AMC10 2004 A

AMC10 2004 A · Q5

AMC10 2004 A · Q5. It mainly tests Combinations, Geometry misc.

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?

从所示网格中随机选择三个点。每个三点集被选择的概率相同。点位于同一条直线上的概率是多少？

(A) 1/21 $\frac{1}{21}$

(B) 1/14 $\frac{1}{14}$

(D) 1/7 $\frac{1}{7}$

(E) 2/7 $\frac{2}{7}$

Answer

Correct choice: (C)

正确答案：（C）

Solution

(C) The number of three-point sets that can be chosen from the nine grid points is \[ \binom{9}{3}=\frac{9!}{3!\cdot 6!}=84. \] Eight of these sets consist of three collinear points: 3 sets of points lie on vertical lines, 3 on horizontal lines, and 2 on diagonals. Hence the probability is $8/84=2/21$.

（C）从九个网格点中选取三个点的组合数为 \[ \binom{9}{3}=\frac{9!}{3!\cdot 6!}=84。 \] 其中有 8 组由三点共线组成：有 3 组在竖直线上，3 组在水平线上，2 组在对角线上。因此概率为 $8/84=2/21$。

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