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 $N,$ $0 < N < 10,$ 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?

$\dfrac{32}{79}$

$\dfrac{161}{384}$

$\dfrac{63}{146}$

$\dfrac{7}{16}$

$\dfrac{1}{2}$

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 $p_2=\frac15p_1+\frac45p_3=\frac3{10}p_1+\frac7{30}$ . Finally $p_1=\frac9{10}p_2$ .

Substituting the expression for $p_2$ gives $p_1=\frac9{10}\left(\frac3{10}p_1+\frac7{30}\right)$ , so $p_1=\frac{63}{146}$ .

Thus, the correct answer is C .

2014 AMC 10B Problem 25

Solution:

Problem 25 in Other Years