Difference between revisions of "2014 AMC 12B Problems/Problem 22"
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We therefore seek to rewrite P(1) in terms of P(5), using the fact that | We therefore seek to rewrite P(1) in terms of P(5), using the fact that | ||
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P(N) = <math>\frac {N} {10}P(N - 1) + \frac {10 - N} {N}P(N + 1)</math> | P(N) = <math>\frac {N} {10}P(N - 1) + \frac {10 - N} {N}P(N + 1)</math> | ||
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as said in the problem. | as said in the problem. | ||
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Hence P(1) = <math>\frac {1} {10}P(0) + \frac {9} {10}P(2) = \frac {9} {10}P(2)</math> | Hence P(1) = <math>\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> | <math>\Rightarrow P(2) = \frac {10} {9}P(1)</math> | ||
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Returning to our original equation: | 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> | <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> | <math>= \frac {9} {50}P(1) + \frac {18} {25}P(3) \Rightarrow P(1) - \frac {9} {50}P(1)</math> | ||
<math>= \frac {18} {25}P(3)</math> | <math>= \frac {18} {25}P(3)</math> | ||
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<math>\Rightarrow P(3) = \frac {41} {36}P(1)</math> | <math>\Rightarrow P(3) = \frac {41} {36}P(1)</math> | ||
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Returning to our original equation: | 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> | <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> | <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> | <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: | 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> | <math>= \frac {21} {50}P(1) + \frac {126} {625}P(3) + \frac {189} {625}P(5)</math> | ||
| − | <math>= \frac {21} {50}P(1) + \frac {126} {625}(\frac {41} {36}P(1)) + \frac {189} {625}(\frac {1} {2})</math> | + | |
| + | <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, P(1) = <math>\frac {525} {1250}P(1) + \frac {287} {1250}P(1) + \frac {189} {1250}</math> | Hence, P(1) = <math>\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> | <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> | = <math>\boxed{\frac {63} {146}\, (C)}</math> | ||
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| + | -Dukejukem | ||
Revision as of 20:50, 20 February 2014
A long, but straightforward bash:
Define P(N) to be the probability that the frog survives starting from pad N.
Then note that by symmetry, P(5) = 1/2, since the probabilities of the frog moving subsequently in either direction from pad 5 are equal.
We therefore seek to rewrite P(1) in terms of P(5), using the fact that
P(N) = 
as said in the problem.
Hence P(1) = 
Returning to our original equation:
Returning to our original equation:
Cleaing up the coefficients, we have:
Hence, P(1) = 
= 
-Dukejukem
