1986 AMC 8 第 24 题

先试着解答 1986 AMC 8 第 24 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 1986 AMC 8 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。

24.

金氏初中的 600600 名学生午餐时被平均分成三组。每组在不同时间吃午餐。电脑随机把每名学生分到三组之一。三个朋友艾尔、鲍勃和卡罗尔被分到同一午餐组的概率大约是
The 600600 students at King Middle School are divided into three groups of equal size for lunch. Each group has lunch at a different time. A computer randomly assigns each student to one of the three lunch groups. The probability that three friends, Al, Bob, and Carol, will be assigned to the same lunch group is approximately
127\dfrac{1}{27}
19\dfrac19
18\dfrac18
16\dfrac16
13\dfrac13
答案:B
知识点:条件概率无放回抽样
小提示:
固定艾尔的组;再求鲍勃也进同一组的概率
Fix Al’s group; then find the chance that Bob lands in that same group
大提示:
固定艾尔的组后,鲍勃进入该组的概率是 199599\frac{199}{599},接着卡罗尔进入该组的概率是 198598\frac{198}{598}
After Al’s group is fixed, Bob’s chance is 199599\frac{199}{599} and then Carol’s is 198598\frac{198}{598}
解答:
不论艾尔在哪一组,鲍勃进入同一组的概率为 199599\dfrac{199}{599}。如果鲍勃进入该组,卡罗尔也进入该组的概率为 198598\dfrac{198}{598}。
所以三人在同一组的概率约为 199599⋅198598≈0.110\dfrac{199}{599} \cdot \dfrac{198}{598} \approx 0.110,最接近 19\dfrac19。
所以正确答案是 B。
Whatever group Al is in, Bob joins that same group with probability 199599.\dfrac{199}{599}. If Bob does, Carol joins them with probability 198598.\dfrac{198}{598}.
Thus all three share a group with probability 199599⋅198598≈0.110,\dfrac{199}{599} \cdot \dfrac{198}{598} \approx 0.110, closest to 19.\dfrac19.
Thus, the correct answer is B .
LIVE by Po-Shen Loh
LIVE项目

LIVE直播课程

Self-Paced录播课程

启明星Star计划

奖学金项目

关于我们

罗博深是谁?

邮件联系

订阅英文简报

加入我们

申请明星导师

For a more Thought Full world.

© 2026 Expii, Inc.