Question

青蛙 Sonya 在坐标平面中的正方形 $[0, 6] \times [0, 6]$ 内均匀随机选择一个点并跳到该点。然后她从 $[0, 1]$ 中均匀随机选择一个距离，并从 $\{\text{north}, \text{south}, \text{east}, \text{west}\}$ 中均匀随机选择一个方向。所有选择相互独立。她沿所选方向跳所选距离。她落在正方形外的概率是多少？

Sonya the frog chooses a point uniformly at random lying within the square $[0, 6] \times [0, 6]$ in the coordinate plane and hops to that point. She then chooses a distance uniformly at random from $[0, 1]$ and a direction uniformly at random from $\{\text{north}, \text{south}, \text{east}, \text{west}\}.$ All her choices are independent. She now hops the distance in the chosen direction. What is the probability that she lands outside the square?

$\dfrac{1}{6}$

$\dfrac{1}{12}$

$\dfrac{1}{4}$

$\dfrac{1}{10}$

$\dfrac{1}{9}$

答案：B

知识点：几何概率对称性

难度评级：1990

解答：

四个方向由对称性相同。假设她向东跳。她跳出正方形当且仅当原来的 $x$ 坐标加跳跃距离 $d$ 超过 $6$ 。固定 $d$ 时， $x$ 在 $[0, 6]$ 上均匀分布，所以横坐标大于 $6 - d$ 的概率为 $\frac{d}{6}$ 。现在对 $d$ 在 $[0, 1]$ 上取平均，得到 $\frac{1}{6} \cdot \frac{1}{2} = \frac{1}{12}$ 。所以正确答案是 B。

The four directions behave the same by symmetry, so say she hops east. She lands outside exactly when her $x$ -coordinate plus the hop distance $d$ tops $6.$ Fix $d.$ Her $x$ -coordinate is uniform on $[0, 6],$ so it beats $6 - d$ with probability $\frac{d}{6}.$ Now average over $d$ uniform on $[0, 1]:$ $\frac{1}{6} \cdot \frac{1}{2} = \frac{1}{12}.$ Thus, B is the correct answer.

2023 AMC 10B 第 19 题

答案：B

难度评级：1990

解答：

其他年份的第 19 题