AMC12 2022 B

AMC12 2022 B · Q22

AMC12 2022 B · Q22. It mainly tests Probability (basic), Conditional probability (basic).

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 $1$?

蚂蚁 Amelia 从数轴上的 $0$ 点开始，以以下方式爬行。对于 $n=1,2,3$，Amelia 独立均匀随机地从区间 $(0,1)$ 中选择时间持续时间 $t_n$ 和增量 $x_n$。在第 $n$ 步过程中，Amelia 向正方向移动 $x_n$ 个单位，耗时 $t_n$ 分钟。如果在第 $n$ 步过程中总经过时间超过 $1$ 分钟，她在该步结束时停止；否则，她继续下一步，总共最多 $3$ 步。她停止时的位置大于 $1$ 的概率是多少？

(A) \frac{1}{3} \frac{1}{3}

(B) \frac{1}{2} \frac{1}{2}

(D) \frac{3}{4} \frac{3}{4}

(E) \frac{5}{6} \frac{5}{6}

Answer

Correct choice: (C)

正确答案：（C）

Solution

Let $x$ and $y$ be random variables that are independently and uniformly distributed in the interval $(0,1).$ Note that \[P(x+y\leq 1)=\frac{\frac12\cdot1^2}{1^2}=\frac12,\] as shown below: Let $x,y,$ and $z$ be random variables that are independently and uniformly distributed in the interval $(0,1).$ Note that \[P(x+y+z\leq 1)=\frac{\frac13\cdot\left(\frac12\cdot1^2\right)\cdot1}{1^3}=\frac16,\] as shown below: We have two cases: 1. Amelia takes exactly $2$ steps. 2. Amelia takes exactly $3$ steps. Together, the answer is $\frac14 + \frac{5}{12} = \boxed{\textbf{(C) }\frac{2}{3}}.$

设 $x$ 和 $y$ 为独立均匀分布在区间 $(0,1)$ 中的随机变量。注意到 \[P(x+y\leq 1)=\frac{\frac12\cdot1^2}{1^2}=\frac12,\] 如图所示：设 $x,y,$ 和 $z$ 为独立均匀分布在区间 $(0,1)$ 中的随机变量。注意到 \[P(x+y+z\leq 1)=\frac{\frac13\cdot\left(\frac12\cdot1^2\right)\cdot1}{1^3}=\frac16,\] 如图所示：我们有两种情况： 1. Amelia 恰好走 $2$ 步。 2. Amelia 恰好走 $3$ 步。总和为 $\frac14 + \frac{5}{12} = \boxed{\textbf{(C) }\frac{2}{3}}$。

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