Question

Three numbers are chosen at random without replacement from $1,2,3,.....,10$. The probability that the minimum of the chosen numbers is $3$ or their maximum is $7$ is

• A. $1/2$
• B. $1/3$
• C. $1/4$
• D. $11/40$

Answer 1

The correct option is D $11/40$

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=$\left(\frac{{}^7{C}_2{+}^6{C}_2{-}^3{C}_1}{{}^{10}{C}_3}\right)$
=$\frac{(21+15-3)}{120}$=$\frac{33}{120}=\frac{11}{40}$