2004 AMC 10A 第 5 题

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所有题目均经美国数学协会(MAA)官方合法授权使用。

5.

从图中网格点随机选取三个点,每组三点被选中的概率相同。这三个点在同一直线上的概率是多少?

A set of three points is chosen randomly from the grid shown. Each three-point set has the same probability of being chosen. What is the probability that the points lie on the same straight line?

121\dfrac{1}{21}

114\dfrac{1}{14}

221\dfrac{2}{21}

17\dfrac{1}{7}

27\dfrac{2}{7}

答案:C
知识点:基本概率组合格点
难度评级:1240
解答:

三点集合总数为 (93)=84. \binom{9}{3} = 84.

共线三点包括 33 行、33 列和 22 条主对角线,共 88 组。

因此所求概率为 884=221. \dfrac{8}{84} = \dfrac{2}{21}.

所以正确答案是 C

The number of three-point sets is (93)=84. \binom{9}{3} = 84.

The collinear triples are the 33 rows, the 33 columns, and the 22 main diagonals, for a total of 8.8.

The probability is therefore 884=221. \dfrac{8}{84} = \dfrac{2}{21}.

Thus, the correct answer is C.

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