Difference between revisions of "2019 AIME II Problems/Problem 2"
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− | + | ==Problem 2== | |
+ | |||
+ | Lily pads <math>1,2,3,\ldots</math> lie in a row on a pond. A frog makes a sequence of jumps starting on pad <math>1</math>. From any pad <math>k</math> the frog jumps to either pad <math>k+1</math> or pad <math>k+2</math> chosen randomly with probability <math>\tfrac{1}{2}</math> and independently of other jumps. The probability that the frog visits pad <math>7</math> is <math>\tfrac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>. | ||
+ | |||
+ | ==Solution== | ||
+ | Let <math>P_n</math> be the probability the frog visits pad <math>7</math> starting from pad <math>n</math>. Then <math>P_7 = 1</math>, <math>P_6 = \frac12</math>, and <math>P_n = \frac12(P_{n + 1} + P_{n + 2})</math> for all integers <math>1 \leq n \leq 5</math>. Working our way down, we find | ||
+ | <cmath>P_5 = \frac{3}{4}</cmath> | ||
+ | <cmath>P_4 = \frac{5}{8}</cmath> | ||
+ | <cmath>P_3 = \frac{11}{16}</cmath> | ||
+ | <cmath>P_2 = \frac{21}{32}</cmath> | ||
+ | <cmath>P_1 = \frac{43}{64}</cmath> | ||
+ | <math>43 + 64 = \boxed{107}</math>. | ||
+ | |||
+ | ==Solution 2(Casework)== | ||
+ | Define a one jump to be a jump from <math>k</math> to <math>k + 1</math> and a two jump to be a jump from <math>k</math> to <math>k + 2</math>. | ||
+ | |||
+ | Case 1: (6 one jumps) <math>\left (\frac{1}{2} \right)^6 = \frac{1}{64}</math> | ||
+ | |||
+ | Case 2: (4 one jumps and 1 two jumps) <math>\binom{5}{1} \cdot \left(\frac{1}{2}\right)^5 = \frac{5}{32}</math> | ||
+ | |||
+ | Case 3: (2 one jumps and 2 two jumps) <math>\binom{4}{2} \cdot \left(\frac{1}{2}\right)^4 = \frac{3}{8}</math> | ||
+ | |||
+ | Case 4: (3 two jumps) <math>\left(\frac{1}{2}\right)^3 = \frac{1}{8}</math> | ||
+ | |||
+ | Summing the probabilities gives us <math>\frac{43}{64}</math> so the answer is <math>\boxed{107}</math>. | ||
+ | |||
+ | - pi_is_3.14 | ||
+ | |||
+ | ==Solution 3 (easiest)== | ||
+ | Let <math>P_n</math> be the probability that the frog lands on lily pad <math>n</math>. The probability that the frog never lands on pad <math>n</math> is <math>\frac{1}{2}P_{n-1}</math>, so <math>1-P_n=\frac{1}{2}P_{n-1}</math>. This rearranges to <math>P_n=1-\frac{1}{2}P_{n-1}</math>, and we know that <math>P_1=1</math>, so we can compute <math>P_7</math> to be <math>\frac{43}{64}</math>, meaning that our answer is <math>\boxed{107}</math> | ||
+ | |||
+ | -Stormersyle | ||
+ | |||
+ | ==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>. 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 + 64 = \boxed{107}</math> | ||
+ | |||
+ | -bradleyguo | ||
+ | |||
+ | == Video Solution (2 Solutions) == | ||
+ | https://youtu.be/wopflrvUN2c?t=652 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | ==See Also== | ||
+ | {{AIME box|year=2019|n=II|num-b=1|num-a=3}} | ||
+ | [[Category: Introductory Combinatorics Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 22:15, 17 January 2021
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 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 (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
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 | |
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.