Question

队伍 $T_1$ 、 $T_2$ 、 $T_3$ 和 $T_4$ 进入季后赛。半决赛中， $T_1$ 对阵 $T_4$ ， $T_2$ 对阵 $T_3$ 。这两场比赛的胜者将在决赛中相遇，决出冠军。当 $T_i$ 对阵 $T_j$ 时， $T_i$ 获胜的概率为 $\frac{i}{i+j}$ ，并且所有比赛结果相互独立。 $T_4$ 获得冠军的概率为 $\frac{p}{q}$ ，其中 $p$ 和 $q$ 是互质的正整数。求 $p + q$ 。

Teams $T_1,$ $T_2,$ $T_3,$ and $T_4$ are in the playoffs. In the semifinal matches, $T_1$ plays $T_4,$ and $T_2$ plays $T_3.$ The winners of those two matches will play each other in the final match to determine the champion. When $T_i$ plays $T_j,$ the probability that $T_i$ wins is $\frac{i}{i+j},$ and the outcomes of all the matches are independent. The probability that $T_4$ will be the champion is $\frac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find $p + q.$

答案

答案：781

知识点：基本概率独立事件分类讨论

难度评级：2070

小提示：

$T_4$ 必须先在半决赛击败 $T_1$ ，再击败 $T_2$ 和 $T_3$ 中赢下另一场半决赛的队伍

$T_4$ must beat $T_1$ in its semifinal, then beat whichever of $T_2$ and $T_3$ wins the other semifinal

大提示：

按另一场半决赛分类： $T_2$ 进入决赛的概率是 $\frac{2}{5}$ ，而 $T_4$ 击败 $T_2$ 或 $T_3$ 的概率分别是 $\frac{4}{6}$ 和 $\frac{4}{7}$

Split into cases by the other semifinal: $T_2$ reaches the final with probability $\frac{2}{5},$ and $T_4$ beats $T_2$ or $T_3$ with probability $\frac{4}{6}$ or $\frac{4}{7}$

解答：

要成为冠军， $T_4$ 首先必须击败 $T_1$ ，概率为 $\frac{4}{4+1} = \frac{4}{5}$ 。另一场半决赛中， $T_2$ 进入决赛的概率为 $\frac{2}{2+3} = \frac{2}{5}$ ， $T_3$ 进入决赛的概率为 $\frac{3}{5}$ ；决赛中， $T_4$ 击败 $T_2$ 的概率为 $\frac{4}{4+2} = \frac{2}{3}$ ，击败 $T_3$ 的概率为 $\frac{4}{4+3} = \frac{4}{7}$ 。

因此 $T_4$ 成为冠军的概率为 $\begin{aligned} &\frac{4}{5}\left(\frac{2}{5} \cdot \frac{2}{3} + \frac{3}{5} \cdot \frac{4}{7}\right) \\ &= \frac{4}{5} \cdot \frac{64}{105} \\ &= \frac{256}{525} \end{aligned}\text{。}$ 因为 $256 = 2^8$ ，且 $525 = 3 \cdot 5^2 \cdot 7$ ，这个分数已经最简，所以 $p + q = 256 + 525 = 781$ 。

To be champion, $T_4$ must first beat $T_1,$ which happens with probability $\frac{4}{4+1} = \frac{4}{5}.$ The other semifinal sends $T_2$ to the final with probability $\frac{2}{2+3} = \frac{2}{5}$ and $T_3$ with probability $\frac{3}{5};$ in the final, $T_4$ beats $T_2$ with probability $\frac{4}{4+2} = \frac{2}{3}$ and beats $T_3$ with probability $\frac{4}{4+3} = \frac{4}{7}.$

The probability that $T_4$ is champion is therefore $\begin{aligned} &\frac{4}{5}\left(\frac{2}{5} \cdot \frac{2}{3} + \frac{3}{5} \cdot \frac{4}{7}\right) \\ &= \frac{4}{5} \cdot \frac{64}{105} \\ &= \frac{256}{525}. \end{aligned}$ Since $256 = 2^8$ and $525 = 3 \cdot 5^2 \cdot 7,$ this is in lowest terms, and $p + q = 256 + 525 = 781.$

2017 AIME II 第 2 题

答案：781

难度评级：2070

小提示：

大提示：

解答：

其他年份的第 2 题