Difference between revisions of "2019 AIME II Problems/Problem 2"
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− | For any point <math>n</math>, let the probability that the frog lands on lily pad <math>n</math> be <math>P_n</math>. | + | For any point <math>n</math>, let the probability that the frog lands on lily pad <math>n</math> be <math>P_n</math>. The frog can land at lily pad <math>n</math> with either a double jump from lily pad <math>n-2</math> or a single jump from lily pad <math>n-1</math>. Since the probability when the frog is at <math>n-2</math> to make a double jump is <math>\frac{1}{2}</math> and same for when it's at <math>n-1</math>, the recursion is just <math>P_n = \frac{P_{n-2}+P_{n-1}}{2}</math>. Using the fact that <math>P_1 = 1</math>, and <math>P_2 = \frac{1}{2}</math>, we find that <math>P_7 = \frac{43}{64}</math>. <math>43 + 63 = \boxed{107}</math> |
-bradleyguo | -bradleyguo |
Revision as of 22:30, 25 May 2019
Contents
Problem 2
Lily pads lie in a row on a pond. A frog makes a sequence of jumps starting on pad . From any pad the frog jumps to either pad or pad chosen randomly with probability and independently of other jumps. The probability that the frog visits pad is , where and are relatively prime positive integers. Find .
Solution
Let be the probability the frog visits pad starting from pad . Then , , and for all integers . Working our way down, we find .
Solution 2(Casework)
Define a one jump to be a jump from k to K + 1 and a two jump to be a jump from k to k + 2.
Case 1: (6 one jumps) (1/2)^6 = 1/64
Case 2: (4 one jumps and 1 two jumps) 5C1 x (1/2)^5 = 5/32
Case 3: (2 one jumps and 2 two jumps) 4C2 x (1/2)^4 = 3/8
Case 4: (3 two jumps) (1/2)^3 = 1/8
Summing the probabilities gives us 43/64 so the answer is 107.
- pi_is_3.14
Solution 3 (easiest)
Let be the probability that the frog lands on lily pad . The probability that the frog never lands on pad is , so . This rearranges to , and we know that , so we can compute to be , meaning that our answer is
-Stormersyle
Solution 4
For any point , let the probability that the frog lands on lily pad be . The frog can land at lily pad with either a double jump from lily pad or a single jump from lily pad . Since the probability when the frog is at to make a double jump is and same for when it's at , the recursion is just . Using the fact that , and , we find that .
-bradleyguo
See Also
2019 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.