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Difference between revisions of "2014 AMC 10B Problems/Problem 25"

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==Solution==
 
==Solution==
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No matter how the movement of the frog might be, the probabilities of its being killed and its surviving are the same. Therefore, the chance, in which the frog escape successfully, is <math>\boxed{\frac{1}{2}{(E)}}</math>
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Using the techniques of a Markov chain, we can eventually arrive to the answer of, is <math>\boxed{\frac{63}{146}{(C)}}</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 14:40, 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 $\boxed{\frac{63}{146}{(C)}}$

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

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