Difference between revisions of "2019 AIME II Problems/Problem 2"
Pi is 3.14 (talk | contribs) m (→Solution 2(Casework)) |
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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>. | ||
− | Case 1: (6 one jumps) <math>(\frac{1}{2})^6 = \frac{1}{64}</math> | + | Case 1: (6 one jumps) <math>\left(\frac{1}{2}\right)^6 = \frac{1}{64}</math> |
− | Case 2: (4 one jumps and 1 two jumps) <math>\binom{5}{1} | + | Case 2: (4 one jumps and 1 two jumps) <math>\binom{5}{1} \cdot \left(\frac{1}{2}/right)^5 = \frac{5}{32}</math> |
− | Case 3: (2 one jumps and 2 two jumps) <math>\binom{4}{2} | + | Case 3: (2 one jumps and 2 two jumps) <math>\binom{4}{2} /cdot /left(\frac{1}{2}/right)^4 = \frac{3}{8}</math> |
− | Case 4: (3 two jumps) <math>(\frac{1}{2})^3 = \frac{1}{8}</math> | + | Case 4: (3 two jumps) <math>\left(\frac{1}{2}\right)^3 = \frac{1}{8}</math> |
Summing the probabilities gives us <math>\frac{43}{64}</math> so the answer is <math>\boxed{107}</math>. | Summing the probabilities gives us <math>\frac{43}{64}</math> so the answer is <math>\boxed{107}</math>. |
Revision as of 22:48, 31 December 2020
Contents
Problem 2
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) $\binom{5}{1} \cdot \left(\frac{1}{2}/right)^5 = \frac{5}{32}$ (Error compiling LaTeX. Unknown error_msg)
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 (easiest)
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
-Stormersyle
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
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.