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

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

@@ Line 5: / Line 5: @@
 ==Solution==
-We only have three cases: 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>.
+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://youtu.be/fW7OGWee31c?si=oq7toGPh2QaksLHE
+==Video Solution2 by SpreadTheMathLove==
+https://www.youtube.com/watch?v=6SQ74nt3ynw
+==See also==
+{{AMC10 box|year=2024|ab=A|num-b=16|num-a=18}}
+{{MAA Notice}}

2024 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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