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In the coordinate plane, rectangular region R has vertices a
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15 Nov 2010, 11:50
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In the coordinate plane, rectangular region R has vertices at (0,0), (0,3), (4,3), and (4,0). If a point in region R is randomly selected, what is the probability that the point's ycoordinate will be greater than its xcoordinate? A. 7/12 B. 5/12 C. 3/8 D. 1/3 E. 1/4
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In the coordinate plane, rectangular region R has vertices a
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15 Nov 2010, 12:06
shrive555 wrote: In the coordinate plane, rectangular region R has vertices at (0,0), (0,3), (4,3), and (4,0). If a point in region R is randomly selected, what is the probability that the point's ycoordinate will be greater than its xcoordinate?
7/12 5/12 3/8 1/3 1/4 See the diagram below. Now, rectangle R has an area of 3*4=12. All point that has ycoordinate greater than xcoordinate lie above the line \(y=x\), so in yellow triangle, which has an area of 1/2*3*3=4.5. So, the probability equals to favorable outcomes/total=yellow triangle/rectangle R=4.5/12=3/8. Answer: C. Attachment:
graph.PNG [ 15.18 KiB  Viewed 22471 times ]
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15 Nov 2010, 12:21
i took that y intercept 3 to be greater than x all coordinates and divided by 12 and got 1/4. which is completely wrong. and according to your solution in that yellow region every point is greater than x not just only y intercept. Great.
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Re: In the coordinate plane, rectangular region R has vertices a
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27 Jun 2013, 23:46
Bumping for review and further discussion*. Get a kudos point for an alternative solution! *New project from GMAT Club!!! Check HERE
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Re: In the coordinate plane, rectangular region R has vertices a
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28 Jun 2013, 04:43
Bunuel wrote: Bumping for review and further discussion*. Get a kudos point for an alternative solution! *New project from GMAT Club!!! Check HERE I think Bunuel's above explanation is most illuminating and detailed; I do not find any alternate way for the explanation. +1 for that great solution.
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Re: In the coordinate plane, rectangular region R has vertices a
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08 Jan 2014, 16:44
Hi, this is how I did it: The probability if X=3 that Y>X is 0 The probability if X=2 that Y>X is 1/4 The probability if X=1 that Y>X is 2/4 The probability if X=0 that Y>X is 3/4 Then: \(1/4*1/4 + 1/4*2/4 + 1/4*3/4 = 6/16 = 3/8\) Hope it helps
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Re: In the coordinate plane, rectangular region R has vertices a
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08 Jan 2014, 23:48
shrive555 wrote: In the coordinate plane, rectangular region R has vertices at (0,0), (0,3), (4,3), and (4,0). If a point in region R is randomly selected, what is the probability that the point's ycoordinate will be greater than its xcoordinate?
A. 7/12 B. 5/12 C. 3/8 D. 1/3 E. 1/4 there is no better solution than this compute the area above y=x line and below y=x line and then proceed to finding probability



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Re: In the coordinate plane, rectangular region R has vertices a
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29 Mar 2014, 06:47
Bunuel wrote: Bumping for review and further discussion*. Get a kudos point for an alternative solution! *New project from GMAT Club!!! Check HEREI calculated that the no. of points in the region including the boundaries of the rectangle=\(16\). Out of that \(6\) points [for \(6\) ordered pairs\((0,1),(0,2),(0,3),(1,2),(1,3)\) and \((2,3)\)],will have \(x\) coordinate less than \(y\) coordinate. So probability=\(6/16=3/8\) Isn't this approach right?



