Difference between revisions of "2024 AMC 10A Problems/Problem 17"
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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 by Pi Academy == | ||
| + | https://youtu.be/fW7OGWee31c?si=oq7toGPh2QaksLHE | ||
| − | + | ==See also== | |
| − | + | {{AMC10 box|year=2024|ab=A|num-b=16|num-a=18}} | |
| − | + | {{MAA Notice}} | |
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Latest revision as of 18:33, 10 November 2024
Problem
Two teams are in a best-two-out-of-three playoff: the teams will play at most 
games, and the winner of the playoff is the first team to win 
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 
chance of winning at home, and its probability of winning when playing away from home is 
. Outcomes of the games are independent. The probability that Team A wins the playoff is 
. Then can be written in the form 
, where 
and 
are positive integers. What is 
?
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 
. Multiplying both sides by 6 yields 
, so 
and we find that 
. Luckily, we know that the answer should contain , so the solution is 
and the answer is 
.
~eevee9406
Another way to see the answer is subtraction and not addition is to realize that is between 
and 
since it is a probability.
~andliu766
Video Solution by Pi Academy
https://youtu.be/fW7OGWee31c?si=oq7toGPh2QaksLHE
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
