Difference between revisions of "2019 AIME II Problems/Problem 2"
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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>. | ||
− | \begin{align*} | + | <math>\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}\ | ||
Line 37: | Line 37: | ||
P_6&=\dfrac{21}{32}\ | P_6&=\dfrac{21}{32}\ | ||
P_7&=\dfrac{43}{64}\ | P_7&=\dfrac{43}{64}\ | ||
− | \end{align*} | + | \end{align*}</math> |
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>. | ||
Revision as of 16:01, 31 December 2022
Contents
[hide]Problem
Lily pads 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
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
.
$
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.