2024 AMC 10A Problems/Problem 17: Difference between revisions

Latest revision as of 19:45, 21 October 2025

Problem

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 $\frac{2}{3}$ 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 $\frac{1}{2}$ . Then $p$ can be written in the form $\frac{1}{2}(m - \sqrt{n})$ , where $m$ and $n$ are positive integers. What is $m + n$ ?

$\textbf{(A) } 10 \qquad \textbf{(B) } 11 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 13 \qquad \textbf{(E) } 14$

Solution

We only have three cases where A wins: AA, ABA, and BAA (A denotes a team A win and B denotes a team B win). Knowing this, we can sum up the probability of each case. Thus the total probability is $\frac{2}{3}p+\frac{2}{3}(1-p)p+\frac{1}{3}p^2=\frac{1}{2}$ . Multiplying both sides by 6 yields $4p+4p(1-p)+2p^2=3$ , so $2p^2-8p+3=0$ and we find that $p=\frac{4\pm\sqrt{10}}{2}$ . Luckily, we know that the answer should contain $\frac{1}{2}(m - \sqrt{n})$ , so the solution is $p=\frac{4-\sqrt{10}}{2}=\frac{1}{2}(4-\sqrt{10})$ and the answer is $4+10=\boxed{\textbf{(E) } 14}$ .

~eevee9406

Another way to see the answer is subtraction and not addition is to realize that $p$ is between $0$ and $1$ since it is a probability. ~andliu766

Video Solution 1 by Pi Academy

https://youtube/fW7OGWee31c?si=oq7toGPh2QaksLHE

Video Solution 2 by SpreadTheMathLove

https://youtu.be/Db5nW_t-iP8?si=Ywz_NKciPRGZqInr

Video Solution 3 by TheNeuralMathAcademy

https://www.youtube.com/watch?v=4b_YLnyegtw&t=3361s

2024 AMC 10A (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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+==Problem==
+Two teams are in a best-two-out-of-three playoff: the teams will play at most <math>3</math> games, and the winner of the playoff is the first team to win <math>2</math> 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 <math>\frac{2}{3}</math> chance of winning at home, and its probability of winning when playing away from home is <math>p</math>. Outcomes of the games are independent. The probability that Team A wins the playoff is <math>\frac{1}{2}</math>. Then <math>p</math> can be written in the form <math>\frac{1}{2}(m - \sqrt{n})</math>, where <math>m</math> and <math>n</math> are positive integers. What is <math>m + n</math>?
+<math>\textbf{(A) } 10 \qquad \textbf{(B) } 11 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 13 \qquad \textbf{(E) } 14</math>
+==Solution==
+We only have three cases where A wins: AA, ABA, and BAA (A denotes a team A win and B denotes a team B win). Knowing this, we can sum up the probability of each case. Thus the total probability is <math>\frac{2}{3}p+\frac{2}{3}(1-p)p+\frac{1}{3}p^2=\frac{1}{2}</math>. Multiplying  both sides by 6 yields <math>4p+4p(1-p)+2p^2=3</math>, so <math>2p^2-8p+3=0</math> and we find that <math>p=\frac{4\pm\sqrt{10}}{2}</math>. Luckily, we know that the answer should contain  <math>\frac{1}{2}(m - \sqrt{n})</math>, so the solution is <math>p=\frac{4-\sqrt{10}}{2}=\frac{1}{2}(4-\sqrt{10})</math> and the answer is <math>4+10=\boxed{\textbf{(E) } 14}</math>.
+~eevee9406
+Another way to see the answer is subtraction and not addition is to realize that <math>p</math> is between <math>0</math> and <math>1</math> since it is a probability.
+~andliu766
+== Video Solution 1 by Pi Academy ==
+https://youtube/fW7OGWee31c?si=oq7toGPh2QaksLHE
+==Video Solution 2 by SpreadTheMathLove==
+https://youtu.be/Db5nW_t-iP8?si=Ywz_NKciPRGZqInr
+== Video Solution 3 by TheNeuralMathAcademy ==
+https://www.youtube.com/watch?v=4b_YLnyegtw&t=3361s
+==See Also==
+{{AMC10 box|year=2024|ab=A|before=[[2023 AMC 10B Problems]]|after=[[2024 AMC 10B Problems]]}}
+* [[AMC 10]]
+* [[AMC 10 Problems and Solutions]]
+* [[Mathematics competitions]]
+* [[Mathematics competition resources]]
+{{MAA Notice}}

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