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Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Location: India
WE:Education (Education)
T is defined as the region in the first quadrant satisfying the inequa
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07 Mar 2022, 00:49
07 Mar 2022, 00:49
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Question Stats:
40% (02:06) correct
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T is defined as the region in the first quadrant satisfying the inequality 3x + 4y < 12.
If a point P with coordinates (x, y) lies within the region T, what is the probability that x > 2?
If a point P with coordinates (x, y) lies within the region T, what is the probability that x > 2?
Show: ::
\(\frac{1}{4}\)
Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Location: India
WE:Education (Education)
T is defined as the region in the first quadrant satisfying the inequa
[#permalink]
25 Mar 2022, 06:39
25 Mar 2022, 06:39
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OA Explanation:
3x + 4y < 12
4y < -3x + 12
y < \(\frac{-3x}{4} + 3\)
Draw a line going down from left to right having slope \(\frac{-3}{4}\) and y-intercept at (0,3).
Since, we have < sign - we need to shade the region below the line
Now, let us find out the y-value on line when x = 2
Plug x = 2 in the above line;
\(y = \frac{-3}{4}(2) + 3\)
\(y = \frac{3}{2}\)
Probability that x > 2 (Green region in the figure below) = \(\frac{Ar. (△DAB)}{Ar. (△COB)} = \frac{(2)(1.5)}{(4)(3)} = \frac{1}{4}\)
_________________
3x + 4y < 12
4y < -3x + 12
y < \(\frac{-3x}{4} + 3\)
Draw a line going down from left to right having slope \(\frac{-3}{4}\) and y-intercept at (0,3).
Since, we have < sign - we need to shade the region below the line
Now, let us find out the y-value on line when x = 2
Plug x = 2 in the above line;
\(y = \frac{-3}{4}(2) + 3\)
\(y = \frac{3}{2}\)
Probability that x > 2 (Green region in the figure below) = \(\frac{Ar. (△DAB)}{Ar. (△COB)} = \frac{(2)(1.5)}{(4)(3)} = \frac{1}{4}\)
Attachments
T is defined as the region in the first quadrant.png [ 9.15 KiB | Viewed 1215 times ]
_________________
I hope this helps!
Regards:
Karun Mendiratta
Founder and Quant Trainer
Prepster Education, Delhi, India
https://www.instagram.com/prepster_education/
Regards:
Karun Mendiratta
Founder and Quant Trainer
Prepster Education, Delhi, India
https://www.instagram.com/prepster_education/
Re: T is defined as the region in the first quadrant satisfying the inequa
[#permalink]
25 Mar 2022, 07:54
25 Mar 2022, 07:54
Expert Reply
Great explanation. Your post "best replies' is on its way
_________________
_________________
Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Location: India
WE:Education (Education)
Re: T is defined as the region in the first quadrant satisfying the inequa
[#permalink]
25 Mar 2022, 08:41
25 Mar 2022, 08:41
Carcass wrote:
Great explanation. Your post "best replies' is on its way
Thanks, just trying to help the community
_________________
I hope this helps!
Regards:
Karun Mendiratta
Founder and Quant Trainer
Prepster Education, Delhi, India
https://www.instagram.com/prepster_education/
Regards:
Karun Mendiratta
Founder and Quant Trainer
Prepster Education, Delhi, India
https://www.instagram.com/prepster_education/
Re: T is defined as the region in the first quadrant satisfying the inequa
[#permalink]
20 Jul 2023, 08:44
20 Jul 2023, 08:44
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Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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