AP EAMCET Engineering 26 Apr 2017 Shift 1 Paper

The total number of possible outcomes is 10C3Let A be the event that minimum of chosen number is 3 and B be the event that maximum of chosen number is 7 .
Number of ways of choosing remaining two numbers from A is,
n(A)=7C2
=21
Number of ways of choosing remaining two numbers from B is,
n(B)=6C2
=15
Therefore, the required probability is,
P=

21+15−3

120

=

33

120

=

11

40