Question

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 $m + n?$

$10$

$11$

$12$

$13$

$14$

Solution:

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 Problem 17

Solution:

Problem 17 in Other Years