AoPSWiki
Art of Problem Solving celebrates the many
accomplishments of its students and community members.

1995 AIME Problems/Problem 3

From AoPSWiki

Problem

Starting at (0,0), an object moves in the coordinate plane via a sequence of steps, each of length one. Each step is left, right, up, or down, all four equally likely. Let p be the probability that the object reaches (2,2) in six or fewer steps. Given that p can be written in the form m/n, where m and n are relatively prime positive integers, find m+n.

Solution

It takes an even number of steps for the object to reach (2,2), so the number of steps the object may have taken is either 4 or 6.

If the object took 4 steps, then it must have gone two steps N and two steps E, in some permutation. There are \frac{4!}{2!2!} = 6 ways for these four steps of occuring, and the probability is \frac{6}{4^{4}}.

If the object took 6 steps, then it must have gone two steps N and two steps E, and an additional pair of moves that would cancel out, either N/S or W/E. The sequences N,N,N,E,E,S can be permuted in \frac{6!}{3!2!1!} = 60 ways. However, if the first four steps of the sequence are N,N,E,E in some permutation, it would have already reached the point (0,0) in four moves. There are \frac{4!}{2!2!} ways to order those four steps and 2! ways to determine the order of the remaining two steps, for a total of 12 sequences that we have to exclude. This gives 60-12=48 sequences of steps. There are the same number of sequences for the steps N,N,E,E,E,W, so the probability here is \frac{2 \times 48}{4^6}.

The total probability is \frac{6}{4^4} + \frac{96}{4^6} = \frac{3}{64}, and m+n= \boxed{067}.

See also

1995 AIME (ProblemsResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Try our innovative online adaptive learning system, Alcumus.
Over 1100 problems and 60+ video lessons. FREE!
© Copyright 2008 AoPS Incorporated. All Rights Reserved. • FoundationPrivacyContact Us