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

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

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 [ 15.18 KiB | Viewed 7463 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.

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. _________________

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. _________________

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

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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

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

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08 Jan 2014, 23:48

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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

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?

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.

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

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30 Jul 2015, 11:12

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

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13 Oct 2015, 11:08

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?

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

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13 Oct 2015, 11:18

Expert's post

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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