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

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

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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