# 2014 AMC 10B Problems/Problem 25

The following problem is from both the 2014 AMC 12B #22 and 2014 AMC 10B #25, so both problems redirect to this page.

## Problem

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, 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 without being eaten by the snake? $\textbf {(A) } \frac{32}{79} \qquad \textbf {(B) } \frac{161}{384} \qquad \textbf {(C) } \frac{63}{146} \qquad \textbf {(D) } \frac{7}{16} \qquad \textbf {(E) } \frac{1}{2}$

## Solution 1

A long, but straightforward bash:

Define $P(N)$ to be the probability that the frog survives starting from pad N.

Then note that by symmetry, $P(5) = 1/2$, since the probabilities of the frog moving subsequently in either direction from pad 5 are equal.

We therefore seek to rewrite $P(1)$ in terms of $P(5)$, using the fact that $P(N) = \frac {N} {10}P(N - 1) + \frac {10 - N} {10}P(N + 1)$

as said in the problem.

Hence $P(1) = \frac {1} {10}P(0) + \frac {9} {10}P(2) = \frac {9} {10}P(2)$ $\Rightarrow P(2) = \frac {10} {9}P(1)$

Returning to our original equation: $P(1) = \frac {9} {10}P(2) = \frac {9} {10}\left(\frac{2} {10}P(1) + \frac{8} {10}P(3)\right)$ $= \frac {9} {50}P(1) + \frac {18} {25}P(3) \Rightarrow P(1) - \frac {9} {50}P(1)$ $= \frac {18} {25}P(3)$ $\Rightarrow P(3) = \frac {41} {36}P(1)$

Returning to our original equation: $P(1) = \frac {9} {50}P(1) + \frac {18} {25}\left(\frac {3} {10}P(2) + \frac {7} {10}P(4)\right)$ $= \frac {9} {50}P(1) + \frac {27} {125}P(2) + \frac {63} {125}P(4)$ $= \frac {9} {50}P(1) + \frac {27} {125}\left(\frac {10} {9}P(1)\right) + \frac {63} {125}\left(\frac {4} {10}P(3) + \frac {6} {10}P(5)\right)$

Cleaing up the coefficients, we have: $= \frac {21} {50}P(1) + \frac {126} {625}P(3) + \frac {189} {625}P(5)$ $= \frac {21} {50}P(1) + \frac {126} {625}\left(\frac {41} {36}P(1)\right) + \frac {189} {625}\left(\frac {1} {2}\right)$

Hence, $P(1) = \frac {525} {1250}P(1) + \frac {287} {1250}P(1) + \frac {189} {1250}$ $\Rightarrow P(1) - \frac {812} {1250}P(1) = \frac {189} {1250} \Rightarrow P(1) = \frac {189} {438}$ $= \boxed{\frac {63} {146}\, (C)}$

Or set $P(1)=a,P(2)=b,P(3)=c,P(4)=d,P(5)=e=1/2$: $$a=0.1\emptyset+0.9b,b=0.2a+0.8c,c=0.3b+0.7d,d=0.4c+0.6e$$ $$10a=\emptyset+9b,10b=2a+8c,10c=3b+7d,10d=4c+6e$$ $$\implies b=\frac{10a-\emptyset}{9},c=\frac{5b-a}{4},d=\frac{10c-3b}{7},e=\frac{5d-2c}{3}=1/2$$ $b=\frac{10a}{9}$ $c=\frac{5\left(\frac{10a}{9}\right)-a}{4}=\frac{\frac{50a}{9}-a}{9}=\frac{41a}{36}$ $d=\frac{10\left(\frac{41a}{36}\right)-3\left(\frac{30a}{9}\right)}{7}=\frac{\frac{205a}{18}-\frac{10a}{3}}{7}=\frac{145a}{126}$ $e=\frac{5\left(\frac{145a}{126}\right)-2\left(\frac{41a}{36}\right)}{3}=\frac{\frac{725a}{126}-\frac{41a}{18}}{3}=\frac{73a}{63}$

Since $e=\frac{1}{2}$, $\frac{73a}{63}=\frac{1}{2}\implies a=\boxed{\textbf{(C) }\frac{63}{146}}$.

## Solution 2

Notice that the probabilities are symmetrical around the fifth lily pad. If the frog is on the fifth lily pad, there is a $\frac{1}{2}$ chance that it escapes and a $\frac{1}{2}$ that it gets eaten. Now, let $P_k$ represent the probability that the frog escapes if it is currently on pad $k$. We get the following system of $5$ equations: $$P_1=\frac{9}{10}\cdot P_2$$ $$P_2=\frac{2}{10}\cdot P_1 + \frac{8}{10}\cdot P_3$$ $$P_3=\frac{3}{10}\cdot P_2 + \frac{7}{10}\cdot P_4$$ $$P_4=\frac{4}{10}\cdot P_3 + \frac{6}{10}\cdot P_5$$ $$P_5=\frac{5}{10}$$ We want to find $P_1$, since the frog starts at pad $1$. Solving the above system yields $P_1=\frac{63}{146}$, so the answer is $\boxed{(C)}$.

## Video Solution 2

~IceMatrix

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 