2019 AIME II Problems/Problem 2
Contents
Problem
Lilypads 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 (Probability States)
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 to and a two jump to be a jump from to .
Case 1: (6 one jumps)
Case 2: (4 one jumps and 1 two jumps)
Case 3: (2 one jumps and 2 two jumps)
Case 4: (3 two jumps)
Summing the probabilities gives us so the answer is .
- pi_is_3.14
Solution 3
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 . We calculate to be , meaning that our answer is .
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
Video Solution (2 Solutions)
https://youtu.be/wopflrvUN2c?t=652
~ pi_is_3.14
See Also
2019 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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