AMC 12 B, 2024

Let $r$, $s$, and $t$ be the roots of $x^3 + a x^2 + b x + 6$. Then
$$x^3 + a x^2 + b x + 6 = (x - r)(x - s)(x - t) = x^3 - (r + s + t)x^2 + (rs + st + tr)x - rst,$$
so $rst = -6$, $r + s + t = -a$, and $rs + st + tr = b$. The only triples of distinct integers that satisfy the first of these three equations are $(6, 1, -1)$, $(3, 2, -1)$, $(3, -2, 1)$, $(-3, 2, 1)$, and $(-3, -2, -1)$, together with their permutations. The corresponding values of $a$ and $b$ are $(-6, -1)$, $(-4, 1)$, $(-2, -5)$, $(0, -7)$, and $(6, 11)$, respectively. Notice that these ordered pairs are distinct, but in only $4$ of them do both $a$ and $b$ have absolute value not exceeding $10$. There are $21 \cdot 20$ equally likely choices for $a$ and $b$, so the required probability is $\frac{4}{21 \cdot 20} = \frac{1}{105}$.