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The figure shown represents a board with 4 rows of pegs, and [#permalink]
13 Oct 2006, 09:33

00:00

A

B

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D

E

Difficulty:

45% (medium)

Question Stats:

44% (02:14) correct
56% (01:25) wrong based on 75 sessions

Attachment:

Pegs.PNG [ 9.87 KiB | Viewed 1530 times ]

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?

I had a dificulty to count the 8 ways during the test.
I tried to put in a formula, though I'm not sure it makes some sense
4!/2*3 =8
any shortcut?
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Re: The figure shown represents a board with 4 rows of pegs, and [#permalink]
19 Feb 2012, 12:13

Expert's post

ugo_castelo wrote:

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

When ball passes two pegs it will be either on the left route or on the right route. Now, if ball is on the left route, number of possible scenarios will be 4, out of which two will lead to cell 2, if ball is on the right route, number of scenarios will also be 4, but out of these four only one will lead to cell 2. So, total 8 scenarios out of which 2+1=3 lead to cell 2, hence probability 3/8.

Answer: D.

morya003 wrote:

Does anyone know which OG question this is ? I haven't come across this in either 10th or 12th edition or Quant Review 2nd edition !

I think it's from GMAT Prep, not OG.
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Re: The figure shown represents a board with 4 rows of pegs, and [#permalink]
19 Feb 2012, 15:35

If a question is accompanied by a helpful picture, you can sometimes arrive at the answer simply by visualizing what's said to be happening in the picture. Concentrate and picture the ball dropping through those openings, landing on the peg and then taking a right or a left direction. Note the direction as it is dropping- LLR, for example. You will notice that the ball can turn left and left and right OR left and right and left OR right and left and left in order to land on Cell 2. For "AND" you know you have to multiply and for "OR" you know you have to add. So (1/2*1/2*1/2) + (1/2*1/2*1/2)+ (1/2*1/2*1/2). That's 3/8!

The figure shown represents a board with 4 rows of pegs, and [#permalink]
10 May 2012, 03:33

Probability of going left of peg = 1/2 probability of going right of peg = 1/2

After passing the first row, ball will hit a peg. so the probability of going left or right is 1/2. Assume that ball goes to left:

Then the favorable outcome can be ball first going left then going right or ball first going right and then going left. Hence probability of ball hitting cell 2 after taking first left is = 1/2*1/2 = 1/4.

Now assume that ball goes to right:

The favorable outcome after taking right will be to take 2 left on next two pegs. hence probability of hitting cell 2 after taking first right is 1/4 *1/2 =1/8

So the total probability = 1/8+1/4 =3/8
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Re: The figure shown represents a board with 4 rows of pegs, and [#permalink]
18 Jul 2013, 03:18

2

This post received KUDOS

Just a couple of generalization. As you have already known making generalization is important in order not to solve every GMAT question from the very beginning, in other words, to solve GMAT questions quicker.

The quickest way to calculate all possible outcomes is keeping in mind that there are 3 rows of pegs to pass, each row gives 2 ways. Thus, all possible outcomes = 2*2*2 = 2^3=8 How to calculate desirable (or preferred) outcomes the above posts have already showed.

Looking deeper into the question we can find out the Galton board (or Bean machine) that demonstrates normal distribution. Watching the below video sticks you to this principle!

gmatclubot

Re: The figure shown represents a board with 4 rows of pegs, and
[#permalink]
18 Jul 2013, 03:18