2014 AMC 12B Problems/Problem 22
- The following problem is from both the 2014 AMC 12B #22 and 2014 AMC 10B #25, so both problems redirect to this page.
Contents
Problem
In a small pond there are eleven lily pads in a row labeled 0 through 10. A frog is sitting on pad 1. When the frog is on pad , , it will jump to pad with probability and to pad with probability . Each jump is independent of the previous jumps. If the frog reaches pad 0 it will be eaten by a patiently waiting snake. If the frog reaches pad 10 it will exit the pond, never to return. What is the probability that the frog will escape without being eaten by the snake?
Solution 1
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,
Or set :
Since , .
Solution 2
Notice that the probabilities are symmetrical around the fifth lily pad. If the frog is on the fifth lily pad, there is a chance that it escapes and a chance that it gets eaten. Now, let represent the probability that the frog escapes if it is currently on pad . We get the following system of equations: We want to find , since the frog starts at pad . Solving the above system yields , so the answer is .
Solution 3
Assign each lily pad a value, with pad having value and pad having value . If we treat the process as a random walk over these values, note that for all . This is purely by construction, as the expected gain at pad is . Therefore the process is a martingale. Since the stopping time has finite expectation and the increments are bounded, the Optional Stopping Thoerem applies and the expected final outcome is . This gives the probability of reaching safety as , which is choice .
Remark (Markov Chain)
We can represent this problem with the following State Transition Digram of Absorbing Markov Chain. State and state are the absorbing states. This problem asks for the Absorbing Probability of state .
Let , the probability that state transits to state on the next step.
Let be the absorbing probability of being absorbed in the absorbing state when starting from transient state .
is the sum of the products of and of all the next state .
Video Solution
https://www.youtube.com/watch?v=DMdgh2mMiWM
Video Solution by TheBeautyofMath
~IceMatrix
See also
2014 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Question | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
2014 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 21 |
Followed by Problem 23 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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