ps 10

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ps 10

by dunkin77 » Sun Jun 17, 2007 8:48 am
The figure shown represents a board with 4 rows of pegs, and at the bottom of the board are 4 cells numbered 1 to 4. Whenever the ball shown passes through the opening between two adjacent pegs in the same row, it will hit the peg directly beneath the opening. The ball then has the probability 1/2 of passing through the opening immediately to the left of that peg and probability 1/2 of passing through the opening immediately to the right. What is the probability that when the ball passes through the first two pegs at the top it will end in Cell 2?

A. 1/16
B. 1/8
C. 1/4
D. 3/8
E. 1/2
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by jayhawk2001 » Sun Jun 17, 2007 1:37 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.