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SPatel1992
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A grid of light bulbs measures x bulbs by x bulbs, where x > 2. If 4 Updated on: 25 Jun 2019, 15:49
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.
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83% (01:47) correct 17% (02:14) wrong based on 50 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).

Show Spoiler
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!

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A grid of light bulbs measures x bulbs by x bulbs, where x > 2. If 4 25 Jun 2019, 15:36
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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
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).

Show Spoiler
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!
Show more
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SPatel1992
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A grid of light bulbs measures x bulbs by x bulbs, where x > 2. If 4 25 Jun 2019, 15:48
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Thanks for pointing it out. I am confused. How did you get (x^2-1)? Can you solve this using numbers?


nick1816
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
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).

Show Spoiler
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!
Show more
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sudha3510
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A grid of light bulbs measures x bulbs by x bulbs, where x > 2. If 4 18 Jul 2019, 22:20
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nick1816
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
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).

Show Spoiler
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!
Show more


Hi, Can you explain how the no of ways to choose 2 by 2 grid is (x-1)(x-1).
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A grid of light bulbs measures x bulbs by x bulbs, where x > 2. If 4 18 Jul 2019, 22:55
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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
nick1816
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
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).

Show Spoiler
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).
Show more

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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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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!
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A grid of light bulbs measures x bulbs by x bulbs, where x > 2. If 4 07 Dec 2023, 20:49
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