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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>
 
Using the techniques of a Markov chain, we can eventually arrive to the answer of, is <math>\boxed{{(C)}\frac{63}{146}}</math>
 +
 +
OR
 +
 +
Notice that the probabilities are symmetrical around the fifth lily pad. So if the frog is on the fifth lily pad, there is a <math>\frac{1}{2}</math> chance that it escapes and a <math>\frac{1}{2}</math> that it gets eaten. Now, let <math>N_k</math> represent the probability that the frog escapes if it is currently on pad <math>k</math>. We get the following system of 5 equations:
 +
<cmath>N_1=\frac{9}{10}\cdot N_2</cmath>
 +
<cmath>N_2=\frac{1}{5}\cdot N_1 + \frac{4}{5}\cdot N_3</cmath>
 +
<cmath>N_3=\frac{3}{10}\cdot N_2 + \frac{7}{10}\cdot N_4</cmath>
 +
<cmath>N_4=\frac{2}{5}\cdot N_3 + \frac{3}{5}\cdot N_5</cmath>
 +
<cmath>N_5=\frac{1}{2}</cmath>
 +
We want to find <math>N_1</math>, since the frog starts at pad 1. Solving the above system yields <math>N_1=\frac{63}{146}</math>, so the answer is <math>\boxed{(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 17:03, 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 $\boxed{{(C)}\frac{63}{146}}$

OR

Notice that the probabilities are symmetrical around the fifth lily pad. So if the frog is on the fifth lily pad, there is a $\frac{1}{2}$ chance that it escapes and a $\frac{1}{2}$ that it gets eaten. Now, let $N_k$ represent the probability that the frog escapes if it is currently on pad $k$. We get the following system of 5 equations: \[N_1=\frac{9}{10}\cdot N_2\] \[N_2=\frac{1}{5}\cdot N_1 + \frac{4}{5}\cdot N_3\] \[N_3=\frac{3}{10}\cdot N_2 + \frac{7}{10}\cdot N_4\] \[N_4=\frac{2}{5}\cdot N_3 + \frac{3}{5}\cdot N_5\] \[N_5=\frac{1}{2}\] We want to find $N_1$, since the frog starts at pad 1. Solving the above system yields $N_1=\frac{63}{146}$, so the answer is $\boxed{(C)}$.

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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