AMC12 2003 A

AMC12 2003 A · Q22

AMC12 2003 A · Q22. It mainly tests Basic counting (rules of product/sum), Probability (basic).

Objects $A$ and $B$ move simultaneously in the coordinate plane via a sequence of steps, each of length one. Object $A$ starts at $(0,0)$ and each of its steps is either right or up, both equally likely. Object $B$ starts at $(5,7)$ and each of its steps is either to the left or down, both equally likely. Which of the following is closest to the probability that the objects meet?

物体 $A$ 和 $B$ 同时在坐标平面中通过一系列步长为 1 的步子移动。物体 $A$ 从 $(0,0)$ 出发，每一步要么向右要么向上，两种等可能。物体 $B$ 从 $(5,7)$ 出发，每一步要么向左要么向下，两种等可能。以下哪个最接近两物体相遇的概率？

(A) 0.10 0.10

(B) 0.15 0.15

(D) 0.25 0.25

(E) 0.30 0.30

Answer

Correct choice: (C)

正确答案：（C）

Solution

If $A$ and $B$ meet, their paths connect $(0,0)$ and $(5,7).$ There are $\binom{12}{5}=792$ such paths. Since the path is $12$ units long, they must meet after each travels $6$ units, so the probability is $\frac{792}{2^{6}\cdot 2^{6}} \approx 0.20 \Rightarrow \boxed{C}$. Note: The number of paths, $\binom{12}{5}$ comes from the fact that there must be 5 ups/downs and 7 lefts/rights in one path. WLOG, for Object A, the number of paths would be the amount of combinations of the sequence of letters with 5 "U"s 7 "R"s (i.e. UUUUURRRRRRR). This is $\frac{12!}{5!7!}$, which is equivalent to $\binom{12}{5}$.

若 $A$ 与 $B$ 相遇，则它们的路径连接 $(0,0)$ 与 $(5,7).$ 这样的路径共有 $\binom{12}{5}=792$ 条。由于该路径长度为 $12$，它们必须在各自走了 $6$ 个单位后相遇，因此所求概率为 $\frac{792}{2^{6}\cdot 2^{6}} \approx 0.20 \Rightarrow \boxed{C}$。注：路径条数 $\binom{12}{5}$ 来自于一条路径中必须有 5 次上/下与 7 次左/右。无损失一般性地，对物体 $A$，路径数等于由 5 个 "U" 与 7 个 "R" 组成的序列的排列数（例如 UUUUURRRRRRR）的组合数，即 $\frac{12!}{5!7!}$，这等于 $\binom{12}{5}$。

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