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A grid of light bulbs measures x bulbs by x bulbs, where x > 2. If 4

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Bunuel
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A grid of light bulbs measures x bulbs by x bulbs, where x > 2. If 4 [#permalink] New post   22 Jul 2019, 23:31
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A grid of light bulbs measures x bulbs by x bulbs, where x > 2. If 4 light bulbs are illuminated at random, what is the probability, in terms of x, that the 4 bulbs form a 2 bulb by 2 square?

A. \(\frac{4(x - 1)}{x^2}\)

B. \(\frac{24(x - 1)}{x^2(x^2 - 2)(x^2 - 3)(x + 1)}\)

C. \(\frac{24(x + 1)}{x^2(x^2 - 2)(x - 1)}\)

D. \(\frac{4(x + 1)}{x^2(x^2 - 2)(x - 1)}\)

E. \(\frac{4x - 1}{x^2(x^2 - 2)}\)

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Re: A grid of light bulbs measures x bulbs by x bulbs, where x > 2. If 4 [#permalink] New post   23 Jul 2019, 00:33
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Bunuel wrote:
A grid of light bulbs measures x bulbs by x bulbs, where x > 2. If 4 light bulbs are illuminated at random, what is the probability, in terms of x, that the 4 bulbs form a 2 bulb by 2 square?

A. \(\frac{4(x - 1)}{x^2}\)

B. \(\frac{24(x - 1)}{x^2(x^2 - 2)(x^2 - 3)(x + 1)}\)

C. \(\frac{24(x + 1)}{x^2(x^2 - 2)(x - 1)}\)

D. \(\frac{4(x + 1)}{x^2(x^2 - 2)(x - 1)}\)

E. \(\frac{4x - 1}{x^2(x^2 - 2)}\)


Total Ways of selecting 4 bulbs from a grid of x*x bulbs = x^2C4 = x^2(x^2-1)(x^2-2)(x^2-3)/24
Now ways of selecting 4 continuous bulbs can be done in (x-1)^2 <arrived at by looking at numerator of options and checked for 3 and 4 grids>.

SO probability = (x-1)(x-1)*24/(x^2(x-1)(x+1)(x^2-2)(x^2-3)) <factorized x^2-1 into (x+1)(x-1)>
=24(x-1)/(x^2(x+1)(x^2-2)(x^2-3) IMO Option B
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Re: A grid of light bulbs measures x bulbs by x bulbs, where x > 2. If 4 [#permalink] New post   25 Jul 2019, 03:15
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Bunuel wrote:
A grid of light bulbs measures x bulbs by x bulbs, where x > 2. If 4 light bulbs are illuminated at random, what is the probability, in terms of x, that the 4 bulbs form a 2 bulb by 2 square?

A. \(\frac{4(x - 1)}{x^2}\)

B. \(\frac{24(x - 1)}{x^2(x^2 - 2)(x^2 - 3)(x + 1)}\)

C. \(\frac{24(x + 1)}{x^2(x^2 - 2)(x - 1)}\)

D. \(\frac{4(x + 1)}{x^2(x^2 - 2)(x - 1)}\)

E. \(\frac{4x - 1}{x^2(x^2 - 2)}\)


GMATinsight ; sir could you please help in providing solution to this probability question ..
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Re: A grid of light bulbs measures x bulbs by x bulbs, where x > 2. If 4 [#permalink] New post   25 Jul 2019, 05:25
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Bunuel wrote:
A grid of light bulbs measures x bulbs by x bulbs, where x > 2. If 4 light bulbs are illuminated at random, what is the probability, in terms of x, that the 4 bulbs form a 2 bulb by 2 square?

A. \(\frac{4(x - 1)}{x^2}\)

B. \(\frac{24(x - 1)}{x^2(x^2 - 2)(x^2 - 3)(x + 1)}\)

C. \(\frac{24(x + 1)}{x^2(x^2 - 2)(x - 1)}\)

D. \(\frac{4(x + 1)}{x^2(x^2 - 2)(x - 1)}\)

E. \(\frac{4x - 1}{x^2(x^2 - 2)}\)


Let, we have a grid of 3x3 i.e. x = 3 (where a, b, c... etc are the bulbs)

a . b . c

d . e . f

g . h . i


Now, 2x2 grids that are available in 3x3 grid = 4
Grid 1: abde
Grid 2: degh
Grid 3: bcef
Grid 4: efhi


So the probability = 4 / 9C4 = 4/126 = 2/63

Check options

Answer: Option B


Alternatively

The 2x2 grids in x*x grid = ((x-1)*(x-1)

Total ways of picking 4 bulbs out of \(x^2\) = \(x^2\)C4

Required probability = \((x-1)^2\) / \(x^2\)C4

Answer: Option B

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Re: A grid of light bulbs measures x bulbs by x bulbs, where x > 2. If 4 [#permalink] New post   25 Feb 2021, 01:54
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I am having a headache just by reading this question

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Re: A grid of light bulbs measures x bulbs by x bulbs, where x > 2. If 4 [#permalink] New post   27 Feb 2021, 22:37
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Saransh123 wrote:
I am having a headache just by reading this question

Posted from my mobile device


Saransh123

You read the question is the first step towards improvement. :D

So headache or no headache, now you should read the explanation as well. I have given two explanations here and first one is soooo simple. check it out.

