Random Point In Triangle

The probability is the ratio of the area of triangle A to the area of the triangular region enclosed by the lines $3y + 2x = 6, x = 0$ and $y = 0$ , where triangle A is bounded by the lines $3y + 2x = 6, x = 0$ and $x = y$ .

Now the lines $x = y$ and $3y + 2x = 6$ intersect at $(\frac{6}{5}, \frac{6}{5})$ . Thus triangle A has a base length (along the y-axis) of $2$ and a height of $\frac{6}{5}$ , giving an area of $\frac{1}{2} * 2 * \frac{6}{5} = \frac{6}{5}$ . The area of the triangle described in the question is $3$ , so the desired probability is $\dfrac{\frac{6}{5}}{3} = \frac{2}{5} = \boxed{0.4}$ .

The line 3y + 2x = 6 will intersect at (0,2) and (3,0). x=0 and y = 0 are the y and x axis respectively. Thus the enclosed area will be a triangle.

Now the required set of points will be all the points that are above the line x=y, which will pass through the origin and divide the triangle in two smaller parts

Thus the required probability will be the area of triangle above the line x=y divided by the area of the whole triangle. And it comes out to be 2/5

a much faster visual solution is attained by reading the graph generated by WolframAlpha

plot {3y+2x=6, y>x, x>0, y>0}

Area Triangle / Area Large triangle = 1 - Area base tri./ Area large tri. =

1 - (3×1.2/2) ÷ (3×2/2) = 1-0.6=0.4 = Probability

*read 1.2 by grades of 0.2