Difference between revisions of "2019 AIME I Problems/Problem 5"
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==Solution 2== | ==Solution 2== | ||
Alternatively, one could recursively compute the probabilities of reaching <math>(0,0)</math> as the first axes point from any point <math>(x,y)</math> as <cmath>P(x,y) = \frac{1}{3} P(x-1,y) + \frac{1}{3} P(x,y) + \frac{1}{3} P(x-1,y-1)</cmath> for <math>x,y \geq 1,</math> and the base cases are | Alternatively, one could recursively compute the probabilities of reaching <math>(0,0)</math> as the first axes point from any point <math>(x,y)</math> as <cmath>P(x,y) = \frac{1}{3} P(x-1,y) + \frac{1}{3} P(x,y) + \frac{1}{3} P(x-1,y-1)</cmath> for <math>x,y \geq 1,</math> and the base cases are | ||
− | <math>P(0,0) = 1, P(x,0) = P(y,0)</math> for any <math>x,y | + | <math>P(0,0) = 1, P(x,0) = P(y,0) = 0</math> for any <math>x,y</math> not equal to one. |
We then recursively find <math>P(4,4) = \frac{245}{2187}</math> so the answer is <math>245 + 7 = \boxed{252}</math>. | We then recursively find <math>P(4,4) = \frac{245}{2187}</math> so the answer is <math>245 + 7 = \boxed{252}</math>. | ||
Revision as of 00:47, 16 March 2019
Contents
Problem 5
A moving particle starts at the point and moves until it hits one of the coordinate axes for the first time. When the particle is at the point
, it moves at random to one of the points
,
, or
, each with probability
, independently of its previous moves. The probability that it will hit the coordinate axes at
is
, where
and
are positive integers. Find
.
Solution
A move from to
is labeled as down (
), from
to
is labeled as left (
), and from
to
is labeled as slant (
). To arrive at
without arriving at an axis first, the particle must first go to
then do a slant move. The particle can arrive at
through any permutation of the following 4 different cases:
,
,
, and
.
There is only permutation of
. Including the last move, there are
possible moves, making the probability of this move
.
There are permutations of
, as the ordering of the two slants do not matter. There are
possible moves, making the probability of this move
.
There are permutations of
, as the ordering of the two downs and two lefts do not matter. There are
possible moves, making the probability of this move
.
There are permutations of
, as the ordering of the three downs and three lefts do not matter. There are
possible moves, making the probability of this move
.
Adding these, the total probability is . Therefore, the answer is
.
Solution by Zaxter22
Solution 2
Alternatively, one could recursively compute the probabilities of reaching as the first axes point from any point
as
for
and the base cases are
for any
not equal to one.
We then recursively find
so the answer is
.
See Also
2019 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
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.