Difference between revisions of "2024 AMC 10A Problems/Problem 17"

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You make me sad because you wanna cheat.
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==Problem==
Do you wanna make me sad???? 😢 😭 🙁
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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>?
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<math>\textbf{(A) } 10 \qquad \textbf{(B) } 11 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 13 \qquad \textbf{(E) } 14</math>
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==Solution==
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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>.
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~eevee9406
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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.
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~andliu766
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== Video Solution 1 by Pi Academy ==
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https://youtu.be/fW7OGWee31c?si=oq7toGPh2QaksLHE
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==Video Solution2 by SpreadTheMathLove==
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https://www.youtube.com/watch?v=6SQ74nt3ynw
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==See also==
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{{AMC10 box|year=2024|ab=A|num-b=16|num-a=18}}
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{{MAA Notice}}

Latest revision as of 21:49, 13 November 2024

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://youtu.be/fW7OGWee31c?si=oq7toGPh2QaksLHE

Video Solution2 by SpreadTheMathLove

https://www.youtube.com/watch?v=6SQ74nt3ynw


See also

2024 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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