008 · Q19. It mainly tests Basic counting (rules of product/sum), Probability (basic).

Eight points are spaced at intervals of one unit around a $2\times 2$ square, as shown. Two of the $8$ points are chosen at random. What is the probability that the points are one unit apart?

八个点围绕 $2\times 2$ 正方形以一单位间隔等距放置，如图所示。从 8 个点中随机选择两个点。这两个点相距一单位的概率是多少？

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

(B) \frac{2}{7} \frac{2}{7}

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

(E) \frac{4}{7} \frac{4}{7}

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): Choose two points. Any of the 8 points can be the first choice, and any of the 7 other points can be the second choice. So there are $8 \times 7 = 56$ ways of choosing the points in order. But each pair of points is counted twice, so there are $\frac{56}{2} = 28$ possible pairs. Label the eight points as shown. Only segments $\overline{AB}$, $\overline{BC}$, $\overline{CD}$, $\overline{DE}$, $\overline{EF}$, $\overline{FG}$, $\overline{GH}$ and $\overline{HA}$ are 1 unit long. So 8 of the 28 possible segments are 1 unit long, and the probability that the points are one unit apart is $\frac{8}{28}=\frac{2}{7}$.

答案（B）：选取两点。8 个点中的任意一个都可以作为第一个选择，其余 7 个点中的任意一个都可以作为第二个选择。所以按顺序选点共有 $8 \times 7 = 56$ 种方法。但每一对点被计算了两次，因此共有 $\frac{56}{2} = 28$ 对可能的点对。按图所示给八个点标记。只有线段 $\overline{AB}$、$\overline{BC}$、$\overline{CD}$、$\overline{DE}$、$\overline{EF}$、$\overline{FG}$、$\overline{GH}$ 和 $\overline{HA}$ 的长度为 1 个单位。因此，在 28 条可能的线段中有 8 条长度为 1 个单位，两点相距 1 个单位的概率为 $\frac{8}{28}=\frac{2}{7}$。

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