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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here. ### Request Expert Reply # A grid of light bulbs measures x bulbs by x bulbs, where x > 2. If 4

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Manager  B
Joined: 23 Jul 2015
Posts: 63
A grid of light bulbs measures x bulbs by x bulbs, where x > 2. If 4  [#permalink]

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Question Stats: 100% (00:19) correct 0% (00:00) wrong based on 23 sessions

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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 bulb square?

1. 4(x-1)/x^2

2. 24(x−1)/x^2∗(x+1)∗(x^2−2)∗(x^2−3)

3. 24(x+1)/ (x^2)(x^2-2)(x+1)

4. 4(x+1)/(x^2)(x2-2)(x-1)

5. 4(x-1)/(x^2)(x2-2).

Hi Guys! Came by this question and I couldn't find a solution for it:
This is from MGMAT test papers. Couldn't find the official solution.

This what I did:
I assumed:
x=4
Total no. of bulbs =16.
4 bulbs can by chosen 16C4 ways = 4x5x7x13.

And theN I don't know how to narrow the desired possibilities.

Thanks in advance!

Originally posted by SPatel1992 on 25 Jun 2019, 14:43.
Last edited by SPatel1992 on 25 Jun 2019, 15:49, edited 2 times in total.
Director  P
Joined: 19 Oct 2018
Posts: 769
Location: India
Re: A grid of light bulbs measures x bulbs by x bulbs, where x > 2. If 4  [#permalink]

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I don't think any option is correct.

Number of ways to select 4 bulbs out of x^2 bulbs= $$x^2_ C_4$$ = $$\frac{x^2*(x^2-1)*(x^2-2)*(x^2-3)}{4!}$$

Number of ways those selected 4 bulbs form a 2 bulb by 2 bulb square= $$(x-1)^2$$

Probability= $$(x-1)^2$$/$$\frac{x^2*(x^2-1)*(x^2-2)*(x^2-3)}{4!}$$

probability=$$\frac{24(x-1)}{x^2*(x+1)*(x^2-2)*(x^2-3)}$$

You gotta edit option B

SPatel1992 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 bulb square?

1. 4(x-1)/x^2

2. 24(x-1)/(x^2)(x^2-3)(x+1)

3. 24(x+1)/ (x^2)(x^2-2)(x+1)

4. 4(x+1)/(x^2)(x2-2)(x-1)

5. 4(x-1)/(x^2)(x2-2).

Hi Guys! Came by this question and I couldn't find a solution for it:
This is from MGMAT test papers. Couldn't find the official solution.

This what I did:
I assumed:
x=4
Total no. of bulbs =16.
4 bulbs can by chosen 16C4 ways = 4x5x7x13.

And theN I don't know how to narrow the desired possibilities.

Thanks in advance!
Manager  B
Joined: 23 Jul 2015
Posts: 63
Re: A grid of light bulbs measures x bulbs by x bulbs, where x > 2. If 4  [#permalink]

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1
Thanks for pointing it out. I am confused. How did you get (x^2-1)? Can you solve this using numbers?

nick1816 wrote:
I don't think any option is correct.

Number of ways to select 4 bulbs out of x^2 bulbs= $$x^2_ C_4$$ = $$\frac{x^2*(x^2-1)*(x^2-2)*(x^2-3)}{4!}$$

Number of ways those selected 4 bulbs form a 2 bulb by 2 bulb square= $$(x-1)^2$$

Probability= $$(x-1)^2$$/$$\frac{x^2*(x^2-1)*(x^2-2)*(x^2-3)}{4!}$$

probability=$$\frac{24(x-1)}{x^2*(x+1)*(x^2-2)*(x^2-3)}$$

You gotta edit option B

SPatel1992 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 bulb square?

1. 4(x-1)/x^2

2. 24(x-1)/(x^2)(x^2-3)(x+1)

3. 24(x+1)/ (x^2)(x^2-2)(x+1)

4. 4(x+1)/(x^2)(x2-2)(x-1)

5. 4(x-1)/(x^2)(x2-2).

Hi Guys! Came by this question and I couldn't find a solution for it:
This is from MGMAT test papers. Couldn't find the official solution.

