2025 AMC 12B 第 20 题

Question

一只青蛙按如下规则在数轴上跳跃。

• 它从

0

开始。

• 如果它在

0

，那么它移动到

1

的概率为

\dfrac{1}{2}

，并以概率

\dfrac{1}{2}

消失。

• 对于

n = 1

，

2

或

3

，如果它在

n

，那么它移动到

n+1

的概率为

\dfrac{1}{4}

，移动到

n-1

的概率为

\dfrac{1}{4}

，并以概率

\dfrac{1}{2}

消失。

青蛙到达

4

的概率是多少？

A frog hops along the number line according to the following rules.

• It starts at

0.

• If it is at

0,

then it moves to

1

with probability

\dfrac{1}{2}

and it disappears with probability

\dfrac{1}{2}.

• For

n = 1,

2,

or

3,

if it is at

n,

then it moves to

n+1

with probability

\dfrac{1}{4},

it moves to

n-1

with probability

\dfrac{1}{4},

and it disappears with probability

\dfrac{1}{2}.

What is the probability that the frog reaches

4?

\dfrac{1}{101}

\dfrac{1}{100}

\dfrac{1}{99}

\dfrac{1}{98}

\dfrac{1}{97}

答案：E

知识点：随机游走递推概率方程组

难度评级：2110

小提示：

令

f(n)

为从

n

出发最终到达

4

的概率；对每个状态写出第一步方程

Let

f(n)

be the probability of eventually reaching

4

starting from

n;

write a first-step equation at each state

大提示：

取

f(4) = 1

，并把青蛙消失后的值看作

0

，解这个线性方程组。

Solve the linear system with

f(4) = 1

and value

0

whenever the frog disappears

解答：

令

f(n)

为从

n

出发最终到达

4

的概率。则

f(0) = \tfrac{1}{2} f(1)

，且对

n = 1, 2, 3

有

f(n) = \tfrac{1}{4} f(n+1) + \tfrac{1}{4} f(n-1)

，并且

f(4) = 1

。依次求解得

f(2) = \tfrac{7}{2} f(1)

和

f(3) = 13 f(1)

；再由

4 \cdot 13 f(1) = 1 + \tfrac{7}{2} f(1)

得

f(1) = \tfrac{2}{97}

，从而

f(0) = \tfrac{1}{97}

。

所以正确答案是 E。

Let

f(n)

be the probability of reaching

4

from

n.

Then

f(0) = \tfrac{1}{2} f(1),

and for

n = 1, 2, 3,

f(n) = \tfrac{1}{4} f(n+1) + \tfrac{1}{4} f(n-1),

with

f(4) = 1.

Solving upward gives

f(2) = \tfrac{7}{2} f(1)

and

f(3) = 13 f(1);

then

4 \cdot 13 f(1) = 1 + \tfrac{7}{2} f(1)

yields

f(1) = \tfrac{2}{97},

so

f(0) = \tfrac{1}{97}.

Thus, the correct answer is E.

2025 AMC 12B 第 20 题

答案：E

难度评级：2110

小提示：

大提示：

解答：

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