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> |
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.