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The figure shown represents a board with 4 rows of pegs, and [#permalink]

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13 Oct 2006, 10:33

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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?

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19 Feb 2012, 16:35

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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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18 Jul 2013, 04:18

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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!

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.
_________________

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
_________________

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

Hi Karishma,

I understand your calculation and the logic behind it except for the concept of symmetry mentioned above. I've read your post in symmetry http://www.veritasprep.com/blog/2011/10/quarter-wit-quarter-wisdom-linear-arrangement-constraints-part-ii/, but the post discusses mainly about left and right positions in linear arrangement. So, I didn't get it here why cell 1 and cell 4 are symmetrical. Could you please throw some light on it?

Thank you very much!

Say you flip the image along a vertical axis - what do you get? The same image but now 1 is in 4's spot and 4 is in 1's spot. 2 is in 3's spot and 3 is in 2's spot.

The figure is still exactly the same though. The pathways in which you could reach slot 1 are the pathways in which you can reach slot 4 now.

OR think about it like this: To reach slot 1, the ball needs to turn left-left-left. Probability = 1/2 * 1/2 * 1/2 To reach slot 4, the ball needs to turn right-right-right. Probability = 1/2 * 1/2 * 1/2 Since the probability of turning left or right is the same, the situations are symmetrical.

Same will be the case for slots 2 and 3.
_________________

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

Hi another way to look at it is..

there are 4 rows.. the ball has 2 ways of moving-- to left or to right.. again when it falls in the peg in lower row, it can move to either direction- left or right .. the final row again can have two ways so TOTAL ways = 2*2*2 = 8 ways..

now ways the ball will fall in 1 = ways it will fall in 4.. and ways the ball will fall in 2 = ways it will fall in 3..

If you look at ways it can go in 1- at every PEG, the ball has to move ONLY to left, so ways = 1*1*1 = 1 way

similarly in 4, the ball has to move ONLY to RIGHT so 1 way..

total ways for 3 and 2 = 8-2=6.. ways for each = 6/2 = 3.. prob of ball falling in 2 = 3/8

The symmetry here being talked about is both visual and prob wise.. if you look at the number of pegs on 1 and compare it to 4, it is symmetrical.. similarily if you look at the distribution of pegs over 2 and 3, they are same..

SAy you put a mirror right in center and take a view if what is in left side.. the image in mirror will be exactly what is on the right side..

Hi, So, it is true even when we have odd numbers of objects, right? In the example above, there are 4 cells, hence cell 1 and cell 4 is symmetrical, and cell 2 and cell 3 are symmetrical. If we have 5 cells, the third cell (i.e. cell 3) will not be symmetrical to any other cell, right? But cell 1 and cell 5 are symmetrical, and cell 2 and cell 4 are symmetrical. Correct?

Thank you!

Yes it will be correct, as there will be same set up of pegs on top of symmetrical pegs.. in ODD, it will be as you have found, except the middle, all other will have a pair..
_________________

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?
_________________

The figure shown represents a board with 4 rows of pegs, and [#permalink]

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10 May 2012, 04:33

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
_________________

The figure shown represents a board with 4 rows of pegs, and [#permalink]

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12 Jul 2014, 00:32

Bunuel wrote:

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.

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.

Attachments

File comment: Please find attached what i was trying to do..

Ball Problem.png [ 22.38 KiB | Viewed 7072 times ]

The figure shown represents a board with 4 rows of pegs, and [#permalink]

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01 Feb 2015, 15:37

Another way to look at it is:

no matter which path the ball takes, it has to pass three balls to reach a cell, one from each level. At each level, it has a 1/2 probability to take left or right. So for any path we have the probability : 1/2 * 1/2 * 1/2 = 1/8.

Now of all available paths that a ball can take, 3 of them lead to the Cell 2. so we have 3 * 1/8 = 3/8.

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

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10 Apr 2015, 18:05

It seems to me that this a clasical pascal triangle, and you can notice to come to the position in the forth row there are 8 ways, 1+3+3+1=8 and teh posotiion that is asked in teh problem is by teh number 3, which means there are 3 ways to get there, so it is 3/8 as final answer. teh pis I attched you can figure out how teh triangle is made so it will be more clear how to come to the solutiomn to the problem.

Hope it helps

Attachments

construct.pascal.gif [ 2.15 KiB | Viewed 5142 times ]

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

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23 Jan 2016, 10:31

The ball can go either Left or Right for 3 stages. So Total Out Comes is 2*2*2 = 8 By symetry P(1) = P(4) and P(2) = P(3) => P(1) + P(2) = 1/2 For 1; the pattern can only be LLL so P(1) = 1/8 P(2) = 1/2 - 1/8 = 3/8

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

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25 Apr 2016, 22:59

VeritasPrepKarishma wrote:

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

Hi Karishma,

I understand your calculation and the logic behind it except for the concept of symmetry mentioned above. I've read your post in symmetry http://www.veritasprep.com/blog/2011/10/quarter-wit-quarter-wisdom-linear-arrangement-constraints-part-ii/, but the post discusses mainly about left and right positions in linear arrangement. So, I didn't get it here why cell 1 and cell 4 are symmetrical. Could you please throw some light on it?

Thank you very much!

gmatclubot

Re: The figure shown represents a board with 4 rows of pegs, and
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25 Apr 2016, 22:59

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