2014 AMC 10B Problems/Problem 25

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 before 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

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 (really long process) yields $P_1=\frac{63}{146}$ , so the answer is $\boxed{(C)}$ .

Video Solution

https://m.youtube.com/watch?v=dQw4w9WgXcQ

THIS IS A RICKROLL

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2014 AMC 10B Problems/Problem 25

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Problem

Solution

Video Solution

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