SOLUTION: If the random variable X has uniform distribution on the interval [0,a] then what is the probability that the random variable greater than its square i.e. P(x > x^2)?

Uniform distribution means the distribution is a rectangle over
the interval [0,a] whose area is 1. The base of the rectangle is
the interval from 0 to a, so the base of the rectangle is "a" units.
Let its height be h.




Area = base∙height
   1 = a∙h
 1/a = h


  
We need to locate a place on that graph where






Since x is positive, x-1 must be negative,




So we locate 1 somewhere on that graph:



and we shade the area under the rectangle to the left of 1, since it's x < 1:



The area of that shaded part, which has base 1 and height 1/a is

Area = base × height = 1 × 1/a = 1/a

That's the answer: the probability is 1/a.

Edwin

1)  First, let's consider the case  a < 1.


    Then for all values of x in the segment [0,a]  x > x^2  (which is ABSOLUTELY OBVIOUS).


    So, if a < 1, then  P(x > x^2) = 1.




2)  Next, consider the case a >= 1.


    Then x > x^2 if and only if  0 <= x < 1.


    Therefore, P(x > x^2) =   ( since the random variable x has uniform distribution on the interval  [0,a]).




3)  Thus the final ANSWER / (the final formula) is THIS :


    if the the random variable X has uniform distribution on the interval [0,a] then

    
        the probability that the random variable is greater than its square, i.e.  P(x > x^2),  is equal to  1,  if a < 1

    and

        the probability that the random variable is greater than its square, i.e.  P(x > x^2),  is equal to  ,  if a >= 1.