2023 AMC 12A Problem 17

Question

Flora the frog starts at

0

on the number line and makes a sequence of jumps to the right. In any one jump, independent of previous jumps, Flora leaps a positive integer distance

m

with probability

\dfrac{1}{2^m}.

What is the probability that Flora will eventually land at

10?

\dfrac{5}{512}

\dfrac{45}{1024}

\dfrac{127}{1024}

\dfrac{511}{1024}

\dfrac{1}{2}

Answer: E

Concepts:recursive probability recursion

Difficulty rating: 1910

Small hint:

Let

a_n

be the probability Flora ever lands on

n;

condition on the size of the first jump

Big hint:

The recursion

a_n=\sum_{k=1}^{n}\dfrac{a_{n-k}}{2^k}

gives the same value for every

n\ge 1

Solution:

Let

a_n

be the probability that Flora ever lands exactly on

n,

with

a_0=1.

Conditioning on the first jump,

a_n=\sum_{k=1}^{n}\dfrac{1}{2^k}\,a_{n-k}.

We prove by induction that

a_n=\tfrac12

for every

n\ge 1.

The case

n=1

is immediate. If the claim holds through

n-1,

then

\begin{aligned} a_n&=\frac{1}{2^n}a_0 +\sum_{k=1}^{n-1}\frac{1}{2^k}\cdot\frac12\\ &=\frac{1}{2^n} +\frac12\left(1-\frac{1}{2^{n-1}}\right)\\ &=\frac12. \end{aligned}

Hence the probability of landing on

10

is

\dfrac12.

Thus, the correct answer is E.