Difference between revisions of "2019 AIME II Problems/Problem 2"
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==Solution 4== | ==Solution 4== | ||
For any point <math>n</math>, let the probability that the frog lands on lily pad <math>n</math> be <math>P_n</math>. If the frog is at lily pad <math>n-2</math>, it can either double jump with probability <math>\frac{1}{2}</math> or single jump twice with probability <math>\frac{1}{4}</math> to get to lily pad <math>n</math>. Now consider if the frog is at lily pad <math>n-3</math>. It has a probability of landing on lily pad <math>n</math> without landing on lily pad <math>n-2</math> with probability <math>\frac{1}{4}</math>, double jump then single jump. Therefore the recursion is <math>P_n = \frac{3}{4}P_{n-2} + \frac{1}{4}P_{n-3}</math>. Note that all instances of the frog landing on lily pad <math>n-1</math> has been covered. After calculating a few values of <math>P_n</math> using the fact that <math>P_1 = 1</math>, <math>P_2 = \frac{1}{2}</math>, and <math>P_3 = \frac{3}{4}P_1 = \frac{3}{4}</math> we find that <math>P_7 = \frac{43}{64}</math>. <math>43 + 63 = \boxed{107}</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>. If the frog is at lily pad <math>n-2</math>, it can either double jump with probability <math>\frac{1}{2}</math> or single jump twice with probability <math>\frac{1}{4}</math> to get to lily pad <math>n</math>. Now consider if the frog is at lily pad <math>n-3</math>. It has a probability of landing on lily pad <math>n</math> without landing on lily pad <math>n-2</math> with probability <math>\frac{1}{4}</math>, double jump then single jump. Therefore the recursion is <math>P_n = \frac{3}{4}P_{n-2} + \frac{1}{4}P_{n-3}</math>. Note that all instances of the frog landing on lily pad <math>n-1</math> has been covered. After calculating a few values of <math>P_n</math> using the fact that <math>P_1 = 1</math>, <math>P_2 = \frac{1}{2}</math>, and <math>P_3 = \frac{3}{4}P_1 = \frac{3}{4}</math> we find that <math>P_7 = \frac{43}{64}</math>. <math>43 + 63 = \boxed{107}</math> | ||
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-bradleyguo | -bradleyguo | ||
Revision as of 15:39, 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 . If the frog is at lily pad , it can either double jump with probability or single jump twice with probability to get to lily pad . Now consider if the frog is at lily pad . It has a probability of landing on lily pad without landing on lily pad with probability , double jump then single jump. Therefore the recursion is . Note that all instances of the frog landing on lily pad has been covered. After calculating a few values of 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 |
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