AMC12 2019 B

AMC12 2019 B · Q16

AMC12 2019 B · Q16. It mainly tests Probability (basic), Markov / process probability (rare).

Lily pads numbered from 0 to 11 lie in a row on a pond. Fiona the frog sits on pad 0, a morsel of food sits on pad 10, and predators sit on pads 3 and 6. At each unit of time the frog jumps either to the next higher numbered pad or to the pad after that, each with probability $\frac{1}{2}$, independently from previous jumps. What is the probability that Fiona skips over pads 3 and 6 and lands on pad 10?

池塘上有一排从0到11编号的睡莲。青蛙Fiona坐在0号睡莲上，一块食物在10号睡莲上，捕食者在3号和6号睡莲上。每单位时间，青蛙以概率$\frac{1}{2}$跳到下一个编号的睡莲或其后一个睡莲，且每次跳跃独立于之前跳跃。Fiona跳过3号和6号睡莲并落在10号睡莲上的概率是多少？

(A) $\frac{15}{256}$ $\frac{15}{256}$

(B) $\frac{1}{16}$ $\frac{1}{16}$

(D) $\frac{1}{8}$ $\frac{1}{8}$

(E) $\frac{1}{4}$ $\frac{1}{4}$

Answer

Correct choice: (A)

正确答案：（A）

Solution

Answer (A): In order to clear the first hurdle, the sequence of jump lengths must be 1-1-2 or 2-2. These occur with probabilities $(\frac{1}{2})^3$ and $(\frac{1}{2})^2$, respectively, so the probability of landing safely on pad 4 is $\frac{1}{8}+\frac{1}{4}=\frac{3}{8}$. In order for Fiona to make it safely to pad 7 once she has arrived at pad 4, she must jump 1-2, which has probability $\frac{1}{4}$. Once on pad 7, in order to reach the food the frog must jump 1-2 or 2-1 or 1-1-1, with probability $\frac{1}{4}+\frac{1}{4}+\frac{1}{8}=\frac{5}{8}$; otherwise she skips over the food. Therefore the requested probability is $\frac{3}{8}\cdot\frac{1}{4}\cdot\frac{5}{8}=\frac{15}{256}$.

答案（A）：为了越过第一个障碍，跳跃长度的序列必须是 1-1-2 或 2-2。它们发生的概率分别为 $(\frac{1}{2})^3$ 和 $(\frac{1}{2})^2$，因此安全落在第 4 块垫子上的概率为 $\frac{1}{8}+\frac{1}{4}=\frac{3}{8}$。当 Fiona 到达第 4 块垫子后，为了安全到达第 7 块垫子，她必须跳 1-2，其概率为 $\frac{1}{4}$。到达第 7 块垫子后，为了到达食物，青蛙必须跳 1-2 或 2-1 或 1-1-1，其概率为 $\frac{1}{4}+\frac{1}{4}+\frac{1}{8}=\frac{5}{8}$；否则她会跳过食物。因此所求概率为 $\frac{3}{8}\cdot\frac{1}{4}\cdot\frac{5}{8}=\frac{15}{256}$。

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