AMC12 2015 B

AMC12 2015 B · Q9

AMC12 2015 B · Q9. It mainly tests Linear equations, Probability (basic).

Larry and Julius are playing a game, taking turns throwing a ball at a bottle sitting on a ledge. Larry throws first. The winner is the first person to knock the bottle off the ledge. At each turn the probability that a player knocks the bottle off the ledge is $\frac{1}{2}$, independently of what has happened before. What is the probability that Larry wins the game?

Larry和Julius在玩一个游戏，轮流向搁架上瓶子投球。Larry先投。获胜者是第一个把瓶子打下搁架的人。每次投球将瓶子打下搁架的概率为$\frac{1}{2}$，且独立于之前发生的事。求Larry获胜的概率。

(A) $\frac{1}{2}$ $\frac{1}{2}$

(B) $\frac{3}{5}$ $\frac{3}{5}$

(D) $\frac{3}{4}$ $\frac{3}{4}$

(E) $\frac{4}{5}$ $\frac{4}{5}$

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): Let $x$ be the probability that Larry wins the game. Then $x=\frac{1}{2}+\frac{1}{2}\cdot\frac{1}{2}\cdot x$. To see this, note that Larry can win by knocking the bottle off the ledge on his first throw; if he and Julius both miss, then it is as if they started the game all over. Thus $x=\frac{1}{2}+\frac{1}{4}x$, so $\frac{3}{4}x=\frac{1}{2}$ or $x=\frac{2}{3}$.

答案（C）：设 $x$ 为 Larry 赢得比赛的概率。则 $x=\frac{1}{2}+\frac{1}{2}\cdot\frac{1}{2}\cdot x$。为说明这一点，注意 Larry 可以在第一次投掷时把瓶子从台沿上击落从而获胜；如果他和 Julius 都没投中，那么就相当于他们重新开始了游戏。因此 $x=\frac{1}{2}+\frac{1}{4}x$，所以 $\frac{3}{4}x=\frac{1}{2}$，即 $x=\frac{2}{3}$。

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