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Probability + Geometry + Prime Number

Expert replies
by harsh.champ » Thu Feb 04, 2010 6:18 am
There are two ants on opposite corners of a cube. On each move, they can travel along an edge to an adjacent vertex. If the probability that they both return to their starting position after 4 moves is m/n , where m and
n are relatively prime integers, find m+n. (NOTE:They do not stop if they collide.)


(1)17
(2)65
(3)73
(4)85
(5)None of these
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Source: — Problem Solving |

by ajith » Thu Feb 04, 2010 6:58 am
harsh.champ wrote:There are two ants on opposite corners of a cube. On each move, they can travel along an edge to an adjacent vertex. If the probability that they both return to their starting position after 4 moves is m/n , where m and
n are relatively prime integers, find m+n. (NOTE:They do not stop if they collide.)


(1)17
(2)65
(3)73
(4)85
(5)None of these
for the moves the Ant A has got 3 possibilities

so for ants have 81 ways (3*3*3*3) to complete 4 moves

Now the ways in which an ant is successful in coming back

1. He goes AB, comes back Goes AB and comes back - 3 Options
2. He goes AB comes back goes CD and comes back - 3*2 options
3. He goes AB BC CD DA and comes back to A = 3*2 =6 options

total successful options = 15

probability = 15/81 = 5/27

Probability for two ants to do this is = 25/729

I do not know where I went wrong
Always borrow money from a pessimist, he doesn't expect to be paid back.
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by Ian Stewart » Thu Feb 04, 2010 11:09 am
harsh.champ wrote:There are two ants on opposite corners of a cube. On each move, they can travel along an edge to an adjacent vertex. If the probability that they both return to their starting position after 4 moves is m/n , where m and
n are relatively prime integers, find m+n. (NOTE:They do not stop if they collide.)


(1)17
(2)65
(3)73
(4)85
(5)None of these
The question should make clear that the ants choose, randomly, which path to follow at each stage. It should also make clear whether the ant will ever retrace a path already followed. Assuming they can retrace their steps, we have the following for ant number 1:

* at each stage the ant has 3 choices for where to go
* no matter where the ant goes first, on step two, there is a 1/3 probability the ant returns to its starting point, and a 2/3 probability the ant is two steps away from its starting point; we can look at each case in turn:
* In the first case, in which the ant returns to its starting point after two steps, there is again a 1/3 chance it will return to its starting point in the next two steps; so we have a (1/3)(1/3) = 1/9 chance the ant leaves, comes back, leaves and comes back.
* In the second case, in which the ant is two steps away after two steps, there is a 1/3 chance the ant then moves three steps away -- i.e. to the opposite corner from its starting point -- and cannot possibly get back. There is a 2/3 chance the ant moves to a point one step away from the starting point, and then a 1/3 chance on the fourth step the ant returns to where it started. Multiplying all of those probabilities, we have a (2/3)(2/3)(1/3) = 4/27 chance the ant returns to where it started after four steps without returning after two steps.
* We need to add the probabilities from each of the two cases above: there is a (1/9) + (4/27) = 7/27 chance the first ant returns home.
* The same is true for the second ant, so there is a (7/27)(7/27) = 49/729 chance both ants return home. That's a reduced fraction, so the answer is 'none of the above'.

ajith - you've undercounted the second case; I think you've left out the possibility, using your notation, that the ant follows a path: AB, BC, CB, BA.

Given how large m and n turn out to be, my suspicion is that the question designer intends that the ants not be permitted to retrace their steps, but it's impossible to say. It's not a well-worded question, and involves too many cases to be a realistic GMAT question.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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by ajith » Thu Feb 04, 2010 11:25 am
Ian Stewart wrote: ajith - you've undercounted the second case; I think you've left out the possibility, using your notation, that the ant follows a path: AB, BC, CB, BA.
Yes that's another 6, Makes sense!

Thanks Ian
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