2021 AMC 10A Spring 第 23 题

Question

青蛙 Frieda 在一个

3 \times 3

方格网中开始一串跳跃，每次跳一格，并随机选择跳跃方向：上、下、左、右。她不斜着跳。如果某次跳跃方向会让 Frieda 跳出方格网，她会“绕回”并跳到相对的边。例如，如果 Frieda 从中心格开始并连续向上跳两次，第一次会到达上排中间格，第二次会让她跳到相对边，落在下排中间格。

假设 Frieda 从中心格开始，最多随机跳四次，并且一旦落在角格就停止。她在四次跳跃中的某一次到达角格的概率是多少？

Frieda the frog begins a sequence of hops on a

3 \times 3

grid of squares, moving one square on each hop and choosing at random the direction of each hop—up, down, left, or right. She does not hop diagonally. When the direction of a hop would take Frieda off the grid, she “wraps around” and jumps to the opposite edge. For example if Frieda begins in the center square and makes two hops “up”, the first hop would place her in the top row middle square, and the second hop would cause Frieda to jump to the opposite edge, landing in the bottom row middle square.

Suppose Frieda starts from the center square, makes at most four hops at random, and stops hopping if she lands on a corner square. What is the probability that she reaches a corner square on one of the four hops?

\dfrac{9}{16}

\dfrac{5}{8}

\dfrac{3}{4}

\dfrac{25}{32}

\dfrac{13}{16}

答案：D

知识点：随机游走分类讨论

难度评级：1720

小提示：

把位置分为中心格、边格和角格。

Classify positions as center, edge, or corner

大提示：

计算第

2

、

3

、

4

次跳跃首次到达角格的方式。

Count the ways to first hit a corner by hop

2,

3,

or

4

解答：

用

M

表示中心格，

E

表示非角的边格，

C

表示角格。Frieda 从

M

出发，第一次跳一定到

E

。

从一个边格出发，到

C,E,M

的概率分别为

\frac12,\frac14,\frac14

。从

M

出发下一步一定到

E

。

现在列出四次以内首次到达角格的状态模式及其概率。

EC:\quad 1\cdot\frac12=\frac12\text{，}

EEC:\quad 1\cdot\frac14\cdot\frac12=\frac18\text{，}

EEEC:\quad 1\cdot\frac14\cdot\frac14\cdot\frac12=\frac1{32}\text{，}

EMEC:\quad 1\cdot\frac14\cdot1\cdot\frac12=\frac18\text{。}

把这些概率相加：

\frac12+\frac18+\frac1{32}+\frac18=\frac{25}{32}\text{。}

所以正确答案是 D。

Classify a square as

M

for the center,

E

for a non-corner edge square, and

C

for a corner. Frieda starts at

M,

and the first hop always takes her to an

E.

From an edge square, the probabilities of moving to

C,E,M

are

\frac12,\frac14,\frac14,

respectively. From

M,

the next hop always goes to an

E.

Now count the possible first-hit patterns within four hops:

EC:\quad 1\cdot\frac12=\frac12,

EEC:\quad 1\cdot\frac14\cdot\frac12=\frac18,

EEEC:\quad 1\cdot\frac14\cdot\frac14\cdot\frac12=\frac1{32},

EMEC:\quad 1\cdot\frac14\cdot1\cdot\frac12=\frac18.

Adding gives

\frac12+\frac18+\frac1{32}+\frac18=\frac{25}{32}.

Thus, D is the correct answer.

2021 AMC 10A Spring 第 23 题

答案：D

难度评级：1720

小提示：

大提示：

解答：

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