2019 AMC 12B Problem 16

Problem 16 of 25IntermediateCounting & Probability

There are lily pads in a row numbered

0

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?

Show solution

Solution

Let

p(n)

be the probability of landing on pad

n

without first landing on pad

3

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

\begin{gathered} p(10)=\dfrac12 p(8)+\dfrac12 p(9) \\ =\dfrac{3}{128}+\dfrac{9}{256} \\ =\dfrac{15}{256}. \end{gathered}

Thus, A is the correct answer.