Two MUST join YouTube channels for GMAT aspirant GMATinsight (1000+ FREE Videos) and GMATclub :)
For a Comprehensive Topicwise Quant Video course at an affordable price CLICK HERE
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Re: A grid of light bulbs measures x bulbs by x bulbs, where x > 2. If 4 [#permalink] New post   29 Mar 2024, 23:42
GMATinsight wrote:
Bunuel wrote:
A grid of light bulbs measures x bulbs by x bulbs, where x > 2. If 4 light bulbs are illuminated at random, what is the probability, in terms of x, that the 4 bulbs form a 2 bulb by 2 square?

A. \(\frac{4(x - 1)}{x^2}\)

B. \(\frac{24(x - 1)}{x^2(x^2 - 2)(x^2 - 3)(x + 1)}\)

C. \(\frac{24(x + 1)}{x^2(x^2 - 2)(x - 1)}\)

D. \(\frac{4(x + 1)}{x^2(x^2 - 2)(x - 1)}\)

E. \(\frac{4x - 1}{x^2(x^2 - 2)}\)

Let, we have a grid of 3x3 i.e. x = 3 (where a, b, c... etc are the bulbs)

a . b . c

d . e . f

g . h . i


Now, 2x2 grids that are available in 3x3 grid = 4
Grid 1: abde
Grid 2: degh
Grid 3: bcef
Grid 4: efhi


So the probability = 4 / 9C4 = 4/126 = 2/63

Check options

Answer: Option B


Alternatively

The 2x2 grids in x*x grid = ((x-1)*(x-1)

Total ways of picking 4 bulbs out of \(x^2\) = \(x^2\)C4

Required probability = \((x-1)^2\) / \(x^2\)C4

Answer: Option B

Archit3110

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Hi GMATinsight , Why haven't you considered ''DBFH'' combination? That too is a possibility, right?
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Re: A grid of light bulbs measures x bulbs by x bulbs, where x > 2. If 4 [#permalink] New post   30 Mar 2024, 04:26
Contropositive wrote:
GMATinsight wrote:
Bunuel wrote:
A grid of light bulbs measures x bulbs by x bulbs, where x > 2. If 4 light bulbs are illuminated at random, what is the probability, in terms of x, that the 4 bulbs form a 2 bulb by 2 square?

A. \(\frac{4(x - 1)}{x^2}\)

B. \(\frac{24(x - 1)}{x^2(x^2 - 2)(x^2 - 3)(x + 1)}\)

C. \(\frac{24(x + 1)}{x^2(x^2 - 2)(x - 1)}\)

D. \(\frac{4(x + 1)}{x^2(x^2 - 2)(x - 1)}\)

E. \(\frac{4x - 1}{x^2(x^2 - 2)}\)

Let, we have a grid of 3x3 i.e. x = 3 (where a, b, c... etc are the bulbs)

a . b . c

d . e . f

g . h . i


Now, 2x2 grids that are available in 3x3 grid = 4
Grid 1: abde
Grid 2: degh
Grid 3: bcef
Grid 4: efhi


So the probability = 4 / 9C4 = 4/126 = 2/63

Check options

Answer: Option B


Alternatively

The 2x2 grids in x*x grid = ((x-1)*(x-1)

Total ways of picking 4 bulbs out of \(x^2\) = \(x^2\)C4

Required probability = \((x-1)^2\) / \(x^2\)C4

Answer: Option B

Archit3110

­
Hi GMATinsight , Why haven't you considered ''DBFH'' combination? That too is a possibility, right?



Yes.

And if you consider a 4×4 set of lights the additional squares are at least 4 beyond that implied by the answer.

This question is faulty.

Posted from my mobile device
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Re: A grid of light bulbs measures x bulbs by x bulbs, where x > 2. If 4 [#permalink] New post   31 Mar 2024, 08:28
This forum is going to lose its legitimacy if moderators continue to delete without explanation or correction reports on faulty posts.

Posted from my mobile device
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Re: A grid of light bulbs measures x bulbs by x bulbs, where x > 2. If 4 [#permalink] New post   01 Apr 2024, 23:36
Is this really even a GMAT question ?

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Re: A grid of light bulbs measures x bulbs by x bulbs, where x > 2. If 4 [#permalink]
01 Apr 2024, 23:36
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