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

Revision as of 15:39, 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

@@ Line 34: / Line 34: @@
 ==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>
 -bradleyguo

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

Page

Toolbox

Search