AMC12 2023 A

AMC12 2023 A · Q5

AMC12 2023 A · Q5. It mainly tests Basic counting (rules of product/sum), Probability (basic).

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 $3$?

Janet 掷一次标准的 $6$ 面骰子 $4$ 次，并保持掷骰数字的累加总和。她的累加总和某时刻等于 $3$ 的概率是多少？

(A) \frac{2}{9} \frac{2}{9}

(B) \frac{49}{216} \frac{49}{216}

(D) \frac{17}{72} \frac{17}{72}

(E) \frac{13}{54} \frac{13}{54}

Answer

Correct choice: (B)

正确答案：（B）

Solution

There are $3$ cases where the running total will equal $3$: one roll; two rolls; or three rolls: Case 1: The chance of rolling a running total of $3$, namely $(3)$ in exactly one roll is $\frac{1}{6}$. Case 2: The chance of rolling a running total of $3$ in exactly two rolls, namely $(1, 2)$ and $(2, 1)$ is $\frac{1}{6}\cdot\frac{1}{6}\cdot2=\frac{1}{18}$. Case 3: The chance of rolling a running total of 3 in exactly three rolls, namely $(1, 1, 1)$ is $\frac{1}{6}\cdot\frac{1}{6}\cdot\frac{1}{6}=\frac{1}{216}$. Using the rule of sum we have $\frac{1}{6}+\frac{1}{18}+\frac{1}{216}=\boxed{\textbf{(B) }\frac{49}{216}}$.

累加总和等于 $3$ 有 $3$ 种情况：一次掷骰；两次掷骰；或三次掷骰：情况 1：恰好一次掷骰得到累加总和 $3$，即 $(3)$，概率 $\frac{1}{6}$。情况 2：恰好两次掷骰得到累加总和 $3$，即 $(1, 2)$ 和 $(2, 1)$，概率 $\frac{1}{6}\cdot\frac{1}{6}\cdot2=\frac{1}{18}$。情况 3：恰好三次掷骰得到累加总和 $3$，即 $(1, 1, 1)$，概率 $\frac{1}{6}\cdot\frac{1}{6}\cdot\frac{1}{6}=\frac{1}{216}$。用加法原理，总概率 $\frac{1}{6}+\frac{1}{18}+\frac{1}{216}=\boxed{\textbf{(B) }\frac{49}{216}}$。

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