This what I did:
I assumed:
x=4
Total no. of bulbs =16.
4 bulbs can by chosen 16C4 ways = 4x5x7x13.

And theN I don't know how to narrow the desired possibilities.

Thanks in advance!
Intern  B
Joined: 22 Jun 2017
Posts: 4
Re: A grid of light bulbs measures x bulbs by x bulbs, where x > 2. If 4  [#permalink]

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nick1816 wrote:
I don't think any option is correct.

Number of ways to select 4 bulbs out of x^2 bulbs= $$x^2_ C_4$$ = $$\frac{x^2*(x^2-1)*(x^2-2)*(x^2-3)}{4!}$$

Number of ways those selected 4 bulbs form a 2 bulb by 2 bulb square= $$(x-1)^2$$

Probability= $$(x-1)^2$$/$$\frac{x^2*(x^2-1)*(x^2-2)*(x^2-3)}{4!}$$

probability=$$\frac{24(x-1)}{x^2*(x+1)*(x^2-2)*(x^2-3)}$$

You gotta edit option B

SPatel1992 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 bulb square?

1. 4(x-1)/x^2

2. 24(x-1)/(x^2)(x^2-3)(x+1)

3. 24(x+1)/ (x^2)(x^2-2)(x+1)

4. 4(x+1)/(x^2)(x2-2)(x-1)

5. 4(x-1)/(x^2)(x2-2).

Hi Guys! Came by this question and I couldn't find a solution for it:
This is from MGMAT test papers. Couldn't find the official solution.

This what I did:
I assumed:
x=4
Total no. of bulbs =16.
4 bulbs can by chosen 16C4 ways = 4x5x7x13.

And theN I don't know how to narrow the desired possibilities.

Thanks in advance!

Hi, Can you explain how the no of ways to choose 2 by 2 grid is (x-1)(x-1).
Director  P
Joined: 19 Oct 2018
Posts: 769
Location: India
Re: A grid of light bulbs measures x bulbs by x bulbs, where x > 2. If 4  [#permalink]

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Number of ways you can select the top-left corner bulb of the 2*2 grid in any row is x-1
Number of ways you can select the top-left corner bulb of the 2*2 grid in any column is x-1

Total number of ways= (x-1)*(x-1)

I'm attaching the diagram for 3*3 and 4*4 grids. Number of bulbs that can be selected as top-left corner are highlighted with red color.

sudha3510 wrote:
nick1816 wrote:
I don't think any option is correct.

Number of ways to select 4 bulbs out of x^2 bulbs= $$x^2_ C_4$$ = $$\frac{x^2*(x^2-1)*(x^2-2)*(x^2-3)}{4!}$$

Number of ways those selected 4 bulbs form a 2 bulb by 2 bulb square= $$(x-1)^2$$

Probability= $$(x-1)^2$$/$$\frac{x^2*(x^2-1)*(x^2-2)*(x^2-3)}{4!}$$

probability=$$\frac{24(x-1)}{x^2*(x+1)*(x^2-2)*(x^2-3)}$$

You gotta edit option B

SPatel1992 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 bulb square?

1. 4(x-1)/x^2

2. 24(x-1)/(x^2)(x^2-3)(x+1)

3. 24(x+1)/ (x^2)(x^2-2)(x+1)

4. 4(x+1)/(x^2)(x2-2)(x-1)

5. 4(x-1)/(x^2)(x2-2).

Hi Guys! Came by this question and I couldn't find a solution for it:
This is from MGMAT test papers. Couldn't find the official solution.

This what I did:
I assumed:
x=4
Total no. of bulbs =16.
4 bulbs can by chosen 16C4 ways = 4x5x7x13.

And theN I don't know how to narrow the desired possibilities.

Thanks in advance!

Hi, Can you explain how the no of ways to choose 2 by 2 grid is (x-1)(x-1).

Attachments e.png [ 15.96 KiB | Viewed 200 times ] Re: A grid of light bulbs measures x bulbs by x bulbs, where x > 2. If 4   [#permalink] 18 Jul 2019, 22:55
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