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by knight247 » Thu Sep 08, 2011 1:13 pm
In the xy- plane, a triangle has vertices (0,0), (4,0) and (4,5). If a point (x,y) is selected at random from the triangular region, What is the probabilty that x-y>0 ?

A. 1/5
B. 1/3
c. 1/2
D. 2/3
E. 4/5

OA is E. Detailed explanations would be appreciated. Thanks
Last edited by knight247 on Fri Sep 09, 2011 4:44 am, edited 1 time in total.
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by GMATGuruNY » Thu Sep 08, 2011 1:37 pm
knight247 wrote:In the xy- plane, a triangle has vertexes (0,0), (4,0) and (4,5). If a point (x,y) is selected at random from the triangular region, What is the probabilty that x-y>0 ?

A. 1/5
B. 1/3
c. 1/2
D. 2/3
E. 4/5

OA is E. Detailed explanations would be appreciated. Thanks
Question rephrased: In what portion of the triangle is y<x?

Image

Area of the whole triangle = (1/2)*4*5 = 10.
The portion below y=x is where y<x.
This region is the triangle with vertices at (0,0), (4,0) and (4,4).
Area of this triangle = (1/2)*4*4 = 8.
(Portion below y=x)/(Total area) = 8/10 = 4/5.

The correct answer is E.
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by n@resh » Thu Sep 08, 2011 3:08 pm
GMATGuruNY wrote:
knight247 wrote:In the xy- plane, a triangle has vertexes (0,0), (4,0) and (4,5). If a point (x,y) is selected at random from the triangular region, What is the probabilty that x-y>0 ?

A. 1/5
B. 1/3
c. 1/2
D. 2/3
E. 4/5

OA is E. Detailed explanations would be appreciated. Thanks
Question rephrased: In what portion of the triangle is y<x?

Image

Area of the whole triangle = (1/2)*4*5 = 10.
The portion below y=x is where y<x.
This region is the triangle with vertices at (0,0), (4,0) and (4,4).
Area of this triangle = (1/2)*4*4 = 8.
(Portion below y=x)/(Total area) = 8/10 = 4/5.

The correct answer is E.
Beautiful strategy!
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by saketk » Thu Sep 08, 2011 8:43 pm
I agree. That's a simple and clear explanation. Thanks a lot, Mitch.
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by ziko » Sun Jul 15, 2012 9:09 pm
Truly nice explanation Mitch! Many thanks!
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