Three numbers are chosen at random without replacement from 1, 2,...

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

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

$P\left(A\right) = \dfrac{C2 5}{C3 8}$ (number less than $6$ is $5$)

$P\left(B\right) = \dfrac{C2 5}{C3 8}$ (number greater than $3$ is $5$)

$P\left(A \cap B\right) = \dfrac{C1 2}{C3 8}$

$∴$ Required probability

$= P\left(B / A\right) = \dfrac{P\left(A \cap B\right)}{P\left(A\right)} = \dfrac{C1 2}{C2 5} = \dfrac{2}{10} = \dfrac{1}{5}$