Difference between revisions of "2019 AIME II Problems/Problem 2"
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− | ==Problem | + | ==Problem== |
− | + | 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>. | |
− | ==Solution== | + | ==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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<math>43 + 64 = \boxed{107}</math>. | <math>43 + 64 = \boxed{107}</math>. | ||
− | ==Solution 2(Casework)== | + | ==Solution 2 (Casework)== |
Define a one jump to be a jump from <math>k</math> to <math>k + 1</math> and a two jump to be a jump from <math>k</math> to <math>k + 2</math>. | Define a one jump to be a jump from <math>k</math> to <math>k + 1</math> and a two jump to be a jump from <math>k</math> to <math>k + 2</math>. | ||
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- pi_is_3.14 | - pi_is_3.14 | ||
− | ==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> 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>. |
− | + | <cmath> | |
− | + | \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*} | ||
+ | </cmath> | ||
+ | We calculate <math>P_7</math> to be <math>\frac{43}{64}</math>, meaning that our answer is <math>\boxed{107}</math>. | ||
==Solution 4== | ==Solution 4== | ||
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-bradleyguo | -bradleyguo | ||
+ | |||
+ | == Video Solution (2 Solutions) == | ||
+ | https://youtu.be/wopflrvUN2c?t=652 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
==See Also== | ==See Also== |
Latest revision as of 21:27, 24 October 2023
Contents
Problem
Lilypads lie in a row on a pond. A frog makes a sequence of jumps starting on pad
. From any pad
the frog jumps to either pad
or pad
chosen randomly with probability
and independently of other jumps. The probability that the frog visits pad
is
, where
and
are relatively prime positive integers. Find
.
Solution (Probability States)
Let be the probability the frog visits pad
starting from pad
. Then
,
, and
for all integers
. Working our way down, we find
.
Solution 2 (Casework)
Define a one jump to be a jump from to
and a two jump to be a jump from
to
.
Case 1: (6 one jumps)
Case 2: (4 one jumps and 1 two jumps)
Case 3: (2 one jumps and 2 two jumps)
Case 4: (3 two jumps)
Summing the probabilities gives us so the answer is
.
- pi_is_3.14
Solution 3
Let be the probability that the frog lands on lily pad
. The probability that the frog never lands on pad
is
, so
. This rearranges to
, and we know that
, so we can compute
.
We calculate
to be
, meaning that our answer is
.
Solution 4
For any point , let the probability that the frog lands on lily pad
be
. The frog can land at lily pad
with either a double jump from lily pad
or a single jump from lily pad
. Since the probability when the frog is at
to make a double jump is
and same for when it's at
, the recursion is just
. Using the fact that
, and
, we find that
.
-bradleyguo
Video Solution (2 Solutions)
https://youtu.be/wopflrvUN2c?t=652
~ pi_is_3.14
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.