Question

从一个立方体的顶点中随机选择三个不同顶点。由这三个顶点确定的平面包含立方体内部点的概率是多少？

Three distinct vertices of a cube are chosen at random. What is the probability that the plane determined by these three vertices contains points inside the cube?

$\dfrac{1}{4}$

$\dfrac{3}{8}$

$\dfrac{4}{7}$

$\dfrac{5}{7}$

$\dfrac{3}{4}$

答案：C

知识点：补集计数正方体组合

难度评级：1860

解答：

三个顶点确定的平面会穿过内部，除非这三个顶点都在同一个面上。

$6$ 个面中每个面给出 $\binom{4}{3} = 4$ 组三顶点，所以同一面上的共有 $6 \cdot 4 = 24$ 组；总数为 $\binom{8}{3} = 56$ 。

穿过内部的概率为 $1 - \dfrac{24}{56} = \dfrac{4}{7}.$

所以正确答案是 C。

Three vertices determine a plane that cuts through the interior unless all three lie on a single face.

Each of the $6$ faces gives $\binom{4}{3} = 4$ triples, so $6 \cdot 4 = 24$ triples lie on a face out of $\binom{8}{3} = 56$ total.

The probability of hitting the interior is $1 - \dfrac{24}{56} = \dfrac{4}{7}.$

Thus, the correct answer is C.

2009 AMC 10A 第 24 题

答案：C

难度评级：1860

解答：

其他年份的第 24 题