The event occupies the lower triangle of the unit square: $0\le x\le1$ and $0\le y\le1-x$. Integrate the joint density there: $$P=\int_0^1\left[xy+\frac{y^2}{2}\right]_0^{1-x}dx.$$ The inner integral is $x(1-x)+(1-x)^2/2=(1-x^2)/2$. Its integral is $1/2-1/6=$ **1/3≈0.33333**, within 0.32–0.34.
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