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In the coordinate plane, rectangular region R has vertices a

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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 y-coordinate will be greater than its x-coordinate?

A. 7/12
B. 5/12
C. 3/8
D. 1/3
E. 1/4
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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 y-coordinate will be greater than its x-coordinate?

7/12
5/12
3/8
1/3
1/4
See the diagram below.
Attachment:
graph.PNG
graph.PNG [ 15.18 KiB | Viewed 8385 times ]
Now, rectangle R has an area of 3*4=12. All point that has y-coordinate greater than x-coordinate 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.
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New post 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 [#permalink]

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New post 28 Jun 2013, 04:43
Bunuel wrote:
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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:) 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 [#permalink]

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New post 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 [#permalink]

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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 y-coordinate will be greater than its x-coordinate?

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 [#permalink]

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New post 29 Mar 2014, 06:47
Bunuel wrote:
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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I 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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Bunuel wrote:
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I 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 [#permalink]

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New post 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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New post 09 Jul 2014, 14:50
ah ha, that clears it up, thanks a bunch
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In the coordinate plane, rectangular region R has vertices [#permalink]

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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 y-coordinate will be greater than its x-coordinate?

(A) 7/12
(B)5/12
(C) 3/8
(D) 1/3
(E) 1/4
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Re: In the coordinate plane, rectangular region R has vertices a [#permalink]

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New post 13 Oct 2015, 11:18
tuanquang269 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 y-coordinate will be greater than its x-coordinate?

(A) 7/12
(B)5/12
(C) 3/8
(D) 1/3
(E) 1/4


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Re: In the coordinate plane, rectangular region R has vertices a [#permalink]

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New post 31 Jul 2016, 04:47
Bunuel 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 y-coordinate will be greater than its x-coordinate?

7/12
5/12
3/8
1/3
1/4
See the diagram below.
Attachment:
graph.PNG
Now, rectangle R has an area of 3*4=12. All point that has y-coordinate greater than x-coordinate 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.

Quote:
Bunuel,
How do you arrive at the point where the line "y=x" cuts the rectangular region?

Thanks again!

Regards,
Yosita
Re: In the coordinate plane, rectangular region R has vertices a   [#permalink] 31 Jul 2016, 04:47
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