AMC10 2023 B

AMC10 2023 B · Q19

AMC10 2023 B · Q19. It mainly tests Probability (basic), Geometric probability (basic).

Sonya the frog chooses a point uniformly at random lying within the square $[0, 6]$ $\times$ $[0, 6]$ in the coordinate plane and hops to that point. She then randomly chooses a distance uniformly at random from $[0, 1]$ and a direction uniformly at random from {north, south, east, west}. All of her choices are independent. She now hops the distance in the chosen direction. What is the probability that she lands outside the square?

青蛙Sonya在坐标平面上的正方形$[0, 6]$ $\times$ $[0, 6]$内均匀随机选择一个点并跳到该点。然后她均匀随机从$[0, 1]$选择一个距离，并从{north, south, east, west}均匀随机选择一个方向。她的所有选择相互独立。她现在沿选择的方跳跃该距离。求她落在正方形外的概率？

(A) \frac{1}{6} \frac{1}{6}

(B) \frac{1}{12} \frac{1}{12}

(D) \frac{1}{10} \frac{1}{10}

(E) \frac{1}{9} \frac{1}{9}

Answer

Correct choice: (B)

正确答案：（B）

Solution

WLOG, we assume Sonya jumps $0.5$ units every time, since that is her expected value. If Sonya is within $0.5$ blocks of an edge, she can jump off the board. Let us examine the region that is at most $0.5$ blocks from exactly one edge. If Sonya starts in this region, she has a $\dfrac14$ chance of landing outside (there's exactly one direction she can hop to get out). The total area of this region is $4\cdot0.5\cdot5=10.$ For this region, Sonya has a $\dfrac14$ chance, so we multiply $10$ by $\dfrac14$ to get $2.5.$ If Sonya is in one of the corner squares, she can go two directions to get out, so she has a $\dfrac24=\dfrac12$ chance to get out. The total area is $0.5\cdot0.5\cdot4=1$, so this region yields $\dfrac12\cdot1=\dfrac12.$ Adding the two, we get $3$. This is out of $36$ square units of area, so our answer is thus $\boxed{\textbf{(B) }\tfrac{1}{12}}.$ Note: When Sonya is within 0 to 1 units away from the border in a given direction, the probability that she will jump outside the border can be thought of as a function of distance from the border. It is easy to see that if Sonya is x units away from the border, a jump distance in the interval [x,1] will move her outside. Thus the probability is a function P(x)=1−x. The probability in an entire region 0 to 1 units away from the border is an integral of the function, ∫01(1−x)dx=[x−12x2]01=1/2. This is why the probability is 1/2.

不失一般性，我们假设Sonya每次跳$0.5$单位，因为那是她的期望值。如果Sonya距离边缘在$0.5$块以内，她可以跳出棋盘。考察距离恰好一个边缘至多$0.5$块的区域。如果Sonya从这个区域开始，她有$\frac{1}{4}$的概率落在外面（恰好有一个方向能出去）。该区域总面积为$4\cdot0.5\cdot5=10$。对于这个区域，Sonya有$\frac{1}{4}$概率，所以乘$10$得$2.5$。如果Sonya在角落方块之一，她有两个方向能出去，所以有$\frac{2}{4}=\frac{1}{2}$概率出去。该区域总面积为$0.5\cdot0.5\cdot4=1$，所以该区域贡献$\frac{1}{2}\cdot1=\frac{1}{2}$。相加得$3$。总面积$36$平方单位，因此答案为$\boxed{\textbf{(B) }\tfrac{1}{12}}$。注：当Sonya距离边界在$0$到$1$单位时，在给定方向跳出边界的概率可视为距离边界的函数。易见如果Sonya距离边界$x$单位，跳跃距离在$[x,1]$区间会让她出去。因此概率函数$P(x)=1-x$。距离边界$0$到$1$整个区域的概率是积分$\int_0^1(1-x)\,dx=[x-\frac{1}{2}x^2]_0^1=\frac{1}{2}$。这就是概率为$1/2$的原因。

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