Last visit was: 21 Apr 2026, 19:16 It is currently 21 Apr 2026, 19:16
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Close
Request Expert Reply
Confirm Cancel
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 21 Apr 2026
Posts: 109,728
Own Kudos:
Given Kudos: 105,800
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,728
Kudos: 810,476
 [32]
1
Kudos
Add Kudos
31
Bookmarks
Bookmark this Post
User avatar
Lampard42
Joined: 22 Nov 2018
Last visit: 12 Dec 2020
Posts: 424
Own Kudos:
561
 [5]
Given Kudos: 292
Location: India
GMAT 1: 640 Q45 V35
GMAT 2: 740 Q49 V41
GMAT 2: 740 Q49 V41
Posts: 424
Kudos: 561
 [5]
2
Kudos
Add Kudos
2
Bookmarks
Bookmark this Post
User avatar
Archit3110
User avatar
Major Poster
Joined: 18 Aug 2017
Last visit: 21 Apr 2026
Posts: 8,626
Own Kudos:
5,190
 [1]
Given Kudos: 243
Status:You learn more from failure than from success.
Location: India
Concentration: Sustainability, Marketing
GMAT Focus 1: 545 Q79 V79 DI73
GMAT Focus 2: 645 Q83 V82 DI81
GPA: 4
WE:Marketing (Energy)
Products:
GMAT Focus 2: 645 Q83 V82 DI81
Posts: 8,626
Kudos: 5,190
 [1]
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
User avatar
GMATinsight
User avatar
Major Poster
Joined: 08 Jul 2010
Last visit: 21 Apr 2026
Posts: 6,976
Own Kudos:
16,891
 [6]
Given Kudos: 128
Status:GMAT/GRE Tutor l Admission Consultant l On-Demand Course creator
Location: India
GMAT: QUANT+DI EXPERT
Schools: IIM (A) ISB '24
GMAT 1: 750 Q51 V41
WE:Education (Education)
Products:
Expert
Expert reply
Schools: IIM (A) ISB '24
GMAT 1: 750 Q51 V41
Posts: 6,976
Kudos: 16,891
 [6]
2
Kudos
Add Kudos
4
Bookmarks
Bookmark this Post
Bunuel
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
avatar
Saransh123
Joined: 11 Dec 2019
Last visit: 23 Mar 2021
Posts: 65
Own Kudos:
53
 [2]
Given Kudos: 117
Location: India
Concentration: Technology
GPA: 4
Products:
Posts: 65
Kudos: 53
 [2]
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I am having a headache just by reading this question

Posted from my mobile device
User avatar
GMATinsight
User avatar
Major Poster
Joined: 08 Jul 2010
Last visit: 21 Apr 2026
Posts: 6,976
Own Kudos:
16,891
 [2]
Given Kudos: 128
Status:GMAT/GRE Tutor l Admission Consultant l On-Demand Course creator
Location: India
GMAT: QUANT+DI EXPERT
Schools: IIM (A) ISB '24
GMAT 1: 750 Q51 V41
WE:Education (Education)
Products:
Expert
Expert reply
Schools: IIM (A) ISB '24
GMAT 1: 750 Q51 V41
Posts: 6,976
Kudos: 16,891
 [2]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Saransh123
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
User avatar
Contropositive
Joined: 21 Oct 2023
Last visit: 17 Feb 2025
Posts: 54
Own Kudos:
Given Kudos: 21
Location: India
GMAT Focus 1: 635 Q86 V81 DI77
GMAT Focus 1: 635 Q86 V81 DI77
Posts: 54
Kudos: 19
Kudos
Add Kudos
Bookmarks
Bookmark this Post
GMATinsight

Bunuel
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?
User avatar
Regor60
Joined: 21 Nov 2021
Last visit: 19 Apr 2026
Posts: 529
Own Kudos:
Given Kudos: 462
Posts: 529
Kudos: 419
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Contropositive
GMATinsight

Bunuel
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
User avatar
Regor60
Joined: 21 Nov 2021
Last visit: 19 Apr 2026
Posts: 529
Own Kudos:
Given Kudos: 462
Posts: 529
Kudos: 419
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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
User avatar
rmahe11
Joined: 13 Oct 2023
Last visit: 15 Aug 2025
Posts: 110
Own Kudos:
Given Kudos: 99
Posts: 110
Kudos: 29
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Is this really even a GMAT question ?

Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
Moderator:
Math Expert
109728 posts