2014 AMC 10B Problem 25

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,

where

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 without being eaten by the snake?

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Solution

Let

p_i

be the probability that the frog eventually escapes starting from pad

i

. Then

p_0=0

p_{10}=1

, and by symmetry

p_5=\frac12

. For

1\le i\le 4

p_i=\frac{i}{10}p_{i-1}+\frac{10-i}{10}p_{i+1}

. Working downward from

p_5=\frac12

, we get

p_4=\frac25p_3+\frac3{10}

, then

p_3=\frac3{10}p_2+\frac7{10}p_4=\frac5{12}p_2+\frac7{24}

. Next

p_2=\frac15p_1+\frac45p_3=\frac3{10}p_1+\frac7{20}

. Finally

p_1=\frac9{10}p_2

. Substituting the expression for

p_2

gives

p_1=\frac9{10}\left(\frac3{10}p_1+\frac7{20}\right)

, so

p_1=\frac{63}{146}

. Thus, the correct answer is C .