please help

AJWILL wrote:in the (x,y) plane, a triangle have vertex (0,0), (0,10), (5,0). if point (x, y) is selected at random from a triangular region, what is the probability that y<x ?

(If you can't see the figure, click here)

Points that are inside the triangular region with x-coordinate greater than y-coordinate must reside below the line x = y, i.e. inside the blue triangle.

Now, number of points inside a two-dimensional region is directly proportional to the area of the region. If we consider k as the proportionality constant,

• Total number of points inside the triangular region = k*(area of the triangle)
  Total number of points satisfying the given criteria = k*(area of the blue triangle)

Now, both triangle have same base, OB.
But their heights are OA and y-coordinate of D, respectively.

The equation of the line AB is (2x + y) = 10
Hence, the coordinate of point D is (10/3, 10/3)

Hence, required probability = [k*(10/3)]/[k*10] = 1/3