Lets number the "holes" between the pegs as follows
Row1: ____1
Row2: ___1_2
Row3: __1_2_3
Row4: _1_2_3_4
We are asked to find the number of paths that can lead to 2.
The total number of paths = 8
1111 1112
1122 1123
1222 1223
1233 1234
From the above, we can see that only 3 paths end in 2.
So probability = 3/8.
One thing to note is that we can do all of the above only because there
is equal probability of going either left or right at a peg.
I'm sure there is a standard binary-tree with linked nodes formula
that can be used here. GMAT hopefully doesn't expect someone to
remember this :-)
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Q from GMATPrep - Prob, ball in board with rows of pegs
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