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

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==Problem==
 
==Problem==
  
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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>.
+
Lilypads <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>.
  
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==Solution==
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==Solution (Probability States)==
 
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
 
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_5 = \frac{3}{4}</cmath>
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==Solution 3==
 
==Solution 3==
 
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>.
 
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>.
 +
<cmath>
 
\begin{align*}
 
\begin{align*}
 
P_1&=1\\
 
P_1&=1\\
 
P_2&=1-\dfrac{1}{2} \cdot 1=\dfrac{1}{2}\\
 
P_2&=1-\dfrac{1}{2} \cdot 1=\dfrac{1}{2}\\
−
P_3&=1-\dfrac{1}{2} \cdot \dfrac{1}{2}=\dfrac{3}{4}
+
P_3&=1-\dfrac{1}{2} \cdot \dfrac{1}{2}=\dfrac{3}{4}\\
−
P_4&=\dfrac{5}{8}
+
P_4&=\dfrac{5}{8}\\
−
P_5&=\dfrac{11}{16}
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P_5&=\dfrac{11}{16}\\
−
P_6&=\dfrac{21}{32}
+
P_6&=\dfrac{21}{32}\\
−
P_7&=\dfrac{43}{64}
+
P_7&=\dfrac{43}{64}\\
 +
\end{align*}
 +
</cmath>
 
We calculate <math>P_7</math> to be <math>\frac{43}{64}</math>, meaning that our answer is <math>\boxed{107}</math>.
 
We calculate <math>P_7</math> to be <math>\frac{43}{64}</math>, meaning that our answer is <math>\boxed{107}</math>.
  

Latest revision as of 20:27, 24 October 2023

Problem

Lilypads $1,2,3,\ldots$ 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 (Probability States)

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

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

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

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

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

- pi_is_3.14

Solution 3

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 $P_7$. \begin{align*} P_1&=1\\ P_2&=1-\dfrac{1}{2} \cdot 1=\dfrac{1}{2}\\ P_3&=1-\dfrac{1}{2} \cdot \dfrac{1}{2}=\dfrac{3}{4}\\ P_4&=\dfrac{5}{8}\\ P_5&=\dfrac{11}{16}\\ P_6&=\dfrac{21}{32}\\ P_7&=\dfrac{43}{64}\\ \end{align*} We calculate $P_7$ to be $\frac{43}{64}$, meaning that our answer is $\boxed{107}$.

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

Video Solution (2 Solutions)

https://youtu.be/wopflrvUN2c?t=652

~ pi_is_3.14

See Also

2019 AIME II (Problems • Answer Key • Resources)
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All AIME Problems and Solutions

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