2019 AIME II Problems/Problem 2

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) $(\frac{1}{2})^6 = \frac{1}{64}$

Case 2: (4 one jumps and 1 two jumps) $\binom{5}{1} * (\frac{1}{2})^5 = \frac{5}{32}$

Case 3: (2 one jumps and 2 two jumps) $\binom{4}{2} * (\frac{1}{2})^4 = \frac{3}{8}$

Case 4: (3 two jumps) $(\frac{1}{2})^3 = \frac{1}{8}$

Summing the probabilities gives us $\frac{43}{64}$ so the answer is $\boxed{107}$.

- 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 $P_n$. The frog can land at lily pad $n$ with either a double jump from lily pad $n-2$ or a single jump from lily pad $n-1$. Since the probability when the frog is at $n-2$ to make a double jump is $\frac{1}{2}$ and same for when it's at $n-1$, the recursion is just $P_n = \frac{P_{n-2}+P_{n-1}}{2}$. Using the fact that $P_1 = 1$, and $P_2 = \frac{1}{2}$, we find that $P_7 = \frac{43}{64}$. $43 + 64 = \boxed{107}$

-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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