Question

两支队伍进行三局两胜制季后赛：最多打 $3$ 场，先赢 $2$ 场的队伍获胜。第一场在 A 队主场进行，剩余场次在 B 队主场进行。A 队主场获胜概率为 $\tfrac23$ ，客场获胜概率为 $p$ 。各场结果相互独立。A 队赢得系列赛的概率为 $\tfrac12$ 。此时 $p$ 可写成 $\tfrac12\!\left(m - \sqrt{n}\right)$ ，其中 $m$ 、 $n$ 为正整数。求 $m + n$ ？

Two teams are in a best-two-out-of-three playoff: the teams will play at most $3$ games, and the winner of the playoff is the first team to win $2$ games. The first game is played on Team A's home field, and the remaining games are played on Team B's home field. Team A has a $\tfrac23$ chance of winning at home, and its probability of winning when playing away from home is $p.$ Outcomes of the games are independent. The probability that Team A wins the playoff is $\tfrac12.$ Then $p$ can be written in the form $\tfrac12\!\left(m - \sqrt{n}\right),$ where $m$ and $n$ are positive integers. What is $m + n?$

$10$

$11$

$12$

$13$

$14$

答案：E

知识点：独立事件二次方程分类讨论

难度评级：1800

解答：

A 队第 $1$ 场主场获胜概率为 $\tfrac23$ ，后两场客场每场获胜概率为 $p$ 。A 队赢得系列赛有三种互斥方式：赢第 $1, 2$ 场；赢第 $1$ 场、输第 $2$ 场、赢第 $3$ 场；输第 $1$ 场、赢第 $2, 3$ 场。因此 $\tfrac23 p + \tfrac23(1 - p)p + \tfrac13 p^2 = \tfrac12$ 。化简得 $2p^2 - 8p + 3 = 0$ ，合理根为 $p = \tfrac{4 - \sqrt{10}}{2} = \tfrac12\!\left(4 - \sqrt{10}\right)$ 。所以 $m = 4$ 、 $n = 10$ ， $m + n = 14$ ，正确答案是 E。

Team A takes game $1$ at home with probability $\tfrac23,$ and each away game with probability $p.$ It can win the playoff three disjoint ways: win games $1, 2;$ win $1,$ lose $2,$ win $3;$ lose $1,$ win $2, 3.$ Adding those, $\tfrac23 p + \tfrac23(1 - p)p + \tfrac13 p^2 = \tfrac12.$ This cleans up to $2p^2 - 8p + 3 = 0,$ so $p = \tfrac{4 - \sqrt{10}}{2} = \tfrac12\!\left(4 - \sqrt{10}\right).$ Then $m = 4,$ $n = 10,$ and $m + n = 14.$ Thus, E is the correct answer.

2024 AMC 10A 第 17 题

答案：E

难度评级：1800

解答：

其他年份的第 17 题