2024 AMC 10A Problems/Problem 17

Problem

nugget bucket 67 41 monster

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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