3. Let $x, y, z$ be non-negative real numbers not less than 1. If $k=x+y(1-x)+z(1-x)(1-y)$, then the value range of $k$ is $\qquad$ .
3. Prompt: From the given, we have
$$
k=1-(1-x)(1-y)(1-z) \text { . }
$$

And $0 \leqslant 1-x \leqslant 1,0 \leqslant 1-y \leqslant 1,0 \leqslant 1-z \leqslant 1$.
Thus $0 \leqslant 1-(1-x)(1-y)(1-z) \leqslant 1$. That is, $0 \leqslant k \leqslant 1$.