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

m (→‎Solution)
(→‎Solution 2)
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<cmath>P_1 = \frac{43}{64}</cmath>
 
<cmath>P_1 = \frac{43}{64}</cmath>
 
<math>43 + 64 = \boxed{107}</math>.
 
<math>43 + 64 = \boxed{107}</math>.
 +
 +
==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 \boxed{107}$.
  
 
==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 20:18, 22 March 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 \boxed{107}$.

See Also

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

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