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An ant is standing at the origin of a coordinate grid. The ant will take four steps, each 1 unit in length. Each step taken is either forward, backward, right or left, chosen at random. What is the probability that the ant’s fourth step places the ant back at the origin? Express your answer as a common fraction.

dgfgrafgdfge111  Nov 22, 2018
 #1
avatar+357 
+1

Let's think about this. Forward, backward, right, and left is 4 possible movements. 

If the ant goes the same direction twice, the only path it can take is back the opposite direction twice.

If the ant goes a direction once and goes back to the origin again, the ant can move any four ways but then must come back.

If the ant goes a direction once and goes a different direction, there are two ways, come back the original way or go in a square-like shape.

 

Translating this to numbers, we have 

\(4\times 1 \times 1 \times 1=4\)

\(4 \times 1 \times 4 \times 1 = 16\)

\(4\times 2\times 2 \times 1 = 16\)

\(16+16+4=36\)

The total number of pathways are \(4^4=256\)

So, \(\frac{36}{256}=\frac{9}{64}\).

(by the way, can anyone check this, preferably one of the moderators? i'm not 100% sure. thanks!)

 

You are very welcome!
:P

CoolStuffYT  Nov 22, 2018
 #2
avatar+3152 
+2

Moves are \((1,0),~(0,1),~(-1,0),~(0,-1)\)

 

We have a set of 4 moves and the sum of those must equal (0,0)

 

There are 36 of these sets of 4 moves that sum to (0,0)

 

There are a total of \(4^4 = 256\) moves.

 

P[returning to 0] = \(\dfrac{36}{256} = \dfrac{9}{64}\)

 

CoolStuffYT is correct.

Rom  Nov 22, 2018

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