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Re: In the coordinate plane, rectangular region R has vertices a
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29 Mar 2014, 10:12
AKG1593 wrote: Bunuel wrote: Bumping for review and further discussion*. Get a kudos point for an alternative solution! *New project from GMAT Club!!! Check HEREI calculated that the no. of points in the region including the boundaries of the rectangle=\(16\). Out of that \(6\) points [for \(6\) ordered pairs\((0,1),(0,2),(0,3),(1,2),(1,3)\) and \((2,3)\)],will have \(x\) coordinate less than \(y\) coordinate. So probability=\(6/16=3/8\) Isn't this approach right? No, this approach is not right. A point has no dimension, hence there are infinitely many points in any area/segment. The problem with your solution is that you assume that the coordinates must be integers, which is nowhere given. Hope it's clear.
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Re: In the coordinate plane, rectangular region R has vertices a
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08 Jul 2014, 20:02
Hi Bunuel...I came across this question and kept getting caught up by the y=x line. If we calculate the isoc triangle's area, wouldn't this include all points on the y=x line (on which y is not > x)?
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Re: In the coordinate plane, rectangular region R has vertices a
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09 Jul 2014, 03:19
m3equals333 wrote: Hi Bunuel...I came across this question and kept getting caught up by the y=x line.
If we calculate the isoc triangle's area, wouldn't this include all points on the y=x line (on which y is not > x)? A line has no area, hence there won't be any difference.
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Re: In the coordinate plane, rectangular region R has vertices a
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09 Jul 2014, 14:50
ah ha, that clears it up, thanks a bunch
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Re: In the coordinate plane, rectangular region R has vertices a
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06 Nov 2016, 04:43
Bunuel wrote: AKG1593 wrote: Bunuel wrote: Bumping for review and further discussion*. Get a kudos point for an alternative solution! *New project from GMAT Club!!! Check HEREI calculated that the no. of points in the region including the boundaries of the rectangle=\(16\). Out of that \(6\) points [for \(6\) ordered pairs\((0,1),(0,2),(0,3),(1,2),(1,3)\) and \((2,3)\)],will have \(x\) coordinate less than \(y\) coordinate. So probability=\(6/16=3/8\) Isn't this approach right? No, this approach is not right. A point has no dimension, hence there are infinitely many points in any area/segment. The problem with your solution is that you assume that the coordinates must be integers, which is nowhere given. Hope it's clear. Hi Brunei, How the rectangle has 16 points (if we calculate the borders ). Isn't it should be 20 . e.g (0,0),(1,0),(2,0),(3,0),(4,0). 5 points in 1 row and 4 rows all together 5*4 = 20 . Am I missing something ?



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Re: In the coordinate plane, rectangular region R has vertices a
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06 Nov 2016, 04:59
Ganganshu wrote: Hi Brunei,
How the rectangle has 16 points (if we calculate the borders ). Isn't it should be 20 . e.g (0,0),(1,0),(2,0),(3,0),(4,0). 5 points in 1 row and 4 rows all together 5*4 = 20 . Am I missing something ? I don't understand why are you concerned about the number of points? The solution talks about the area...
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Re: In the coordinate plane, rectangular region R has vertices a
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06 Nov 2016, 05:14
Bunuel wrote: Ganganshu wrote: Hi Brunei,
How the rectangle has 16 points (if we calculate the borders ). Isn't it should be 20 . e.g (0,0),(1,0),(2,0),(3,0),(4,0). 5 points in 1 row and 4 rows all together 5*4 = 20 . Am I missing something ? I don't understand why are you concerned about the number of points? The solution talks about the area... Brunei , I didnot get your point(in earlier quote) why calculating points is wrong (I know we get an incorrect answer ). Isn't Probability = No of successes(condition)/ Total no of outcomes and if we go by points arenot we just following it .



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Re: In the coordinate plane, rectangular region R has vertices a
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06 Nov 2016, 06:09
Ganganshu wrote: Bunuel wrote: Ganganshu wrote: Hi Brunei,
How the rectangle has 16 points (if we calculate the borders ). Isn't it should be 20 . e.g (0,0),(1,0),(2,0),(3,0),(4,0). 5 points in 1 row and 4 rows all together 5*4 = 20 . Am I missing something ? I don't understand why are you concerned about the number of points? The solution talks about the area... Brunei , I didnot get your point(in earlier quote) why calculating points is wrong (I know we get an incorrect answer ). Isn't Probability = No of successes(condition)/ Total no of outcomes and if we go by points arenot we just following it . (Probability) = (Favorable outcomes)/(Total) = (The area of yellow triangle)/(The area of the rectangle). Notice that we have an area there no the number of poiint whose coordinates are integers only. Check other Probability and Geometry questions in our Special Questions Directory. Hope it helps.
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Re: In the coordinate plane, rectangular region R has vertices a
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18 Nov 2016, 05:26
Narenn wrote: Bunuel wrote: Bumping for review and further discussion*. Get a kudos point for an alternative solution! *New project from GMAT Club!!! Check HERE I think Bunuel's above explanation is most illuminating and detailed; I do not find any alternate way for the explanation. +1 for that great solution. I did graph the region correctly. However, when I looked at the region I thought the selected region is slightly less than half. However, the answer choices are quite close. So you have to calculate the area, no choice.



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In the coordinate plane, rectangular region R has vertices a
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05 Aug 2019, 20:29
Bunuel wrote: shrive555 wrote: In the coordinate plane, rectangular region R has vertices at (0,0), (0,3), (4,3), and (4,0). If a point in region R is randomly selected, what is the probability that the point's ycoordinate will be greater than its xcoordinate?
7/12 5/12 3/8 1/3 1/4
Now, rectangle R has an area of 3*4=12. All point that has ycoordinate greater than xcoordinate lie above the line \(y=x\), so in yellow triangle, which has an area of 1/2*3*3=4.5. So, the probability equals to favorable outcomes/total=yellow triangle/rectangle R=4.5/12=3/8.
Answer: C.
I do not understand or cannot recalculate the 1/2*3*3=4.5. How did you get 1/2 as other side?




In the coordinate plane, rectangular region R has vertices a
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