AMC10 2014 B

AMC10 2014 B · Q25

AMC10 2014 B · Q25. 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 N/10 and to pad N + 1 with probability 1−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个依次排列的百合垫，标号0到10。青蛙初始在垫1上。当青蛙在垫N上时（0<N<10），它以概率N/10跳到N-1，以概率1-N/10跳到N+1。每跳独立。如果到达垫0会被蛇吃掉，到达垫10则逃出池塘。青蛙逃脱被吃掉的概率是多少？

(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 $$ \frac{\frac{72}{100}}{\left(\frac{72}{100}+\frac{10}{100}\right)}=\frac{36}{41}, $$ and the probability that it ultimately makes it to pad 0 is $$ \frac{\frac{10}{100}}{\left(\frac{72}{100}+\frac{10}{100}\right)}=\frac{5}{41}. $$ 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 back 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 $$ \frac{\frac{42}{100}}{\left(\frac{42}{100}+\frac{6}{100}\right)}=\frac{7}{8}, $$ and the probability that it ultimately makes it to pad 1 is $$ \frac{\frac{6}{100}}{\left(\frac{42}{100}+\frac{6}{100}\right)}=\frac{1}{8}. $$ The sequences of pairs of moves by which the frog will advance to pad 5 without being eaten are $$ 1\to 3\to 5,\; 1\to 3\to 1\to 3\to 5,\; 1\to 3\to 1\to 3\to 1\to 3\to 5, $$ and so on. The sum of the respective probabilities of reaching pad 5 is then $$ \frac{36}{41}\cdot\frac{7}{8}+\frac{36}{41}\cdot\frac{1}{8}\cdot\frac{36}{41}\cdot\frac{7}{8}+\frac{36}{41}\cdot\frac{1}{8}\cdot\frac{36}{41}\cdot\frac{1}{8}\cdot\frac{36}{41}\cdot\frac{7}{8}+\cdots $$ $$ =\frac{63}{82}\left(1+\frac{9}{82}+\left(\frac{9}{82}\right)^2+\cdots\right) $$ $$ =\frac{63}{82}\cdot\left(\frac{1}{1-\frac{9}{82}}\right) $$ $$ =\frac{63}{73}. $$ Therefore the requested probability is $\frac{1}{2}\cdot\frac{63}{73}=\frac{63}{146}$.

答案（C）：首先注意，一旦青蛙到达第 5 块荷叶，它最终被蛇吃掉的概率为 $\frac{1}{2}$，而最终未被吃掉并离开池塘的概率也为 $\frac{1}{2}$。因此，只需要确定：位于第 1 块荷叶的青蛙在被吃掉之前到达第 5 块荷叶的概率。将青蛙的跳跃按“两步一组”来考虑。青蛙从第 1 块荷叶出发，两步后前进到第 3 块荷叶的概率为 $\frac{9}{10}\cdot\frac{8}{10}=\frac{72}{100}$；两步后回到第 1 块荷叶的概率为 $\frac{9}{10}\cdot\frac{2}{10}=\frac{18}{100}$；后退到第 0 块荷叶并被吃掉的概率为 $\frac{1}{10}$。由于青蛙最终必然到达第 3 块或到达第 0 块，因此它最终到达第 3 块荷叶的概率为 $$ \frac{\frac{72}{100}}{\left(\frac{72}{100}+\frac{10}{100}\right)}=\frac{36}{41}, $$ 而最终到达第 0 块荷叶的概率为 $$ \frac{\frac{10}{100}}{\left(\frac{72}{100}+\frac{10}{100}\right)}=\frac{5}{41}. $$ 同理，从第 3 块荷叶出发，两步后前进到第 5 块荷叶的概率为 $\frac{7}{10}\cdot\frac{6}{10}=\frac{42}{100}$；两步后回到第 3 块荷叶的概率为 $\frac{7}{10}\cdot\frac{4}{10}+\frac{3}{10}\cdot\frac{8}{10}=\frac{52}{100}$；两步后退回到第 1 块荷叶的概率为 $\frac{3}{10}\cdot\frac{2}{10}=\frac{6}{100}$。由于从第 3 块出发最终必然到达第 5 块或第 1 块，因此它最终到达第 5 块荷叶的概率为 $$ \frac{\frac{42}{100}}{\left(\frac{42}{100}+\frac{6}{100}\right)}=\frac{7}{8}, $$ 最终到达第 1 块荷叶的概率为 $$ \frac{\frac{6}{100}}{\left(\frac{42}{100}+\frac{6}{100}\right)}=\frac{1}{8}. $$ 青蛙在不被吃掉的情况下到达第 5 块荷叶的“两步一组”的路径序列为 $$ 1\to 3\to 5,\; 1\to 3\to 1\to 3\to 5,\; 1\to 3\to 1\to 3\to 1\to 3\to 5, $$ 等等。到达第 5 块荷叶的相应概率之和为 $$ \frac{36}{41}\cdot\frac{7}{8}+\frac{36}{41}\cdot\frac{1}{8}\cdot\frac{36}{41}\cdot\frac{7}{8}+\frac{36}{41}\cdot\frac{1}{8}\cdot\frac{36}{41}\cdot\frac{1}{8}\cdot\frac{36}{41}\cdot\frac{7}{8}+\cdots $$ $$ =\frac{63}{82}\left(1+\frac{9}{82}+\left(\frac{9}{82}\right)^2+\cdots\right) $$ $$ =\frac{63}{82}\cdot\left(\frac{1}{1-\frac{9}{82}}\right) $$ $$ =\frac{63}{73}. $$ 因此所求概率为 $\frac{1}{2}\cdot\frac{63}{73}=\frac{63}{146}$。

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