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?
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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