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2014 AMC 10B Problems/Problem 25

Revision as of 17:07, 20 February 2014 by Mikechen (talk | contribs)

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 $N$, $0<N<10$, 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?

$\textbf {(A) } \frac{32}{79} \qquad \textbf {(B) } \frac{161}{384} \qquad \textbf {(C) } \frac{63}{146} \qquad \textbf {(D) } \frac{7}{16} \qquad \textbf {(E) } \frac{1}{2}$

Troll "Solution"

Using the techniques of a Markov chain, we can eventually arrive to the answer of, is $\boxed{{(C)}\frac{63}{146}}$

Actual Solution

Notice that the probabilities are symmetrical around the fifth lily pad. So 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 $N_k$ represent the probability that the frog escapes if it is currently on pad $k$. We get the following system of 5 equations: \[N_1=\frac{9}{10}\cdot N_2\] \[N_2=\frac{1}{5}\cdot N_1 + \frac{4}{5}\cdot N_3\] \[N_3=\frac{3}{10}\cdot N_2 + \frac{7}{10}\cdot N_4\] \[N_4=\frac{2}{5}\cdot N_3 + \frac{3}{5}\cdot N_5\] \[N_5=\frac{1}{2}\] We want to find $N_1$, since the frog starts at pad 1. Solving the above system yields $N_1=\frac{63}{146}$, so the answer is $\boxed{(C)}$.

See Also

2014 AMC 10B (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions

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