2019 AMC 12B 第 16 题

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

小提示：

每一步是

+1

或

+2

，各自概率为

\dfrac12

。

Each step is

+1

or

+2,

each with probability

\dfrac12

大提示：

追踪到达每片睡莲叶的概率，把第

3

号和第

6

号视为终止点

Track the probability of reaching each pad, treating pads

3

and

6

as dead ends

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}\text{。}

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

小提示：

大提示：

解答：

其他年份的第 16 题