2023 AMC 12A Problems/Problem 17

Problem

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 $\frac{1}{2^m}$ .

What is the probability that Flora will eventually land at 10?

$\textbf{(A)}~\frac{5}{512}\qquad\textbf{(B)}~\frac{45}{1024}\qquad\textbf{(C)}~\frac{127}{1024}\qquad\textbf{(D)}~\frac{511}{1024}\qquad\textbf{(E)}~\frac{1}{2}$

Solution 1

Let's denote $f(n)$ as the probability of reaching $n$ from $0$ . We immediately see that $f(0) = 1$ , and $f(1) = \frac{1}{2}$ , since there's only one way to get to 1 from 0. Just jump!

Now, let's write an expression for $f(10)$ . Suppose we know $f(9),f(8),\dotsc,f(2)$ .

The probability of reaching 10 from some integer $k < 10$ will be $\frac{1}{2^{10-k}}$ (use the formula given in the problem!) The probability of reaching that integer $k$ from $0$ is going to be $f(k)$ . Then, the probability of going from $0 \to \underbrace{ k \to 10}_{\text{one jump}}$ will be $\mathbb{P}(\text{Reaching } 10 \text{ from } 0 \text { while passing through } k) = f(k) \cdot \frac{1}{2^{10-k}}$ We want the probability of reaching 10 from anywhere though, so what we can do is sum over all passing points $k$ , i.e. $f(10) = \sum_{k=0}^9 \mathbb{P}(\text{Reaching } 10 \text{ from } 0 \text { while passing through } k) = \sum_{k=0}^9 \frac{1}{2^{10-k}} \cdot f(k)$

Now that we have expressed our problem formally, we can actually start solving it!

Let's calculate $f(9)$ (our expression is actually a general fact, not just limited to $10$ ). $f(9) = \sum_{k=0}^8 \frac{1}{2^{9-k}} \cdot f(k)$ $\frac{f(9)}{2} = \sum_{k=0}^8 \frac{1}{2^{10-k}} \cdot f(k)$ Hmm, we see that the first 8 terms of $\frac{f(9)}{2}$ are exactly the first 8 terms of $f(10)$ . Let's substitute it in. $f(10) = \frac{1}{2} \cdot f(9) + \color{red} \sum_{k=0}^8 \frac{1}{2^{10-k}} \cdot f(k)$ $f(10) = \frac{1}{2} \cdot f(9) + \color{red} \frac{1}{2} \cdot f(9)$ $f(10) = f(9)$ Isn't that interesting. Turns out, this reasoning can be extended all the way to $f(10) = f(9) = \dotsc = f(2) = f(1)$ .

It breaks at $f(1) \neq f(0)$ , since $f(1) = \frac{1}{2} f(0)$ . Anyway, with this, we see that the answer is just

$f(10) = f(1) = \boxed{\textbf{(E)} \ \frac{1}{2}}$ ~ $\color{magenta} zoomanTV$

2023 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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2023 AMC 12A Problems/Problem 17

Problem

Solution 1

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