GRE Prep 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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
Location: India
WE:Education (Education)
T is defined as the region in the first quadrant satisfying the inequa
[#permalink]
07 Mar 2022, 00:49
07 Mar 2022, 00:49
1
10
Bookmarks
00:00
Question Stats:
50% (02:16) correct
50% (02:16) wrong based on 8 sessions
Hide Show timer Statistics
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: ::
\(\frac{1}{4}\)
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
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
3
2
Bookmarks
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 2106 times ]
General Discussion
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
