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：

其他年份的第 25 题