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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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50% (02:16) 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?
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\(\frac{1}{4}\)
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
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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) = \(\frac{Ar. (△DAB)}{Ar. (△COB)} = \frac{(2)(1.5)}{(4)(3)} = \frac{1}{4}\)
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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
