Difference between revisions of "2019 AIME I Problems/Problem 5"
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There are <math>\frac{6!}{3! \cdot 3!} = 30</math> permutations of <math>DLDLDL</math>, as the ordering of the two slants do not matter. There are <math>7</math> possible moves, making the probability of this move <math>\frac{20}{3^7}</math>. | There are <math>\frac{6!}{3! \cdot 3!} = 30</math> permutations of <math>DLDLDL</math>, as the ordering of the two slants do not matter. There are <math>7</math> possible moves, making the probability of this move <math>\frac{20}{3^7}</math>. | ||
− | Adding these, we get the total probability as <math>\frac{1}{3^4} + \frac{12}{3^5} + \frac{30}{3^6} + \frac{20}{3^7} = \frac{245}{3^7}</math>. Therefore, the answer is <math>245 + 7 = | + | Adding these, we get the total probability as <math>\frac{1}{3^4} + \frac{12}{3^5} + \frac{30}{3^6} + \frac{20}{3^7} = \frac{245}{3^7}</math>. Therefore, the answer is <math>245 + 7 = 252</math>. |
Solution by Zaxter22 | Solution by Zaxter22 |
Revision as of 23:46, 14 March 2019
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
We label a move from to
as down (
), from
to
as left (
), and from
to
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 can be done 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 pairs of do not matter. 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
.
Adding these, we get the total probability as . Therefore, the answer is
.
Solution by Zaxter22
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.
IMO 1999