Three numbers are chosen at random without replacement from {1, 2, 3, ...... 8}. The probability that their minimum is 3, given that their maximum is 6, is

Let A be the event that maximum is 6.
B be event that minimum is 3

$P\left(A\right)=\frac{{ }^5{C}_2}{{ }^8{C}_3}\text{ (the numbers }<6\text{ are }5\text{ ) }$

$P\left(B\right)=\frac{{ }^5{C}_2}{{ }^8{C}_3}\text{ (the numbers }>3\text{ are }5\text{ ) }$

$P(A\cap B)=\frac{{ }^2{C}_1}{{ }^8{C}_3}$

$\text{ Required probability is }P\left(\frac{B}{A}\right)=\frac{P(A\cap B)}{P\left(A\right)}=\frac{{ }^2{C}_1}{{ }^5{C}_2}=\frac{2}{10}=\frac{1}{5}\text{. }$