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| − | {{duplicate|[[2014 AMC 12B Problems|2014 AMC 12B #22]] and [[2014 AMC 10B Problems|2014 AMC 10B #25]]}}
| + | #redirect [[2014 AMC 12B Problems/Problem 22]] |
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| − | ==Problem==
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| − | 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 without 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 1==
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| − | A long, but straightforward bash:
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| − | Define <math>P(N)</math> to be the probability that the frog survives starting from pad N.
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| − | Then note that by symmetry, <math>P(5) = 1/2</math>, since the probabilities of the frog moving subsequently in either direction from pad 5 are equal.
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| − | We therefore seek to rewrite <math>P(1)</math> in terms of <math>P(5)</math>, using the fact that
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| − | <math>P(N) = \frac {N} {10}P(N - 1) + \frac {10 - N} {10}P(N + 1)</math>
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| − | as said in the problem.
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| − | Hence <math>P(1) = \frac {1} {10}P(0) + \frac {9} {10}P(2) = \frac {9} {10}P(2)</math>
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| − | <math>\Rightarrow P(2) = \frac {10} {9}P(1)</math>
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| − | Returning to our original equation:
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| − | <math>P(1) = \frac {9} {10}P(2) = \frac {9} {10}\left(\frac{2} {10}P(1) + \frac{8} {10}P(3)\right)</math>
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| − | <math>= \frac {9} {50}P(1) + \frac {18} {25}P(3) \Rightarrow P(1) - \frac {9} {50}P(1)</math>
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| − | <math>= \frac {18} {25}P(3)</math>
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| − | <math>\Rightarrow P(3) = \frac {41} {36}P(1)</math>
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| − | Returning to our original equation:
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| − | <math>P(1) = \frac {9} {50}P(1) + \frac {18} {25}\left(\frac {3} {10}P(2) + \frac {7} {10}P(4)\right)</math>
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| − | <math>= \frac {9} {50}P(1) + \frac {27} {125}P(2) + \frac {63} {125}P(4)</math>
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| − | <math>= \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)</math>
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| − | Cleaing up the coefficients, we have:
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| − | <math>= \frac {21} {50}P(1) + \frac {126} {625}P(3) + \frac {189} {625}P(5)</math>
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| − | <math>= \frac {21} {50}P(1) + \frac {126} {625}\left(\frac {41} {36}P(1)\right) + \frac {189} {625}\left(\frac {1} {2}\right)</math>
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| − | Hence, <math>P(1) = \frac {525} {1250}P(1) + \frac {287} {1250}P(1) + \frac {189} {1250}</math>
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| − | <math>\Rightarrow P(1) - \frac {812} {1250}P(1) = \frac {189} {1250} \Rightarrow P(1) = \frac {189} {438}</math>
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| − | <math>= \boxed{\frac {63} {146}\, (C)}</math>
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| − | Or set <math>P(1)=a,P(2)=b,P(3)=c,P(4)=d,P(5)=e=1/2</math>:
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| − | <cmath>a=0.1\emptyset+0.9b,b=0.2a+0.8c,c=0.3b+0.7d,d=0.4c+0.6e</cmath>
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| − | <cmath>10a=\emptyset+9b,10b=2a+8c,10c=3b+7d,10d=4c+6e</cmath>
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| − | <cmath>\implies b=\frac{10a-\emptyset}{9},c=\frac{5b-a}{4},d=\frac{10c-3b}{7},e=\frac{5d-2c}{3}=1/2</cmath>
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| − | <math>b=\frac{10a}{9}</math>
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| − | <math>c=\frac{5\left(\frac{10a}{9}\right)-a}{4}=\frac{\frac{50a}{9}-a}{9}=\frac{41a}{36}</math>
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| − | <math>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}</math>
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| − | <math>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}</math>
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| − | Since <math>e=\frac{1}{2}</math>, <math>\frac{73a}{63}=\frac{1}{2}\implies a=\boxed{\textbf{(C) }\frac{63}{146}}</math>.
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| − | ==Solution 2==
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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 yields <math>P_1=\frac{63}{146}</math>, so the answer is <math>\boxed{(C)}</math>.
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| − | ==Video Solution==
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| − | https://www.youtube.com/watch?v=DMdgh2mMiWM
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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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