Difference between revisions of "2014 AMC 12B Problems/Problem 22"
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A long, but straightforward bash: | 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. | |
| − | |||
| + | 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. | ||
| − | P( | + | |
| + | We therefore seek to rewrite <math>P(1)</math> in terms of <math>P(5)</math>, using the fact that | ||
| + | |||
| + | |||
| + | <math>P(N) = \frac {N} {10}P(N - 1) + \frac {10 - N} {N}P(N + 1)</math> | ||
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| − | Hence P(1) = | + | Hence <math>P(1) = \frac {1} {10}P(0) + \frac {9} {10}P(2) = \frac {9} {10}P(2)</math> |
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| − | Hence, P(1) = | + | Hence, <math>P(1) = \frac {525} {1250}P(1) + \frac {287} {1250}P(1) + \frac {189} {1250}</math> |
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| − | + | <math>= \boxed{\frac {63} {146}\, (C)}</math> | |
-Dukejukem | -Dukejukem | ||
Revision as of 20:54, 20 February 2014
A long, but straightforward bash:
Define 
to be the probability that the frog survives starting from pad N.
Then note that by symmetry, 
, since the probabilities of the frog moving subsequently in either direction from pad 5 are equal.
We therefore seek to rewrite 
in terms of 
, using the fact that
as said in the problem.
Hence 
Returning to our original equation:
Returning to our original equation:
Cleaing up the coefficients, we have:
Hence, 
-Dukejukem
