Question 1151395
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<pre>

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) = {{{1/a}}}  ( since the random variable x has uniform distribution on the interval  [0,a]).




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


    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  {{{1/a}}},  if a >= 1.
</pre>

Solved.


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Notice, I edited my post after receiving a note from Edwin.


What you see now in my post is the final version.


Thanks to Edwin for his note.