2014 AMC 12B 第 22 题

Question

一个小池塘里有十一片睡莲叶排成一行，标号为

0

到

10

。一只青蛙坐在标号

1

的睡莲叶上。当青蛙在标号

N

的叶子上，且

0 \lt N \lt 10

时，它会以概率

\dfrac{N}{10}

跳到

N - 1

，并以概率

1 - \dfrac{N}{10}

跳到

N + 1

。每次跳跃都与之前的跳跃独立。如果青蛙到达标号

0

的叶子，它会被一条耐心等待的蛇吃掉；如果到达标号

10

，它就会离开池塘，不再回来。青蛙逃过被蛇吃掉的概率是多少？

In a small pond there are eleven lily pads in a row labeled

0

through

10.

A frog is sitting on pad

1.

When the frog is on pad

N,

0 \lt N \lt 10,

it will jump to pad

N - 1

with probability

\dfrac{N}{10}

and to pad

N + 1

with probability

1 - \dfrac{N}{10}.

Each jump is independent of the previous jumps. If the frog reaches pad

0

it will be eaten by a patiently waiting snake. If the frog reaches pad

10

it will exit the pond, never to return. What is the probability that the frog will escape being eaten by the snake?

\dfrac{32}{79}

\dfrac{161}{384}

\dfrac{63}{146}

\dfrac{7}{16}

\dfrac{1}{2}

答案：C

知识点：随机游走递推概率对称性

难度评级：2450

小提示：

由对称性，从标号

5

的叶子出发的逃脱概率为

\tfrac12

。

By symmetry the escape probability from pad

5

is

\tfrac12

大提示：

设

p_j

为从标号

j

的叶子出发的逃脱概率，并用相邻位置表示每个

p_j

。

Let

p_j

be the escape probability from pad

j

and write each

p_j

in terms of its neighbors

解答：

设

p_j

为从标号

j

的叶子出发最终到达标号

10

的概率。由中心处跳跃规则的对称性，

p_5 = \tfrac12

。

每个内部位置满足

p_j = \tfrac{10-j}{10}\,p_{j+1} + \tfrac{j}{10}\,p_{j-1}

，因此

\begin{gathered} p_4 = \tfrac25 p_3 + \tfrac35 p_5, \\ \quad p_3 = \tfrac{3}{10} p_2 + \tfrac{7}{10} p_4 \end{gathered}\text{，}

p_2 = \tfrac15 p_1 + \tfrac45 p_3,\quad p_1 = \tfrac{9}{10} p_2\text{。}

令

d_j=p_j-p_{j-1}

。递推式等价于

(10-j)d_{j+1}=jd_j

，所以

\begin{gathered} d_2=\dfrac{d_1}{9},\quad d_3=\dfrac{d_1}{36}, \\ d_4=\dfrac{d_1}{84},\quad d_5=\dfrac{d_1}{126} \end{gathered}\text{。}

又因为

p_0=0

且

p_5=\frac{1}{2}

，所以

\begin{aligned} \dfrac12&=d_1\biggl(1+\dfrac19+\dfrac1{36} \\ &\qquad+\dfrac1{84}+\dfrac1{126}\biggr) \\ &=d_1\cdot\dfrac{73}{63} \end{aligned}\text{。}

因此

p_1=d_1=\frac{63}{146}

。

所以正确答案是 C。

Let

p_j

be the probability of eventually reaching pad

10

starting from pad

j.

By the symmetry of the jump rule at the center,

p_5 = \tfrac12.

Each interior pad satisfies

p_j = \tfrac{10-j}{10}\,p_{j+1} + \tfrac{j}{10}\,p_{j-1},

which gives

\begin{gathered} p_4 = \tfrac25 p_3 + \tfrac35 p_5, \\ \quad p_3 = \tfrac{3}{10} p_2 + \tfrac{7}{10} p_4, \end{gathered}

p_2 = \tfrac15 p_1 + \tfrac45 p_3,\quad p_1 = \tfrac{9}{10} p_2.

Put

d_j=p_j-p_{j-1}.

The recurrence is equivalent to

(10-j)d_{j+1}=jd_j,

so

\begin{gathered} d_2=\dfrac{d_1}{9},\quad d_3=\dfrac{d_1}{36}, \\ d_4=\dfrac{d_1}{84},\quad d_5=\dfrac{d_1}{126}. \end{gathered}

Since

p_0=0

and

p_5=\frac{1}{2},

\begin{aligned} \dfrac12&=d_1\biggl(1+\dfrac19+\dfrac1{36} \\ &\qquad+\dfrac1{84}+\dfrac1{126}\biggr) \\ &=d_1\cdot\dfrac{73}{63}. \end{aligned}

Therefore

p_1=d_1=\frac{63}{146}.

Thus, the correct answer is C.

2014 AMC 12B 第 22 题

答案：C

难度评级：2450

小提示：

大提示：

解答：

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