2014 AMC 12B Problems/Problem 22

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

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:

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

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

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

Returning to our original equation:

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

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

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

Cleaing up the coefficients, we have:

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

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

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

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

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

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