1982 AMC 12 第 25 题

Question

右图是城市一部分的地图：小长方形是街区，其间的空隙是街道。每天早晨，一名学生从十字路口

A

沿图示街道步行到十字路口

B

，且始终向东或向南走。为了使路线有所变化，每到一个可以选择方向的路口，他都以

\frac12

的概率选择向东或向南，各次选择相互独立。求他在任意一天早晨经过十字路口

C

的概率。

The adjoining figure is a map of part of a city: the small rectangles are blocks and the spaces in between are streets. Each morning a student walks from intersection

A

to intersection

B,

always walking along streets shown, always going east or south. For variety, at each intersection where he has a choice, he chooses with probability

\frac12

(independent of all other choices) whether to go east or south. Find the probability that, on any given morning, he walks through intersection

C.

\frac{11}{32}

\frac12

\frac47

\frac{21}{32}

\frac34

答案：D

知识点：基本概率格路二项概率

难度评级：2090

小提示：

到达

C

等价于在第四次向南走之前完成第三次向东走

Reaching

C

means making the third eastward move before the fourth southward move

大提示：

按第三次向东走之前向南走的次数

i=0,1,2,3

分类

Condition on the number

i=0,1,2,3

of south moves made before the third east move

解答：

若在第三次向东走之前已经向南走了

i

次，则到达

C

的最后一步是向东走，而此前的

2+i

步中有两步向东。因此

\begin{aligned} \Pr(C) &=\sum_{i=0}^3\binom{2+i}{2}2^{-(3+i)}\\ &=\frac18+\frac3{16}+\frac6{32}\\ &\quad+\frac{10}{64}\\ &=\frac{21}{32} \end{aligned}\text{。}

因此，正确答案为 D。

If

i

south moves occur before the third east move, the final step to

C

is east and the preceding

2+i

steps contain two east moves. Thus

\begin{aligned} \Pr(C) &=\sum_{i=0}^3\binom{2+i}{2}2^{-(3+i)}\\ &=\frac18+\frac3{16}+\frac6{32}\\ &\quad+\frac{10}{64}\\ &=\frac{21}{32}. \end{aligned}

Therefore, the correct answer is D.

1982 AMC 12 第 25 题

答案：D

难度评级：2090

小提示：

大提示：

解答：

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