Duque3476

We can begin by drawing a diagram of triangle XYZ with legs XY = 12 and YZ = 6. Let's label the right angle as angle Z, and the hypotenuse as XZ.

Let $A$ denote the area of triangle $XYZ$. We find that $A = (1/2)(12)(6) = 36$. Since the area of triangle $XYD$ is half the area of triangle $XYZ$, triangle $XYD$ has area at most 20 if and only if $XYD$ has area at most 10.

Let $E$ denote the foot of the altitude from $X$ to line $DZ$. Then triangle $XYE$ is similar to triangle $XYZ$, so the altitude from $X$ to $YZ$ has length $3$. Since $XY=12$, the length of the altitude from $X$ to line $DY$ is $2A/XY=3/2$.

Hence the length of $DE$ is $6-(3/2)=9/2$, so the area of triangle $XYD$ is $(1/2)(12)(9/2)=27$, which is greater than $20$. Therefore, the probability that the area of triangle $XYD$ is at most 20 is $\boxed{0}$.