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?

$\dfrac{32}{79}$

$\dfrac{161}{384}$

$\dfrac{63}{146}$

$\dfrac{7}{16}$

$\dfrac{1}{2}$

Solution:

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 $p_4 = \tfrac25 p_3 + \tfrac35 p_5,\quad p_3 = \tfrac{3}{10} p_2 + \tfrac{7}{10} p_4,$ $p_2 = \tfrac15 p_1 + \tfrac45 p_3,\quad p_1 = \tfrac{9}{10} p_2.$

Substituting downward from $p_5 = \tfrac12$ and solving yields $p_1 = \dfrac{63}{146}.$

Thus, the correct answer is C.

2014 AMC 12B Problem 22

Solution:

Problem 22 in Other Years