2020 AMC 12A 第 11 题

Question

一只青蛙坐在点

(1, 2)

处，开始一系列跳跃。每次跳跃都平行于一条坐标轴，长度为

1

，且每次跳跃的方向（上、下、右、左）都独立随机选择。当青蛙到达顶点为

(0, 0)

、

(0, 4)

、

(4, 4)

和

(4, 0)

的正方形的一条边时，跳跃序列结束。跳跃序列在该正方形的竖直边上结束的概率是多少？

A frog sitting at the point

(1, 2)

begins a sequence of jumps, where each jump is parallel to one of the coordinate axes and has length

1,

and the direction of each jump (up, down, right, or left) is chosen independently at random. The sequence ends when the frog reaches a side of the square with vertices

(0, 0),

(0, 4),

(4, 4),

and

(4, 0).

What is the probability that the sequence of jumps ends on a vertical side of the square?

\dfrac{1}{2}

\dfrac{5}{8}

\dfrac{2}{3}

\dfrac{3}{4}

\dfrac{7}{8}

答案：B

知识点：随机游走方程组对称性

难度评级：1630

小提示：

令

P(x, y)

为在竖直边上结束的概率；它等于四个相邻点处

P

值的平均

Let

P(x, y)

be the probability of ending on a vertical side; it equals the average of

P

at the four neighbors

大提示：

利用对称性：

P(2, 2) = \tfrac12

，而内部点可化为一个小方程组

Use symmetry:

P(2, 2) = \tfrac12,

and the interior points reduce to a small system of equations

解答：

令

P(x, y)

为在竖直边上结束的概率。在竖直边上

P = 1

，在水平边上

P = 0

，在内部点处

P

是四个相邻点概率的平均。

由左右对称性，

P(2, 2) = \tfrac12

。令

a = P(1, 2)

，

b = P(1, 1) = P(1, 3)

，且

c = P(2, 1) = P(2, 3)

。则

a = \tfrac14\left(1 + \tfrac12 + 2b\right)

，

\;b = \tfrac14(1 + c + a)

，且

c = \tfrac14\left(2b + \tfrac12\right)

。

代入可得

b = \tfrac12

，因此

a = \tfrac38 + \tfrac12 b = \tfrac58

。

因此，正确答案是 B。

Let

P(x, y)

be the probability of ending on a vertical side. On a vertical side

P = 1,

on a horizontal side

P = 0,

and at an interior point

P

is the average of its four neighbors.

By left-right symmetry

P(2, 2) = \tfrac12.

Let

a = P(1, 2),

b = P(1, 1) = P(1, 3),

and

c = P(2, 1) = P(2, 3).

Then

a = \tfrac14\left(1 + \tfrac12 + 2b\right),

\;b = \tfrac14(1 + c + a),

and

c = \tfrac14\left(2b + \tfrac12\right).

Substituting gives

b = \tfrac12,

hence

a = \tfrac38 + \tfrac12 b = \tfrac58.

Thus, B is the correct answer.

2020 AMC 12A 第 11 题

答案：B

难度评级：1630

小提示：

大提示：

解答：

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