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

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==Solution==
 
==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). Thus the probability is <math>\frac{2}{3}p+\frac{2}{3}(1-p)p+\frac{1}{3}p^2=\frac{1}{2}</math>. Multiplying on both sides 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 a <math>-\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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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). Thus the probability is <math>\frac{2}{3}p+\frac{2}{3}(1-p)p+\frac{1}{3}p^2=\frac{1}{2}</math>. Multiplying on both sides 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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~eevee9406 ~andliu766
  
 
==See also==
 
==See also==
 
{{AMC10 box|year=2024|ab=A|num-b=16|num-a=18}}
 
{{AMC10 box|year=2024|ab=A|num-b=16|num-a=18}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 23:30, 8 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). Thus the probability is $\frac{2}{3}p+\frac{2}{3}(1-p)p+\frac{1}{3}p^2=\frac{1}{2}$. Multiplying on both sides 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 $4+10=\boxed{\textbf{(E) } 14}$.

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.

~eevee9406 ~andliu766

See also

2024 AMC 10A (Problems • Answer Key • Resources)
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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