AMC10 2024 A

AMC10 2024 A · Q17

AMC10 2024 A · Q17. It mainly tests Probability (basic), Conditional probability (basic).

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 $m + n$?

两支队伍进行三局两胜的季后赛：最多打$3$场，季后赛获胜者是率先赢得$2$场比赛的队伍。第一场比赛在A队主场进行，其余比赛在B队主场进行。A队主场获胜概率为$\frac{2}{3}$，客场获胜概率为$p$。比赛结果独立。A队赢得季后赛的概率为$\frac{1}{2}$。则$p$可写成$\frac{1}{2}(m - \sqrt{n})$的形式，其中$m,n$为正整数。求$m + n$？

(A) 10 10

(B) 11 11

(D) 13 13

(E) 14 14

Answer

Correct choice: (E)

正确答案：（E）

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 $4+10=\boxed{\textbf{(E) } 14}$. 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.

A队获胜仅有三种情况：AA、ABA和BAA（A表示A队胜，B表示B队胜）。因此总概率为$\frac{2}{3}\cdot p + \frac{2}{3}(1-p)p + \frac{1}{3}p^2 = \frac{1}{2}$。两边乘以6得$4p + 4p(1-p) + 2p^2 = 3$，即$2p^2 - 8p + 3 = 0$，解得$p = \frac{4 \pm \sqrt{10}}{2}$。由于答案应为$\frac{1}{2}(m - \sqrt{n})$形式，且$p$为概率介于$0$和$1$之间，故取$p = \frac{4 - \sqrt{10}}{2} = \frac{1}{2}(4 - \sqrt{10})$，答案为$4 + 10 = \boxed{\textbf{(E) } 14}$。另一种理解是减法而非加法的方式，即意识到$p$作为概率介于$0$和$1$之间。

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