AMC12 2025 B

AMC12 2025 B · Q20

AMC12 2025 B · Q20. It mainly tests Probability (basic), Recursion & DP style counting (basic).

A frog hops along the number line according to the following rules: What is the probability that the frog reaches $4?$

一只青蛙沿数轴跳跃，按照以下规则：青蛙到达$4$的概率是多少？

(A) \frac1{101} \frac1{101}

(B) \frac 1{100} \frac 1{100}

(D) \frac1{98} \frac1{98}

(E) \frac1{97} \frac1{97}

Answer

Correct choice: (E)

正确答案：（E）

Solution

We will solve this using states. Let $P_n$ be the probability of reaching $4$, given that you start from $n$. We want to find $P_0$. Of course, $P_4 = 1$. We also know that \[P_3 = \frac{1}{4}P_4 + \frac{1}{4}P_2\] \[P_2 = \frac{1}{4}P_3 + \frac{1}{4}P_1\] \[P_1 = \frac{1}{4}P_2 + \frac{1}{4}P_0\] \[P_0 = \frac{1}{2}P_1.\] Solving the system, we find that $P_1 = \frac{2}{97}$ and $P_0 = \boxed{\frac{1}{97}}.$

使用状态法。设$P_n$为从$n$出发到达$4$的概率。求$P_0$。显然$P_4 = 1$。另 \[P_3 = \frac{1}{4}P_4 + \frac{1}{4}P_2\] \[P_2 = \frac{1}{4}P_3 + \frac{1}{4}P_1\] \[P_1 = \frac{1}{4}P_2 + \frac{1}{4}P_0\] \[P_0 = \frac{1}{2}P_1.\] 解方程组，得$P_1 = \frac{2}{97}$，$P_0 = \boxed{\frac{1}{97}}$。

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