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

Revision as of 17:19, 18 January 2021

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

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 (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 $P_7$ .

$P_1=1$

$P_2=1-\dfrac{1}{2} \cdot 1=\dfrac{1}{2}$

$P_3=1-\dfrac{1}{2} \cdot \dfrac{1}{2}=\dfrac{3}{4}$

$P_4=\dfrac{5}{8}$

$P_5=\dfrac{11}{16}$

$P_6=\dfrac{21}{32}$

$P_7=\dfrac{43}{64}$

We calculate  to be , meaning that our answer is .

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

@@ Line 27: / Line 27: @@
 - pi_is_3.14
-==Solution 3 (easiest)==
+==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>
+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>.
--Stormersyle
+<math>P_1=1</math>
+<math>P_2=1-\dfrac{1}{2} \cdot 1=\dfrac{1}{2}</math>
+<math>P_3=1-\dfrac{1}{2} \cdot \dfrac{1}{2}=\dfrac{3}{4}</math>
+<math>P_4=\dfrac{5}{8}</math>
+<math>P_5=\dfrac{11}{16}</math>
+<math>P_6=\dfrac{21}{32}</math>
+<math>P_7=\dfrac{43}{64}</math>
+ We calculate <math>P_7</math> to be <math>\frac{43}{64}</math>, meaning that our answer is <math>\boxed{107}</math>.
 ==Solution 4==

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