From the given description, I drew a right traingle in XY-grid with the following 3 points: (0,0), (0,10), and (5,0). I suggest you draw this one too so follow my logic. The area of this trainagle, let's call it S1, is

S1 = 0.5*10*5 = 25

Now, all points with y>x will be above y=x line. This line divides original triangle into 2 trinagles, where the "upper" one (let's call it S2) will have all points where y>x. So, if we find the area of S2, then Prob=S2/S1.

To find S2, I first found the intersection point on the original (S1's) hypotenuse by solving for intersection point given two quations for the straight lines: y=-2x+10 and y=x.

From here, -2x+10=x

or, x=10/3.

So, the height of the S2 is 10/3 and the base is 10 => S2=0.5*10*(10/3)=50/3

Finally, Prob(y>x)=S2/S1 = (50/3)/25 = 2/3

Unfortunately, it took me way over 2 minutes to get to this point, because I originally assumed that one of the angles should be 30 degrees, which sent me on a wrong path for a while