2023 AMC 12A 第 5 题

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}

一旦累加和超过

3

，就不可能回到三，所以只有前几次掷骰有关

Once the running total passes

3

it can never return, so only the first few rolls matter

把互斥的开头

3

、

1,2

、

2,1

和

1,1,1

的概率相加

Add the probabilities of the disjoint openings

3;

1,2;

2,1;

and

1,1,1

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}\text{。}

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.

其他年份的第 5 题