AMC12 2014 B

AMC12 2014 B · Q22

AMC12 2014 B · Q22. It mainly tests Probability (basic), Markov / process probability (rare).

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 $N$, $0 < N < 10$, it will jump to pad $N-1$ with probability $\frac{N}{10}$ and to pad $N+1$ with probability $1 - \frac{N}{10}$. 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?

一个小池塘中有11个睡莲垫，依次标记为0到10。一只青蛙坐在垫1上。当青蛙在垫$N$上时，$0 < N < 10$，它以概率$\frac{N}{10}$跳到垫$N-1$，以概率$1 - \frac{N}{10}$跳到垫$N+1$。每次跳跃独立于前一次。如果青蛙到达垫0，它将被耐心等待的蛇吃掉。如果到达垫10，它将离开池塘，再不返回。青蛙逃脱被蛇吃掉的概率是多少？

(A) $\frac{32}{79}$ $\frac{32}{79}$

(B) $\frac{161}{384}$ $\frac{161}{384}$

(D) $\frac{7}{16}$ $\frac{7}{16}$

(E) $\frac{1}{2}$ $\frac{1}{2}$

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): First note that once the frog is on pad 5, it has probability $\frac{1}{2}$ of eventually being eaten by the snake, and a probability $\frac{1}{2}$ of eventually exiting the pond without being eaten. It is therefore necessary only to determine the probability that the frog on pad 1 will reach pad 5 before being eaten. Consider the frog’s jumps in pairs. The frog on pad 1 will advance to pad 3 with probability $\frac{9}{10}\cdot\frac{8}{10}=\frac{72}{100}$, will be back at pad 1 with probability $\frac{9}{10}\cdot\frac{2}{10}=\frac{18}{100}$, and will retreat to pad 0 and be eaten with probability $\frac{1}{10}$. Because the frog will eventually make it to pad 3 or make it to pad 0, the probability that it ultimately makes it to pad 3 is $\frac{72}{100}\div\left(\frac{72}{100}+\frac{10}{100}\right)=\frac{36}{41}$, and the probability that it ultimately makes it to pad 0 is $\frac{10}{100}\div\left(\frac{72}{100}+\frac{10}{100}\right)=\frac{5}{41}$. Similarly, in a pair of jumps the frog will advance from pad 3 to pad 5 with probability $\frac{7}{10}\cdot\frac{6}{10}=\frac{42}{100}$, will be back at pad 3 with probability $\frac{7}{10}\cdot\frac{4}{10}+\frac{3}{10}\cdot\frac{8}{10}=\frac{52}{100}$, and will retreat to pad 1 with probability $\frac{3}{10}\cdot\frac{2}{10}=\frac{6}{100}$. Because the frog will ultimately make it to pad 5 or pad 1 from pad 3, the probability that it ultimately makes it to pad 5 is $\frac{42}{100}\div\left(\frac{42}{100}+\frac{6}{100}\right)=\frac{7}{8}$, and the probability that it ultimately makes it to pad 1 is $\frac{6}{100}\div\left(\frac{42}{100}+\frac{6}{100}\right)=\frac{1}{8}$. The sequences of pairs of moves by which the frog will advance to pad 5 without being eaten are $1\to3\to5,\;1\to3\to1\to3\to5,\;1\to3\to1\to3\to1\to3\to5,$ and so on. The sum of the respective probabilities of reaching pad 5 is then $\frac{36}{41}\cdot\frac{7}{8}+\frac{36}{41}\cdot\frac{1}{8}\cdot\frac{36}{41}\cdot\frac{7}{8}+\frac{36}{41}\cdot\frac{1}{8}\cdot\frac{36}{41}\cdot\frac{1}{8}\cdot\frac{36}{41}\cdot\frac{7}{8}+\cdots$ $=\frac{63}{82}\left(1+\frac{9}{82}+\left(\frac{9}{82}\right)^2+\cdots\right)$ $=\frac{63}{82}\div\left(1-\frac{9}{82}\right)$ $=\frac{63}{73}.$ Therefore the requested probability is $\frac{1}{2}\cdot\frac{63}{73}=\frac{63}{146}.$

答案（C）：首先注意，一旦青蛙在垫子 5 上，它最终被蛇吃掉的概率是 $\frac{1}{2}$，最终未被吃掉而离开池塘的概率也是 $\frac{1}{2}$。因此只需要确定：在被吃掉之前，起始于垫子 1 的青蛙到达垫子 5 的概率。把青蛙的跳跃按“两步一组”来考虑。青蛙从垫子 1 出发，两步后前进到垫子 3 的概率为 $\frac{9}{10}\cdot\frac{8}{10}=\frac{72}{100}$；回到垫子 1 的概率为 $\frac{9}{10}\cdot\frac{2}{10}=\frac{18}{100}$；后退到垫子 0 并被吃掉的概率为 $\frac{1}{10}$。由于青蛙最终必然到达垫子 3 或到达垫子 0，因此它最终到达垫子 3 的概率为 $\frac{72}{100}\div\left(\frac{72}{100}+\frac{10}{100}\right)=\frac{36}{41}$，最终到达垫子 0 的概率为 $\frac{10}{100}\div\left(\frac{72}{100}+\frac{10}{100}\right)=\frac{5}{41}$。同理，从垫子 3 出发，两步后前进到垫子 5 的概率为 $\frac{7}{10}\cdot\frac{6}{10}=\frac{42}{100}$；回到垫子 3 的概率为 $\frac{7}{10}\cdot\frac{4}{10}+\frac{3}{10}\cdot\frac{8}{10}=\frac{52}{100}$；后退到垫子 1 的概率为 $\frac{3}{10}\cdot\frac{2}{10}=\frac{6}{100}$。由于青蛙从垫子 3 出发最终必然到达垫子 5 或垫子 1，因此它最终到达垫子 5 的概率为 $\frac{42}{100}\div\left(\frac{42}{100}+\frac{6}{100}\right)=\frac{7}{8}$，最终到达垫子 1 的概率为 $\frac{6}{100}\div\left(\frac{42}{100}+\frac{6}{100}\right)=\frac{1}{8}$。青蛙在不被吃掉的情况下到达垫子 5 的“两步序列”可以是 $1\to3\to5,\;1\to3\to1\to3\to5,\;1\to3\to1\to3\to1\to3\to5,$ 依此类推。到达垫子 5 的相应概率之和为 $\frac{36}{41}\cdot\frac{7}{8}+\frac{36}{41}\cdot\frac{1}{8}\cdot\frac{36}{41}\cdot\frac{7}{8}+\frac{36}{41}\cdot\frac{1}{8}\cdot\frac{36}{41}\cdot\frac{1}{8}\cdot\frac{36}{41}\cdot\frac{7}{8}+\cdots$ $=\frac{63}{82}\left(1+\frac{9}{82}+\left(\frac{9}{82}\right)^2+\cdots\right)$ $=\frac{63}{82}\div\left(1-\frac{9}{82}\right)$ $=\frac{63}{73}.$ 因此所求概率为 $\frac{1}{2}\cdot\frac{63}{73}=\frac{63}{146}.$

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