Solution for probability Problem for area

Thanks everyone for trying the problem and posting in the comments, and congrats for those of you who got the right answer!


Problem:
Suppose a and b are single-digit positive integers chosen independently and at random. What is the probability that the point (a,b) lies above the parabola y=ax^2-bx.


Solution:
Because the point (a,b) needs to be above the parabola ax^2-bx, the value of y has to be greater than the function itself. Therefore we get y>ax^2-bx. Next, we're finding the probability that point (a,b) is above the parabola, so we just plug in a for x and b for y. Thus, we get b>a*a^2-b*a = a^3-ab. Now we can solve for b in terms of a.


b>a^3-ab 
b+ab>a^3
b(1+a)>a^3
b>(a^3)/(1+a) (We can divide by (1+a) because in the problem it states a is positive)


Now we have the inequality b>(a^3)/(1+a). From this you can see that (a^3)/(1+a) is an increasing function, since a^3 grows faster than 1+a. From this, you see that the largest integer you can plug in for a is 3 because plugging in a 4 for a requires b to be a double digit integer. We have restricted a to 1≤a≤3. Now, we plug in values of a and see what positive single digit integers of b can make the inequality true.


If we plug in 1 for a, we get b>1/4. All numbers 1-9 can be plugged in for b for the inequality to hold. Which gives 9 possible ordered pairs. 
If we plug in 2 for a, we get b>8/3. Numbers 3-9 can be plugged in for b which gives 7 ordered pairs.
If we plug in 3 for a, we get b>27/4. Number 7-9 can be plugged in for b which gives 3 ordered pairs.


In total, the number of possible ordered pairs is 9+7+3=19. This is out of the total number of 9*9=81 ordered pairs. Hence the probability is 19/81.


Here's the next problem:
Problem:
A square of side length 1 and a circle of radius (√3)/3 share the same center. What is the area inside the circle, but outside the square?


Good luck! and remember calculators are not allowed.