With
, we have identical PDFs

for
, and 0 otherwise, where

Since
are independent, the joint PDF is

for points
in the unit square, and 0 otherwise.
1. The distribution is continuous, so
.
2.
is the region in the
plane contained within the unit square and above the line
. This region is empty, because this line lies above the square altogether, so
.
3.
is the region in the same square below the line
. So we have
