Question

蚂蚁 Amelia 从数轴上的 $0$ 出发，并按如下方式爬行。对于 $n = 1, 2, 3$ ，Amelia 独立地从区间 $(0, 1)$ 中均匀随机选取持续时间 $t_n$ 和位移增量 $x_n$ 。在过程的第 $n$ 步，她沿正方向移动 $x_n$ 个单位，用时 $t_n$ 分钟。如果总用时在第 $n$ 步中已超过 $1$ 分钟，她就在该步结束时停止；否则继续下一步，最多共走 $3$ 步。Amelia 停止时的位置大于 $1$ 的概率是多少？

Ant Amelia starts on the number line at $0$ and crawls in the following manner. For $n = 1, 2, 3,$ Amelia chooses a time duration $t_n$ and an increment $x_n$ independently and uniformly at random from the interval $(0, 1).$ During the $n$ th step of the process, Amelia moves $x_n$ units in the positive direction, using up $t_n$ minutes. If the total elapsed time has exceeded $1$ minute during the $n$ th step, she stops at the end of that step; otherwise, she continues with the next step, taking at most $3$ steps in all. What is the probability that Amelia's position when she stops will be greater than $1?$

$\dfrac13$

$\dfrac12$

$\dfrac23$

$\dfrac34$

$\dfrac56$

答案：C

知识点：几何概率独立事件

难度评级：2110

解答：

因为每个 $t_n \lt 1$ ，Amelia 一定至少完成两步。她恰好两步后停止当且仅当 $t_1 + t_2 \gt 1$ ，其概率为 $\tfrac12$ ；否则她走满三步。

位移与时间独立。若走两步，位置为 $x_1 + x_2$ ，且 $P(x_1 + x_2 \gt 1) = \tfrac12$ 。若走三步，位置为 $x_1 + x_2 + x_3$ ，且 $P(x_1 + x_2 + x_3 \gt 1)$ $= 1 - \tfrac16 = \tfrac56$ 。

所求概率为 $\tfrac12 \cdot \tfrac12 + \tfrac12 \cdot \tfrac56 = \tfrac14 + \tfrac{5}{12} = \tfrac23$ 。

所以正确答案是 C。

Because each $t_n \lt 1,$ Amelia always completes at least two steps. She stops after exactly two steps when $t_1 + t_2 \gt 1,$ which happens with probability $\tfrac12;$ otherwise she takes all three steps.

The increments are independent of the times. If she takes two steps, her position is $x_1 + x_2,$ and $P(x_1 + x_2 \gt 1) = \tfrac12.$ If she takes three, her position is $x_1 + x_2 + x_3,$ and $P(x_1 + x_2 + x_3 \gt 1)$ $= 1 - \tfrac16 = \tfrac56.$

The answer is $\tfrac12 \cdot \tfrac12 + \tfrac12 \cdot \tfrac56 = \tfrac14 + \tfrac{5}{12} = \tfrac23.$

Thus, the correct answer is C.

2022 AMC 12B 第 22 题

答案：C

难度评级：2110

解答：

其他年份的第 22 题