Let $A$ be the event that maximum is $6$.

$B$ be the event that minimum is $3$.

$P\left(A\right)=\frac{{}^{5}C_2}{{}^{8}C_3}$ (number less than $6$ is $5$)

$P\left(B\right)=\frac{{}^{5}C_2}{{}^{8}C_3}$ (number greater than $3$ is $5$)

$P\left(A\cap B\right)=\frac{{}^{2}C_1}{{}^{8}C_3}$

$\therefore$ Required probability

$=P\left(B/A\right)=\frac{P\left(A\cap B\right)}{P\left(A\right)}=\frac{{}^{2}C_1}{{}^{5}C_2}=\frac{2}{10}=\frac{1}{5}$