Three numbers are chosen at random without replacement from {1, 2, 3, 4, 5, 6, 7, 8}. The probability that their minimum is 3, (given that their maximum is 6) is :

$\frac{1}{5}$

The correct answer is Option (2) → $\frac{1}{5}$

for 6 to be max

other two numbers can be chosen from subset → $\{1, 2, 3, 4, 5\}$

with 3 as min only $\{4,5\}$ are elements to be chosen 

so probability = $\frac{2}{{^5C}_2}=\frac{2×2}{5×4}=\frac{1}{5}$