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Q from GMATPrep - Prob, ball in board with rows of pegs

Expert replies
by takegmat » Sun Mar 18, 2007 3:44 pm
Can someone solve this please -- See the attached screenshot.

The OA is 3/8
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Source: — Problem Solving |

by jayhawk2001 » Sun Mar 18, 2007 4:07 pm
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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by takegmat » Sun Mar 18, 2007 6:38 pm
Wow, you made it look so simple. Thanks !!!
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