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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.

Logic  Nov 25, 2018
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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

 

 

cool cool cool

CPhill  Nov 25, 2018
edited by CPhill  Nov 25, 2018

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