Question

一所新学校中， $40$ % 的学生是一年级生， $30$ % 是二年级生， $20$ % 是三年级生， $10$ % 是四年级生。所有一年级生都必须上拉丁语课；二年级生中有 $80$ %、三年级生中有 $50$ %、四年级生中有 $20$ % 选择上拉丁语课。随机选一名上拉丁语课的学生，他是二年级生的概率为 $\frac{m}{n}$ ，其中 $m$ 和 $n$ 是互质正整数。求 $m + n$ 。

In a new school $40$ percent of the students are freshmen, $30$ percent are sophomores, $20$ percent are juniors, and $10$ percent are seniors. All freshmen are required to take Latin, and $80$ percent of the sophomores, $50$ percent of the juniors, and $20$ percent of the seniors elect to take Latin. The probability that a randomly chosen Latin student is a sophomore is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

答案

答案：25

知识点：条件概率百分数

难度评级：1750

解答：

假设学校有 $100$ 名学生。上拉丁语课的学生有 $40$ 名一年级生、 $30(0.8) = 24$ 名二年级生、 $20(0.5) = 10$ 名三年级生，以及 $10(0.2) = 2$ 名四年级生，总共 $76$ 人。

随机选一名拉丁语学生是二年级生的概率为 $\frac{24}{76} = \frac{6}{19}$ ，所以 $m + n = 6 + 19 = 25$ 。

Assume the school has $100$ students. The Latin students are then $40$ freshmen, $30(0.8) = 24$ sophomores, $20(0.5) = 10$ juniors, and $10(0.2) = 2$ seniors, for a total of $76.$

The probability that a random Latin student is a sophomore is $\frac{24}{76} = \frac{6}{19},$ so $m + n = 6 + 19 = 25.$

2015 AIME II 第 2 题

答案：25

难度评级：1750

解答：

其他年份的第 2 题