+212 Two numbers, x and y are selected at random from the interval (0, 3). What is the probability that a triangle with sides of length 1, x, and y exists?
+129899 I think this can be solved by a graph as well, Coldplay...here : https://www.desmos.com/calculator/myszb3kghi
The total possible area for (x, y) is a 3 by 3 square = 9 units^2
The feasible region of (x, y) values occur inside the boundary points (0,1), (1,0), (2,3), (3,2) and (3,3)
We have an upper rigtt triangle with an area of (1/2) (product of the leg lengths) = (1/2) (1) (1) = 1/2
And we have a rectangle of length √ [ (3 - 1)^2 + (2 - 0)^2 ] = √ [2^2 + 2^2] = √8
And a height of √[ (2-3)^2 + (3 - 2)^2] = √ [1^2 + 1^2 ] = √2
So...the area of this rectangle is √8 * √2 = √16 = 4
So....the toal area of the feasible region is [ 4 + 1/2] = 4.5 units^2
So...the probability that a triangle exists is 4.5 / 9 = 1/2
