Difference between revisions of "2019 AIME I Problems/Problem 5"
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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-1) + \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-1) + \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) = 0</math> for any <math>x,y</math> not equal to one. | <math>P(0,0) = 1, P(x,0) = P(y,0) = 0</math> for any <math>x,y</math> not equal to one. | ||
Revision as of 21:53, 7 March 2020
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 
.
jaskdfjskdjfksdfj
Solution
Alternatively, one could recursively compute the probabilities of reaching as the first axes point from any point 
as ![\[P(x,y) = \frac{1}{3} P(x-1,y) + \frac{1}{3} P(x,y-1) + \frac{1}{3} P(x-1,y-1)\]](http://latex.artofproblemsolving.com/c/0/c/c0c68668fdf9c80a5d85ca829509a731584b1d24.png)
for 
and the base cases are
for any 
not equal to one.
We then recursively find 
so the answer is 
.
If this algebra seems intimidating, you can watch a nice pictorial explanation of this by On The Spot Stem. https://www.youtube.com/watch?v=XBRuy3_TM9w
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. 
