Question

Amelia 有一枚正面朝上的概率为 $\frac{1}{3}$ 的硬币，Blaine 有一枚正面朝上的概率为 $\frac{2}{5}$ 的硬币。Amelia 和 Blaine 轮流抛自己的硬币，直到有人抛出正面；第一个抛出正面的人获胜。所有抛硬币事件相互独立。Amelia 先抛。Amelia 获胜的概率为 $\frac{p}{q}$ ，其中 $p$ 、 $q$ 是互质正整数。求 $q-p$ 。

Amelia has a coin that lands heads with probability $\frac{1}{3},$ and Blaine has a coin that lands on heads with probability $\frac{2}{5}.$ Amelia and Blaine alternately toss their coins until someone gets a head; the first one to get a head wins. All coin tosses are independent. Amelia goes first. The probability that Amelia wins is $\frac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. What is $q-p?$

$1$

$2$

$3$

$4$

$5$

答案：D

知识点：递推概率几何分布

难度评级：1480

小提示：

设 $x$ 为从 Amelia 回合开始时她最终获胜的概率。

Let $x$ be Amelia’s chance to win from the start of her turn

大提示：

如果两人都抛出反面，局面会回到原样。

After both players toss tails, the same situation repeats

解答：

设 Amelia 获胜的概率为 $x$ 。

Amelia 第一次抛硬币就获胜的概率为 $\frac{1}{3}$ 。

若她抛出反面，只有 Blaine 也抛出反面，游戏才会回到原来的状态。Blaine 抛出反面的概率为 $\frac{3}{5}$ 。

两人都抛出反面的概率为 $\dfrac{2}{3} \cdot \dfrac{3}{5} = \dfrac{2}{5}\text{。}$

此时又轮到 Amelia，她仍以概率 $x$ 获胜。

因此得到方程 $x = \dfrac{1}{3} + \dfrac{2}{5}x$ $\dfrac{3}{5}x = \dfrac{1}{3}$ $x = \dfrac{5}{9}\text{。}$

分母与分子的差为 $9 - 5 = 4$ 。

所以正确答案是 D。

Let $x$ be the probability that Amelia wins.

There is a $\frac{1}{3}$ chance Amelia wins off her first flip.

If she gets tails, Blaine must also get tails for the game to return to the same state; this happens with probability $\frac{3}{5}.$

The total probability of this case is $\dfrac{2}{3} \cdot \dfrac{3}{5} = \dfrac{2}{5}.$

The game then goes back to Amelia, who then again has a $x$ chance of winning.

Therefore, we get the following equation. $x = \dfrac{1}{3} + \dfrac{2}{5}x$ $\dfrac{3}{5}x = \dfrac{1}{3}$ $x = \dfrac{5}{9}.$

The difference between the denominator and numerator is $9 - 5 = 4.$

Thus, D is the correct answer.

2017 AMC 10A 第 18 题

答案：D

难度评级：1480

小提示：

大提示：

解答：

其他年份的第 18 题