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 
with probability 
.
What is the probability that Flora will eventually land at 10?
Solution 1
Let's denote 
as the probability of reaching 
from 
. We immediately see that 
, and 
, since there's only one way to get to 1 from 0. Just jump!
Now, let's write an expression for 
. Suppose we know 
.
The probability of reaching 10 from some integer 
will be 
(use the formula given in the problem!)
The probability of reaching that integer 
from is going to be 
. Then, the probability of going from 
will be ![\[\mathbb{P}(\text{Reaching } 10 \text{ from } 0 \text { while passing through } k) = f(k) \cdot \frac{1}{2^{10-k}}\]](http://latex.artofproblemsolving.com/b/9/d/b9def17eff7aad704caee35c4227e5e9aeb111f0.png)
We want the probability of reaching 10 from anywhere though, so what we can do is sum over all passing points , i.e.
![\[f(10) = \sum_{k=0}^9 \mathbb{P}(\text{Reaching } 10 \text{ from } 0 \text { while passing through } k) = \sum_{k=0}^9 \frac{1}{2^{10-k}} \cdot f(k)\]](http://latex.artofproblemsolving.com/9/c/4/9c456a5ccd6f12b131342dd9df4bd9edcf571181.png)
Now that we have expressed our problem formally, we can actually start solving it!
Let's calculate 
(our expression is actually a general fact, not just limited to 
).
![\[f(9) = \sum_{k=0}^8 \frac{1}{2^{9-k}} \cdot f(k)\]](http://latex.artofproblemsolving.com/2/c/3/2c30bae040188d9feb4154313beefbe4713043be.png)
![\[\frac{f(9)}{2} = \sum_{k=0}^8 \frac{1}{2^{10-k}} \cdot f(k)\]](http://latex.artofproblemsolving.com/9/4/0/9409798d112322862395d3e64ff677a8022d856d.png)
Hmm, we see that the first 8 terms of 
are exactly the first 8 terms of . Let's substitute it in.
![\[f(10) = \frac{1}{2} \cdot f(9) + \color{red} \sum_{k=0}^8 \frac{1}{2^{10-k}} \cdot f(k)\]](http://latex.artofproblemsolving.com/a/5/3/a539a2936e0780d43848aeed64269ffbbeb1211c.png)
![\[f(10) = \frac{1}{2} \cdot f(9) + \color{red} \frac{1}{2} \cdot f(9)\]](http://latex.artofproblemsolving.com/a/5/a/a5a120c0470aa635c024b9793a1eb01f8fb89c61.png)
![\[f(10) = f(9)\]](http://latex.artofproblemsolving.com/c/e/7/ce7b8ba790f7b7d88e026984c8b8a27fe1679c27.png)
Isn't that interesting. Turns out, this reasoning can be extended all the way to 
.
It breaks at 
, since 
. Anyway, with this, we see that the answer is just
~ 
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 
. 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
multiplied by or 
~ShangJ2
Solution 3 (engineer's induction)
The probability frog lands on 1 is trivially $\frac{1}{2}. \$
The probability frog lands on 2 is$ (Error compiling LaTeX. Unknown error_msg)\frac{1}{4} + \frac{1}{2}\cdot \frac{1}{2} = \frac{1}{2},
\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}
\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}
\frac{1}{2}
\boxed{\textbf{(E)} \frac{1}{2}}$
~sirswagger21
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 | |
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