6. Given the function $f(x)=\frac{1}{3} x^{3}-(a-1) x^{2}+b^{2} x$, where $a \in\{1,2,3,4\}, b \in\{1,2,3\}$. Then the probability that the function $f(x)$ is increasing on $\mathbf{R}$ is
6 6 $\frac{3}{4}$ Hint: Since $f(x)$ is an increasing function on $\mathbf{R}$, we have $f^{\prime}(x)=x^{2}-$ $2(a-1) x+b^{2} \geqslant 0$ always holds. Therefore, $\Delta=4(a-1)^{2}-4 b^{2} \leqslant 0$ always holds. That is, $|a-1| \leqslant|b|$. Thus,
$$
\begin{aligned}
(a, b)= & (1,1),(2,1),(1,2),(2,2),(3,2), \\
& (1,3),(2,3),(3,3),(4,3)
\end{aligned}
$$

There are 9 in total. Since all possible $(a, b)$ are $4 \times 3=12$, the probability that $f(x)$ is an increasing function on $\mathbf{R}$ is $\frac{9}{12}=\frac{3}{4}$.