Question

Dave 到达一座机场，机场有十二个登机口排成一直线，相邻登机口之间正好相距 $100$ 英尺。他的出发登机口随机分配。在该登机口等待后，Dave 被告知出发登机口改到了另一个不同的登机口，这个新登机口也随机分配。设 Dave 走到新登机口的距离为 $400$ 英尺或更少的概率为分数 $\frac{m}{n}$ ，其中 $m$ 和 $n$ 是互质的正整数。求 $m + n$ 。

Dave arrives at an airport which has twelve gates arranged in a straight line with exactly $100$ feet between adjacent gates. His departure gate is assigned at random. After waiting at that gate, Dave is told the departure gate has been changed to a different gate, again at random. Let the probability that Dave walks $400$ feet or less to the new gate be a fraction $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

答案

答案：52

知识点：基本概率数对计数分类讨论

难度评级：2170

解答：

将登机口编号为 $1$ 到 $12$ 。所有 $12 \cdot 11 = 132$ 个不同（原登机口，新登机口）的有序对等可能，且 Dave 走 $400$ 英尺或更少，当且仅当登机口编号相差至多 $4$ 。

登机口 $i$ 有 $\min(i + 4, 12) - \max(i - 4, 1)$ 个合格的新登机口：登机口 $1$ 和 $12$ 各有 $4$ 个，登机口 $2$ 和 $11$ 各有 $5$ 个，登机口 $3$ 和 $10$ 各有 $6$ 个，登机口 $4$ 和 $9$ 各有 $7$ 个，登机口 $5$ 到 $8$ 各有 $8$ 个。总数为 $2(4 + 5 + 6 + 7) + 4 \cdot 8 = 76$ 。

概率为 $\frac{76}{132} = \frac{19}{33}$ ，所以 $m + n = 19 + 33 = 52$ 。

Number the gates $1$ through $12.$ All $12 \cdot 11 = 132$ ordered pairs of distinct (old, new) gates are equally likely, and Dave walks $400$ feet or less exactly when the gate numbers differ by at most $4.$

A gate $i$ has $\min(i + 4, 12) - \max(i - 4, 1)$ qualifying new gates: gates $1$ and $12$ have $4$ each, gates $2$ and $11$ have $5,$ gates $3$ and $10$ have $6,$ gates $4$ and $9$ have $7,$ and gates $5$ through $8$ have $8$ each. The total is $2(4 + 5 + 6 + 7) + 4 \cdot 8 = 76.$

The probability is $\frac{76}{132} = \frac{19}{33},$ so $m + n = 19 + 33 = 52.$

2010 AIME II 第 4 题

答案：52

难度评级：2170

解答：

其他年份的第 4 题