+1206 A point (x,y) is randomly selected such that \(0 \le x \le 3\) and \(0 \le y \le 6\). What is the probability that \(x+y \le 4\)? Express your answer as a common fraction.
+128399 See this graph : https://www.desmos.com/calculator/3psnr4vmhw
The area covered by the first two inequalities is 18 units^2
The inequality x + y ≤ 4 forms a trapezoid inside this region
Its area is (1/2)(3) ( 4 + 1) = 15/2 units^2
So....the probability that a point is chosen inside the trapezoid in the feasible region is
(15/2) / 18 = 15/ 36 = 5 / 12
