There is a math question that is bothering me for a long time : given unit square pick two points on adjacent sides of the square , what is the probability that the area of the resulting triangle is smaller that 1/8? I tried to tackle the problem by first sketching a square with the four vertices at (0,0), (1,1), (0,1) and (1,1). The answer seems to be the area under the curve y=0.25/x between 1 and 0, but how can you integrate such a function when ln0 is undefined? any help would be highly appreciated.

Hi Hoda,

In this problem X and Y lie on the side of unit square, that is X and Y vary in the interval [0,1]. If you draw a graph of unit square (as you suggested with vertices (0,0), (0,1), (1,0) and (1,1) ) and the graph of the function Y = 0.25/X (this is hyperbola) you will see that the curse intercepts square at point (0.25,1) and (1,0.25) (the last is not important). So, we need to integrate the function on the interval [.25,1] and after this add the area of rectangle with vertices (0,0), (0.25,0), (0.25,1), (0,1). The function Ln(0.25) and Ln(1) are well defined. My answer to the problem is (0.25)*(Ln4 +1 )

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