Question

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

Solution:

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 $\frac{4}{5}\left(\frac{2}{5} \cdot \frac{2}{3} + \frac{3}{5} \cdot \frac{4}{7}\right) = \frac{4}{5} \cdot \frac{64}{105} = \frac{256}{525}.$ 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 Problem 2

Solution:

Problem 2 in Other Years