Difference between revisions of "2019 AIME II Problems/Problem 2"

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-Stormersyle
 
-Stormersyle
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==Solution 4==
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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>. 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
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2019|n=II|num-b=1|num-a=3}}
 
{{AIME box|year=2019|n=II|num-b=1|num-a=3}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 16:38, 25 May 2019

Problem 2

Lily pads $1,2,3,\ldots$ lie in a row on a pond. A frog makes a sequence of jumps starting on pad $1$. From any pad $k$ the frog jumps to either pad $k+1$ or pad $k+2$ chosen randomly with probability $\tfrac{1}{2}$ and independently of other jumps. The probability that the frog visits pad $7$ is $\tfrac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.

Solution

Let $P_n$ be the probability the frog visits pad $7$ starting from pad $n$. Then $P_7 = 1$, $P_6 = \frac12$, and $P_n = \frac12(P_{n + 1} + P_{n + 2})$ for all integers $1 \leq n \leq 5$. Working our way down, we find \[P_5 = \frac{3}{4}\] \[P_4 = \frac{5}{8}\] \[P_3 = \frac{11}{16}\] \[P_2 = \frac{21}{32}\] \[P_1 = \frac{43}{64}\] $43 + 64 = \boxed{107}$.

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 $P_n$ be the probability that the frog lands on lily pad $n$. The probability that the frog never lands on pad $n$ is $\frac{1}{2}P_{n-1}$, so $1-P_n=\frac{1}{2}P_{n-1}$. This rearranges to $P_n=1-\frac{1}{2}P_{n-1}$, and we know that $P_1=1$, so we can compute $P_7$ to be $\frac{43}{64}$, meaning that our answer is $\boxed{107}$

-Stormersyle

Solution 4

For any point $n$, let the probability that the frog lands on lily pad $n$ be $P_n$. If the frog is at lily pad $n-2$, it can either double jump with probability $\frac{1}{2}$ or single jump twice with probability $\frac{1}{4}$ to get to lily pad $n$. Now consider if the frog is at lily pad $n-3$. It has a probability of landing on lily pad $n$ without landing on lily pad $n-2$ with probability $\frac{1}{4}$, double jump then single jump. Therefore the recursion is $P_n = \frac{3}{4}P_{n-2} + \frac{1}{4}P_{n-3}$. Note that all instances of the frog landing on lily pad $n-1$ has been covered. After calculating a few values of $P_n$ using the fact that $P_1 = 1$, $P_2 = \frac{1}{2}$, and $P_3 = \frac{3}{4}P_1 = \frac{3}{4}$ we find that $P_7 = \frac{43}{64}$. $43 + 63 = \boxed{107}$ -bradleyguo

See Also

2019 AIME II (ProblemsAnswer KeyResources)
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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