Question

Janet 掷一个标准 $6$ 面骰子 $4$ 次，并持续记录掷出点数的累加和。她的累加和在某一时刻等于 $3$ 的概率是多少？

Janet rolls a standard $6$ -sided die $4$ times and keeps a running total of the numbers she rolls. What is the probability that at some point, her running total will equal $3?$

$\dfrac{2}{9}$

$\dfrac{49}{216}$

$\dfrac{25}{108}$

$\dfrac{17}{72}$

$\dfrac{13}{54}$

答案：B

知识点：骰子（概率）分类讨论

难度评级：1270

解答：

累加和递增，所以它恰好到达 $3$ 当且仅当出现以下互斥开头之一：第一次掷出 $3$ ；前两次为 $1,2$ ；前两次为 $2,1$ ；或前三次为 $1,1,1$ 。

它们的概率之和为 $\begin{gathered} \dfrac16+\dfrac1{36}+\dfrac1{36}+\dfrac1{216}\\ {}=\dfrac{36+6+6+1}{216}\\ {}=\dfrac{49}{216}. \end{gathered}$

所以正确答案是 B。

The running total is increasing, so it hits $3$ exactly when one of these disjoint openings occurs: a first roll of $3;$ rolls $1,2;$ rolls $2,1;$ or rolls $1,1,1.$

Their probabilities are $\begin{gathered} \dfrac16+\dfrac1{36}+\dfrac1{36}+\dfrac1{216}\\ {}=\dfrac{36+6+6+1}{216}\\ {}=\dfrac{49}{216}. \end{gathered}$

Thus, the correct answer is B.

2023 AMC 12A 第 5 题

答案：B

难度评级：1270

解答：

其他年份的第 5 题