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The figure shown represents a board with 4 rows of pegs, and

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

  55% (hard)

Question Stats:

44% (02:16) correct 56% (01:20) wrong based on 142 sessions
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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
[Reveal] Spoiler: OA

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Last edited by Bunuel on 18 Jul 2013, 05:35, edited 4 times in total.
Edited the question and added the OA
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 [#permalink] New post 13 Oct 2006, 09:42
think its 3/8.

8 ways the ball lands in any of the cells.
only 3 ways does the ball land in cell two
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 [#permalink] New post 13 Oct 2006, 10:22
not sure i woul have got this without making a mistake, but......

1st gate, left or right

if it goes left it can then go : LR or RL

if it goes right it can then go : LL

so

LLR
LRL
RLL

(1/2 * 1/2 * 1/2) * 3

= 3/8
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 [#permalink] New post 13 Oct 2006, 10:28
The same logice...... must be 3/8

There is only 1 chance each to fall in 1 and 4
There are 3 chances to fall in 2 and 3.

So totally 8 out of which we need 3

So 3/8
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 [#permalink] New post 14 Oct 2006, 02:36
you're right

the OA is 3/8

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: probability toughy one from OG [#permalink] New post 19 Feb 2012, 12:01
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 !
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Re: The figure shown represents a board with 4 rows of pegs, and [#permalink] New post 19 Feb 2012, 12:13
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ugo_castelo wrote:
Image
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] New post 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!
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The figure shown represents a board with 4 rows of pegs, and [#permalink] New post 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] New post 18 Jul 2013, 03:18
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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!
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The figure shown represents a board with 4 rows of pegs, and [#permalink] New post 11 Jul 2014, 23:32
Bunuel wrote:
ugo_castelo wrote:
Image
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.



Hi

Let's say ball be at spot 1 - 2nd peg .. Now from there it can reach Cell 2 through LRR ,RLR,RRL .
Now if ball be at spot 2 - 2nd Peg .. Now from there it can reach Cell 2through LRL,RLL.
Similarly if ball be at spot 3- 2nd Peg .. It can reach cell 2 through LLL.

SO summing up results in 6/8 . Please let me know what i am doing wrong here.
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File comment: Please find attached what i was trying to do..
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The figure shown represents a board with 4 rows of pegs, and [#permalink] New post 15 Jul 2014, 20:49
Expert's post
Manik12345 wrote:


Hi

Let's say ball be at spot 1 - 2nd peg .. Now from there it can reach Cell 2 through LRR ,RLR,RRL .
Now if ball be at spot 2 - 2nd Peg .. Now from there it can reach Cell 2through LRL,RLL.
Similarly if ball be at spot 3- 2nd Peg .. It can reach cell 2 through LLL.

SO summing up results in 6/8 . Please let me know what i am doing wrong here.


Responding to a pm:

Note that the ball does not jump from peg to peg, it falls between two pegs. So when you have 4 different arrows, they show only 2 different paths.

You can use symmetry in this question to solve it. By symmetry, we see that probability of coming to cell 1 = probability of coming to cell 4

Similarly, probability of coming to cell 2 = probability of coming to cell 3

probability of coming to cell 1 + probability of coming to cell 2 + probability of coming to cell 3 + probability of coming to cell 4 = 1
2*(probability of coming to cell 1) + 2*(probability of coming to cell 2) = 1


Finding the probability of coming to cell 1:
After the first slot (not the peg but the slot between pegs 1 and 2 in first row), it moves left (between pegs 1 and 2 in second row) or right (between pegs 2 and 3 in second row). It must move left. Probability = 1/2
After that, it must move left again. Probability = 1/2
Finally it must love left again to come at cell 1. Probability = 1/2

Total probability = (1/2)*(1/2)*(1/2) = 1/8

Putting this in equation above:

2*(1/8) + 2*probability of coming to cell 2 = 1

Probability of coming to cell 2 = 3/8
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