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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 before 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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| − | Notice that the probabilities are symmetrical around the fifth lily pad. 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>P_k</math> represent the probability that the frog escapes if it is currently on pad <math>k</math>. We get the following system of <math>5</math> equations:
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| − | <cmath>P_1=\frac{9}{10}\cdot P_2</cmath>
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| − | <cmath>P_2=\frac{2}{10}\cdot P_1 + \frac{8}{10}\cdot P_3</cmath>
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| − | <cmath>P_3=\frac{3}{10}\cdot P_2 + \frac{7}{10}\cdot P_4</cmath>
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| − | <cmath>P_4=\frac{4}{10}\cdot P_3 + \frac{6}{10}\cdot P_5</cmath>
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| − | <cmath>P_5=\frac{5}{10}</cmath>
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| − | We want to find <math>P_1</math>, since the frog starts at pad <math>1</math>. Solving the above system (really long process) yields <math>P_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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