Selection of 3 no.'s from {1,2,3,4,5,6,7,8,9,10}
$\Rightarrow$ No.of combinations to be made=${}^{10}{C}_3$
Selection of 3 no.'s having 3 as minimum should be selected from the set {3,4,5,6,7,8,9,10} for which one number is fixed i.e.,3
$\Rightarrow$ No.of combinations to be made=${}^7{C}_2$
selection of 3 no.'s having 7 as maximum should be selected from the set {1,2,3,4,5,6,7} for which one number is fixed i.e.,7
$\Rightarrow$ No.of combinations to be made=${}^6{C}_2$
selection of 3 no.'s having 7 as maximum  and 3 as minimum should be selected from the set {3,4,5,6,7} for which two numbers fixed i.e.,3 and 7
$\Rightarrow$ No.of combinations to be made=${}^3{C}_1$
Therefore the probability that the minimum of the chosen numbers is 3 or their maximum is 7 is=$({\displaystyle \frac{{}^7{C}_2{+}^6{C}_2{-}^3{C}_1}{{}^{10}{C}_3}})$
=${\displaystyle \frac{(21+15-3)}{120}}$=${\displaystyle \frac{33}{120}}={\displaystyle \frac{11}{40}}$