AMC12 2023 A

AMC12 2023 A · Q17

AMC12 2023 A · Q17. It mainly tests Probability (basic), Counting & probability misc.

Flora the frog starts at 0 on the number line and makes a sequence of jumps to the right. In any one jump, independent of previous jumps, Flora leaps a positive integer distance $m$ with probability $\frac{1}{2^m}$. What is the probability that Flora will eventually land at 10?

青蛙 Flora 从数轴上的 0 开始，向右进行一系列跳跃。在任意一次跳跃中，与之前的跳跃无关，Flora 以概率 $\frac{1}{2^m}$ 跳跃正整数距离 $m$。 Flora 最终落在 10 上的概率是多少？

(A) \frac{5}{512} \frac{5}{512}

(B) \frac{45}{1024} \frac{45}{1024}

(D) \frac{511}{1024} \frac{511}{1024}

(E) \frac{1}{2} \frac{1}{2}

Answer

Correct choice: (E)

正确答案：（E）

Solution

Initially, the probability of landing at $10$ and landing past $10$ (summing the infinte series) are exactly the same. Landing before 10 repeats this initial condition, with a different irrelevant scaling factor. Therefore, the probability must be $\boxed{\textbf{(E)}~\frac12}$.

最初，落在 10 上的概率和超过 10 的概率（求无穷级数）完全相等。落在 10 之前会重复这个初始条件，只是有一个不同的无关缩放因子。因此，概率一定是 $\boxed{\textbf{(E)}~\frac12}$。

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