Question

There are lily pads in a row numbered $0$ to $11,$ in that order. There are predators on lily pads $3$ and $6,$ and a morsel of food on lily pad $10.$ Fiona the frog starts on pad $0,$ and from any given lily pad, has a $\dfrac12$ chance to hop to the next pad, and an equal chance to jump $2$ pads. What is the probability that Fiona reaches pad $10$ without landing on either pad $3$ or pad $6?$

$\dfrac{15}{256}$

$\dfrac{1}{16}$

$\dfrac{15}{128}$

$\dfrac{1}{8}$

$\dfrac{1}{4}$

Solution:

Let $p(n)$ be the probability of landing on pad $n$ without first landing on pad $3$ or $6.$ Each pad sends probability $\dfrac12$ to the next pad and $\dfrac12$ two pads ahead, and pads $3$ and $6$ pass nothing on.

Then $p(0)=1,\ p(1)=\dfrac12,\ p(2)=\dfrac34,$ and (skipping $3$ ) $p(4)=\dfrac38,\ p(5)=\dfrac{3}{16},$ then (skipping $6$ ) $p(7)=\dfrac{3}{32},\ p(8)=\dfrac{3}{64},\ p(9)=\dfrac{9}{128}.$

Finally $p(10)=\dfrac12 p(8)+\dfrac12 p(9)=\dfrac{3}{128}+\dfrac{9}{256}=\dfrac{15}{256}.$

Thus, A is the correct answer.

2019 AMC 12B Problem 16

Solution:

Problem 16 in Other Years