AMC10 2022 B
AMC10 2022 B · Q23
AMC10 2022 B · Q23. It mainly tests Probability (basic), Counting & probability misc.
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 $0$ 开始,按照以下方式爬行。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 $x_n$。在过程的第 $n$ 步中,Amelia 正向移动 $x_n$ 个单位,使用 $t_n$ 分钟。如果总经过时间在第 $n$ 步期间超过 $1$ 分钟,她在那一步结束时停止;否则,她继续下一步,总共最多 $3$ 步。Amelia 停止时位置大于 $1$ 的概率是多少?
(A)
\frac{1}{3}
\frac{1}{3}
(B)
\frac{1}{2}
\frac{1}{2}
(C)
\frac{2}{3}
\frac{2}{3}
(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 $y$ 是独立均匀分布在区间 $(0,1)$ 中的随机变量。注意 \[P(x+y\leq 1)=\frac{\frac12\cdot1^2}{1^2}=\frac12,\] 如图所示:
设 $x,y,$ 和 $z$ 是独立均匀分布在区间 $(0,1)$ 中的随机变量。注意 \[P(x+y+z\leq 1)=\frac{\frac13\cdot\left(\frac12\cdot1^2\right)\cdot1}{1^3}=\frac16,\] 如图所示:
我们有两个情况:
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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