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| − | ==Problem==
| + | #redirect [[2014 AMC 12B Problems/Problem 22]] |
| − | In a small pond there are eleven lily pads in a row labeled <math>0</math> through <math>10</math>. A frog is sitting on pad <math>1</math>. When the frog is on pad <math>N</math>, <math>0<N<10</math>, it will jump to pad <math>N-1</math> with probability <math>\frac{N}{10}</math> and to pad <math>N+1</math> with probability <math>1-\frac{N}{10}</math>. Each jump is independent of the previous jumps. If the frog reaches pad <math>0</math> it will be eaten by a patiently waiting snake. If the frog reaches pad <math>10</math> it will exit the pond, never to return. what is the probability that the frog will escape being eaten by the snake?
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| − | <math> \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} </math>
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| − | ==Solution==
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| − | Using the techniques of a Markov chain, we can eventually arrive to the answer of, is <math>\boxed{{(C)}\frac{63}{146}}</math>
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| − | OR
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| − | Notice that the probabilities are symmetrical around the fifth lily pad. So if the frog is on the fifth lily pad, there is a <math>\frac{1}{2}</math> chance that it escapes and a <math>\frac{1}{2}</math> that it gets eaten. Now, let <math>N_k</math> represent the probability that the frog escapes if it is currently on pad <math>k</math>. We get the following system of 5 equations:
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| − | <cmath>N_1=\frac{9}{10}\cdot N_2</cmath>
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| − | <cmath>N_2=\frac{1}{5}\cdot N_1 + \frac{4}{5}\cdot N_3</cmath>
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| − | <cmath>N_3=\frac{3}{10}\cdot N_2 + \frac{7}{10}\cdot N_4</cmath>
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| − | <cmath>N_4=\frac{2}{5}\cdot N_3 + \frac{3}{5}\cdot N_5</cmath>
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| − | <cmath>N_5=\frac{1}{2}</cmath>
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| − | We want to find <math>N_1</math>, since the frog starts at pad 1. Solving the above system yields <math>N_1=\frac{63}{146}</math>, so the answer is <math>\boxed{(C)}</math>.
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| − | ==See Also==
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| − | {{AMC10 box|year=2014|ab=B|num-b=24|after=Last Problem}}
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| − | {{MAA Notice}}
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