2014 AMC 12B Problem 22

Question

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 \lt N \lt 10,

it will jump to pad

N - 1

with probability

\dfrac{N}{10}

and to pad

N + 1

with probability

1 - \dfrac{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?

\dfrac{32}{79}

\dfrac{161}{384}

\dfrac{63}{146}

\dfrac{7}{16}

\dfrac{1}{2}

By symmetry the escape probability from pad

5

is

\tfrac12

Let

p_j

be the escape probability from pad

j

and write each

p_j

in terms of its neighbors

Let

p_j

be the probability of eventually reaching pad

10

starting from pad

j.

By the symmetry of the jump rule at the center,

p_5 = \tfrac12.

Each interior pad satisfies

p_j = \tfrac{10-j}{10}\,p_{j+1} + \tfrac{j}{10}\,p_{j-1},

which gives

\begin{gathered} p_4 = \tfrac25 p_3 + \tfrac35 p_5, \\ \quad p_3 = \tfrac{3}{10} p_2 + \tfrac{7}{10} p_4, \end{gathered}

p_2 = \tfrac15 p_1 + \tfrac45 p_3,\quad p_1 = \tfrac{9}{10} p_2.

Put

d_j=p_j-p_{j-1}.

The recurrence is equivalent to

(10-j)d_{j+1}=jd_j,

so

\begin{gathered} d_2=\dfrac{d_1}{9},\quad d_3=\dfrac{d_1}{36}, \\ d_4=\dfrac{d_1}{84},\quad d_5=\dfrac{d_1}{126}. \end{gathered}

Since

p_0=0

and

p_5=\frac{1}{2},

\begin{aligned} \dfrac12&=d_1\biggl(1+\dfrac19+\dfrac1{36} \\ &\qquad+\dfrac1{84}+\dfrac1{126}\biggr) \\ &=d_1\cdot\dfrac{73}{63}. \end{aligned}

Therefore

p_1=d_1=\frac{63}{146}.

Thus, the correct answer is C.

Problem 22 in other years