There are 2 methods that I know of to find this probability: one using double integrals, and the other using geometric probability. When you have a uniform distribution, you can do that and avoid some crazy calculus.
If I could show the graph, I would, because, regardless of which method you use, it's crucial that you begin with the right visual. So I'll need to describe it.
Since this is really two identical and independent random variables, we have a two-dimensional graph. On the xy coordinate plane, the interval (5, 14) is bounded by vertical lines at x = 8 and x = 14. The same interval is represented on the y axis, bounded by y = 5 and y = 14. The two independent observations are the area that these rectangles have in common, namely,. 9 by 9 square bounded by 5 and 14 both horizontally and vertically.
So that's the basic visual for the joint random variable. What we want is the probability that the sum of x and y is greater than 22, represented by the inequality x + y › 22. This graph needs to be overlaid on the square that is our random variable sample space. When we graph y › 22 - x, it intersects the square at (14, 8) and (8, 14). The space that concerns us is the right triangle this forms in the upper right corner of the square.
If we approach this geometrically, we're almost done. All that's needed is to find the area of the triangle (.5 · 6 · 6 = 18) and the square (9 · 9 = 81). The probability of being in the success region with area 18, given the sample space region with area 81. So the probability is 18/81 = 2/9.
I can also step through the double integral method in a later addendum if anyone is interested.
