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 独立且均匀地从区间 $(0,1)$ 中随机选择一个持续时间 $t_n$ 和一个增量 $x_n$ 。在第 $n$ 步中，Amelia 沿正方向移动 $x_n$ 个单位，用时 $t_n$ 分钟。如果总用时在第 $n$ 步期间已经超过 $1$ 分钟，她会在该步结束时停止；否则继续下一步，最多走 $3$ 步。Amelia 停止时位置大于 $1$ 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 $1?$

$\dfrac 13$

$\dfrac 12$

$\dfrac 23$

$\dfrac 34$

$\dfrac 56$

答案：C

知识点：几何概率独立事件分类讨论

难度评级：2150

解答：

停止时间只依赖时间变量，而最终位置只依赖距离变量，所以相应概率可以相乘。

两个独立 $(0,1)$ 数之和小于 $1$ 的概率是单位正方形中一个直角三角形的面积，即 $\frac12$ 。

三个独立 $(0,1)$ 数之和小于 $1$ 的概率是截距为 $1$ 的四面体体积，即 $\frac16$ 。

若 $t_1+t_2>1$ ，Amelia 走两步后停止。其概率为 $\frac12$ ，且独立地有 $x_1+x_2>1$ 的概率为 $\frac12$ ，贡献 $\frac14$ 。

若 $t_1+t_2<1$ ，Amelia 会走第三步。其概率为 $\frac12$ ，且独立地有 $x_1+x_2+x_3>1$ 的概率为 $1-\frac16=\frac56$ ，贡献 $\frac5{12}$ 。

总概率为 $\frac14+\frac5{12}=\frac23$ 。

所以正确答案是 C。

The stopping time depends only on the time variables, while the final position depends only on the distance variables, so the corresponding probabilities multiply.

For two independent numbers in $(0,1),$ the probability that their sum is less than $1$ is the area of a right triangle, namely $\frac12.$

For three independent numbers in $(0,1),$ the probability that their sum is less than $1$ is the volume of a tetrahedron with side intercepts $1,$ namely $\frac16.$

If $t_1+t_2>1,$ Amelia stops after two steps. This has probability $\frac12,$ and independently $x_1+x_2>1$ has probability $\frac12,$ contributing $\frac14.$

If $t_1+t_2<1,$ Amelia takes the third step. This has probability $\frac12,$ and independently $x_1+x_2+x_3>1$ has probability $1-\frac16=\frac56,$ contributing $\frac5{12}.$

The total probability is $\frac14+\frac5{12}=\frac23.$

Thus, the answer is C .

2022 AMC 10B 第 23 题

答案：C

难度评级：2150

解答：

其他年份的第 23 题