AMC10 2023 A

AMC10 2023 A · Q7

AMC10 2023 A · Q7. 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掷一个标准的$6$面骰子$4$次，并保持她掷出的数字的累积总和。累积总和在某个时刻等于$3$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 $\frac{1}{6}+\frac{1}{18}+\frac{1}{216}=\boxed{\textbf{(B) }\frac{49}{216}}$.

累积总和等于$3$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积总和$3$，即$(3)$，概率为$\frac{1}{6}$。情况2：恰好两次掷骰得到累积总和$3$，即$(1, 2)$和$(2, 1)$，概率为$\frac{1}{6}\cdot\frac{1}{6}\cdot2=\frac{1}{18}$。情况3：恰好三次掷骰得到累积总和$3$，即$(1, 1, 1)$，概率为$\frac{1}{6}\cdot\frac{1}{6}\cdot\frac{1}{6}=\frac{1}{216}$。使用加法原理，$\frac{1}{6}+\frac{1}{18}+\frac{1}{216}=\boxed{\textbf{(B) }\frac{49}{216}}$。

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