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{(C)}{\frac{63}{146}}</math>
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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 15:45, 20 February 2014

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 $N$, $0<N<10$, it will jump to pad $N-1$ with probability $\frac{N}{10}$ and to pad $N+1$ with probability $1-\frac{N}{10}$. 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 being eaten by the snake?

$\textbf {(A) } \frac{32}{79} \qquad \textbf {(B) } \frac{161}{384} \qquad \textbf {(C) } \frac{63}{146} \qquad \textbf {(D) } \frac{7}{16} \qquad \textbf {(E) } \frac{1}{2}$

Solution

Using the techniques of a Markov chain, we can eventually arrive to the answer of, is $\boxed{{(C)}\frac{63}{146}}$

See Also

2014 AMC 10B (ProblemsAnswer KeyResources)
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

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