2023 AMC 10A 第 7 题

先试着解答 2023 AMC 10A 第 7 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2023 AMC 10A 解答,或核对答案

所有题目均经美国数学协会(MAA)官方合法授权使用。

7.

Janet 掷一个标准的 66 面骰子 44 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 33 的概率是多少?

Janet rolls a standard 66-sided die 44 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?3?

29\dfrac{2}{9}

49216\dfrac{49}{216}

25108\dfrac{25}{108}

1772\dfrac{17}{72}

1354\dfrac{13}{54}

答案:B
知识点:骰子(概率)分类讨论
难度评级:1340
解答:

累加和只能通过开头几次掷骰恰好到达 33,且这些方式互斥:单独掷出 33(概率 16\frac16),再是 1,21,22,12,1(各为 136\frac1{36}),以及 1,1,11,1,1(概率 1216\frac1{216})。相加得 36216+6216+6216+1216=49216\frac{36}{216} + \frac{6}{216} + \frac{6}{216} + \frac{1}{216} = \frac{49}{216}。因此,正确答案是 B

The total can only reach exactly 33 through the opening rolls, and these ways are disjoint: 33 alone (probability 16\frac16), then 1,21,2 and 2,12,1 (each 136\frac1{36}), and 1,1,11,1,1 (probability 1216\frac1{216}). Add them up: 36216+6216+6216+1216=49216.\frac{36}{216} + \frac{6}{216} + \frac{6}{216} + \frac{1}{216} = \frac{49}{216}. Thus, B is the correct answer.

← 第 6 题#6
完整试卷

其他年份的第 7 题