Question

一排睡莲叶依次编号为 $0$ 到 $11$ 。第 $3$ 号和第 $6$ 号睡莲叶上有捕食者，第 $10$ 号睡莲叶上有一小块食物。青蛙 Fiona 从第 $0$ 号睡莲叶出发；从任意一片睡莲叶出发，她有 $\dfrac12$ 的概率跳到下一片，也有同样的概率向前跳 $2$ 片。Fiona 不落在第 $3$ 号或第 $6$ 号睡莲叶上而到达第 $10$ 号的概率是多少？

There are lily pads in a row numbered $0$ to $11,$ in that order. There are predators on lily pads $3$ and $6,$ and a morsel of food on lily pad $10.$ Fiona the frog starts on pad $0,$ and from any given lily pad, has a $\dfrac12$ chance to hop to the next pad, and an equal chance to jump $2$ pads. What is the probability that Fiona reaches pad $10$ without landing on either pad $3$ or pad $6?$

$\dfrac{15}{256}$

$\dfrac{1}{16}$

$\dfrac{15}{128}$

$\dfrac{1}{8}$

$\dfrac{1}{4}$

答案：A

知识点：递推概率

难度评级：1760

解答：

设 $p(n)$ 为在之前没有落到第 $3$ 号或第 $6$ 号的情况下落到第 $n$ 号睡莲叶的概率。每片睡莲叶把概率 $\dfrac12$ 传到下一片，另一个 $\dfrac12$ 传到再下一片，而第 $3$ 号和第 $6$ 号不再向外传递概率。

因此 $p(0)=1,\ p(1)=\dfrac12,\ p(2)=\dfrac34$ ，并且（跳过 $3$ ） $p(4)=\dfrac38,\ p(5)=\dfrac{3}{16}$ ，然后（跳过 $6$ ） $p(7)=\dfrac{3}{32}$ ， $\ p(8)=\dfrac{3}{64}$ ， $\ p(9)=\dfrac{9}{128}$ 。

最后 $\begin{gathered} p(10)=\dfrac12 p(8)+\dfrac12 p(9) \\ =\dfrac{3}{128}+\dfrac{9}{256} \\ =\dfrac{15}{256}. \end{gathered}$

所以正确答案是 A。

Let $p(n)$ be the probability of landing on pad $n$ without first landing on pad $3$ or $6.$ Each pad sends probability $\dfrac12$ to the next pad and $\dfrac12$ two pads ahead, and pads $3$ and $6$ pass nothing on.

Then $p(0)=1,\ p(1)=\dfrac12,\ p(2)=\dfrac34,$ and (skipping $3$ ) $p(4)=\dfrac38,\ p(5)=\dfrac{3}{16},$ then (skipping $6$ ) $p(7)=\dfrac{3}{32},$ $\ p(8)=\dfrac{3}{64},$ $\ p(9)=\dfrac{9}{128}.$

Finally $\begin{gathered} p(10)=\dfrac12 p(8)+\dfrac12 p(9) \\ =\dfrac{3}{128}+\dfrac{9}{256} \\ =\dfrac{15}{256}. \end{gathered}$

Thus, A is the correct answer.

2019 AMC 12B 第 16 题

答案：A

难度评级：1760

解答：

其他年份的第 16 题