2017 AMC 10A 第 18 题

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：

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