jmc

Call the six numbers selected $x_1>x_2>x_3>x_4>x_5>x_6$. Clearly, $x_1$ must be a dimension of the box, and $x_6$ must be a dimension of the brick. If $x_2$ is a dimension of the box, then any of the other three remaining dimensions will work as a dimension of the box. That gives us $3$ possibilities. If $x_2$ is not a dimension of the box but $x_3$ is, then both remaining dimensions will work as a dimension of the box. That gives us $2$ possibilities. If $x_4$ is a dimension of the box but $x_2,x_3$ aren’t, there are no possibilities (same for $x_5$). The total number of arrangements is $\left(\genfrac{}{}{0ex}{}{6}{3}\right)=20$; therefore, $p=\frac{3+2}{20}=\frac{1}{4}$, and the answer is $1+4=\boxed{5}$.