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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?

ColdplayMX  Aug 15, 2018
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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

 

 

cool cool cool

CPhill  Aug 15, 2018

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