Difference between revisions of "2014 AMC 10B Problems/Problem 25"
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==Solution== | ==Solution== | ||
| − | Using the techniques of a Markov chain, we can eventually arrive to the answer of, is <math>\boxed{\frac{63}{146 | + | Using the techniques of a Markov chain, we can eventually arrive to the answer of, is <math>\boxed{(C)}}{\frac{63}{146}</math> |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2014|ab=B|num-b=24|after=Last Problem}} | {{AMC10 box|year=2014|ab=B|num-b=24|after=Last Problem}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 14:44, 20 February 2014
Problem
In a small pond there are eleven lily pads in a row labeled 
through 
. A frog is sitting on pad 
. 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 it will be eaten by a patiently waiting snake. If the frog reaches pad it will exit the pond, never to return. what is the probability that the frog will escape being eaten by the snake?
Solution
Using the techniques of a Markov chain, we can eventually arrive to the answer of, is $\boxed{(C)}}{\frac{63}{146}$ (Error compiling LaTeX. Unknown error_msg)
See Also
| 2014 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 24 |
Followed by Last Problem | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
