Difference between revisions of "2019 AIME II Problems/Problem 2"
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− | Let <math>P_n</math> be the probability the frog | + | 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> | ||
<cmath>P_4 = \frac{5}{8}</cmath> | <cmath>P_4 = \frac{5}{8}</cmath> |
Revision as of 18:50, 22 March 2019
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
.
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.