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

Let $A$ be the event that the maximum is 6.Let $B$ be the event that the minimum is 3.We want $P(B|A)=\frac{P(A\cap B)}{P(A)}$.Event $A$: Maximum is 6. The other two numbers must be chosen from $\{1,2,3,4,5\}$.Number of ways $n(A)=\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10$.Event $A\cap B$: Maximum is 6 and minimum is 3.The set of numbers is $\{3,x,6\}$. Here $x$ must be from $\{4,5\}$.Number of ways $n(A\cap B)=\left(\genfrac{}{}{0ex}{}{2}{1}\right)=2$.Required probability $=\frac{2}{10}=\frac{1}{5}$.