Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2023 AMC 12A Problems/Problem 17"

m (→‎Solution 3 (engineer's induction))
Line 51: Line 51:
 
~sirswagger21
 
~sirswagger21
  
 +
 +
==Solution 4==
 +
No matter what the probability of getting to the end has a probability of <math>\frac{1}{1024}</math>
 +
 +
So the thing is how many ways to jump from the first spot to the last spot.
 +
 +
Given it requires <math>n</math> steps to reach the end, there are <math>\binom{10-1}{n-1}=\binom{9}{n-1}</math> ways to get the end.
 +
 +
<math>\sum_{n=1}^{10}\binom{9}{n-1}=2^9=512\implies \frac{1}{2}</math>
 +
 +
~bluesoul
 
==See also==
 
==See also==
 
{{AMC12 box|ab=A|year=2023|num-b=16|num-a=18}}
 
{{AMC12 box|ab=A|year=2023|num-b=16|num-a=18}}

Revision as of 22:12, 9 November 2023

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

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

Solution 2

In order to find the probability of landing on 10, we must multiply the amount of successful combinations by the probability of those combinations. Notice for any successful combination of steps, the probability must always be $\frac{1}{2^{10}}$. Now, we only need to find the amount of possibilites for steps since we know the probaility of each combination occuring is the same. This can be done using sticks and stones (9C0+9C1+9C2+9C3+...9C9) = 2^(9). This the final answer is

${2^{9}}$ multiplied by $\frac{1}{2^{10}}$ or $\boxed{\textbf{(E)} \frac{1}{2}}$ ~ShangJ2

Solution 3 (engineer's induction)

The probability frog lands on 1 is trivially $\frac{1}{2}.$

The probability frog lands on 2 is $\frac{1}{4} + \frac{1}{2}\cdot \frac{1}{2} = \frac{1}{2},$ from the two cases 0-2 and 0-1-2.

The probability frog lands on 3 is $\frac{1}{8} + 2\cdot \frac{1}{2} \cdot \frac{1}{4} + \frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2} = \frac{1}{2}$ from the cases 0-3, 0-1-3 and 0-2-3, 0-1-2-3.

The probability frog lands on 4 is $\frac{1}{16} + 2\cdot \frac{1}{2} \cdot \frac{1}{8} + \frac{1}{4}\cdot \frac{1}{4} + 3\cdot \frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{4} + \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{2}$ from the cases 0-4, 0-1-4 and 0-3-4, 0-2-4, 0-1-2-4 and 0-1-3-4 and 0-2-3-4, 0-1-2-3-4.

It looks like the probability is $\frac{1}{2}$ regardless of the ending number. Therefore, we choose $\boxed{\textbf{(E)} \frac{1}{2}}$

~sirswagger21


Solution 4

No matter what the probability of getting to the end has a probability of $\frac{1}{1024}$

So the thing is how many ways to jump from the first spot to the last spot.

Given it requires $n$ steps to reach the end, there are $\binom{10-1}{n-1}=\binom{9}{n-1}$ ways to get the end.

$\sum_{n=1}^{10}\binom{9}{n-1}=2^9=512\implies \frac{1}{2}$

~bluesoul

See also

2023 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_12A_Problems/Problem_17&oldid=201477"