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Difference between revisions of "2014 AMC 12B Problems/Problem 22"

m (Addressed some final formatting issues)
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<math>P(N) = \frac {N} {10}P(N - 1) + \frac {10 - N} {N}P(N + 1)</math>
+
<math>P(N) = \frac {N} {10}P(N - 1) + \frac {10 - N} {10}P(N + 1)</math>
  
  

Revision as of 23:02, 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


$P(N) = \frac {N} {10}P(N - 1) + \frac {10 - N} {10}P(N + 1)$


as said in the problem.


Hence $P(1) = \frac {1} {10}P(0) + \frac {9} {10}P(2) = \frac {9} {10}P(2)$


$\Rightarrow P(2) = \frac {10} {9}P(1)$


Returning to our original equation:


$P(1) = \frac {9} {10}P(2) = \frac {9} {10}\left(\frac{2} {10}P(1) + \frac{8} {10}P(3)\right)$


$= \frac {9} {50}P(1) + \frac {18} {25}P(3) \Rightarrow P(1) - \frac {9} {50}P(1)$ $= \frac {18} {25}P(3)$


$\Rightarrow P(3) = \frac {41} {36}P(1)$


Returning to our original equation:


$P(1) = \frac {9} {50}P(1) + \frac {18} {25}\left(\frac {3} {10}P(2) + \frac {7} {10}P(4)\right)$


$= \frac {9} {50}P(1) + \frac {27} {125}P(2) + \frac {63} {125}P(4)$


$= \frac {9} {50}P(1) + \frac {27} {125}\left(\frac {10} {9}P(1)\right) + \frac {63} {125}\left(\frac {4} {10}P(3) + \frac {6} {10}P(5)\right)$


Cleaing up the coefficients, we have:


$= \frac {21} {50}P(1) + \frac {126} {625}P(3) + \frac {189} {625}P(5)$


$= \frac {21} {50}P(1) + \frac {126} {625}\left(\frac {41} {36}P(1)\right) + \frac {189} {625}\left(\frac {1} {2}\right)$


Hence, $P(1) = \frac {525} {1250}P(1) + \frac {287} {1250}P(1) + \frac {189} {1250}$


$\Rightarrow P(1) - \frac {812} {1250}P(1) = \frac {189} {1250} \Rightarrow P(1) = \frac {189} {438}$


$= \boxed{\frac {63} {146}\, (C)}$


-Dukejukem

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