Difference between revisions of "2023 AMC 12A Problems/Problem 17"
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Now that we have expressed our problem formally, we can actually start solving it! | Now that we have expressed our problem formally, we can actually start solving it! | ||
− | Let's calculate <math>f(9)</math>. | + | Let's calculate <math>f(9)</math> (our expression is actually a general fact, not just limited to <math>10</math>). |
<cmath>f(9) = \sum_{k=0}^8 \frac{1}{2^{9-k}} \cdot f(k)</cmath> | <cmath>f(9) = \sum_{k=0}^8 \frac{1}{2^{9-k}} \cdot f(k)</cmath> | ||
<cmath>\frac{f(9)}{2} = \sum_{k=0}^8 \frac{1}{2^{10-k}} \cdot f(k)</cmath> | <cmath>\frac{f(9)}{2} = \sum_{k=0}^8 \frac{1}{2^{10-k}} \cdot f(k)</cmath> |
Revision as of 16:23, 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 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 We want the probability of reaching 10 from anywhere though, so what we can do is sum over all passing points , i.e.
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 ). Hmm, we see that the first 8 terms of are exactly the first 8 terms of . Let's substitute it in. 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
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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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