Probability

Let's start by drawing a diagram of the triangle XYZ:

```
         X
         |\
         | \
         |  \
         |   \
         |    \
         |     \
         |      \
         |       \
         |        \
         |         \
         |          \
         |__________\
         Y    12     Z
            6
```

The area of triangle XYZ is (1/2) * 12 * 6 = 36.

Let's choose a coordinate system with the origin at point X, and the x-axis along the line XY. Then the coordinates of points Y and Z are (12, 0) and (0, 6), respectively.

Since point D is chosen at random within the triangle, any point in the triangle is equally likely to be chosen. So we can assume that the coordinates of point D are (x, y), where 0 ≤ x ≤ 12 and 0 ≤ y ≤ 6.

The area of triangle XYD is given by the formula:

(1/2) * base * height

where the base is the length of XY, and the height is the perpendicular distance from point D to line XY. The height can be expressed as:

h = (6 - y) * (x/12)

Therefore, the area of triangle XYD is:

A = (1/2) * 12 * (6 - y) * (x/12) = x * (6 - y)

We want the probability that the area of triangle XYD is at most 20. So we need to find the region in the xy-plane where x * (6 - y) ≤ 20.

If we graph this region, we get a trapezoid with vertices at (0, 0), (0, 6), (20/3, 2/3), and (12, 0):

```
       |\
       | \
       |  \
       |   \
       |    \
       |     \
       |      \
       |       \
       |        \
       |         \
       |__________\
```

The area of this trapezoid is:

(1/2) * (6) * (20/3) + (1/2) * (12 - 20/3) * (6 - 2/3) = 44/3

Therefore, the probability that the area of triangle XYD is at most 20 is:

P(A ≤ 20) = (area of trapezoid) / (area of triangle XYZ) = (44/3) / 36 = 11/27

So the probability that the area of triangle XYD is at most 20 is 11/27.