AP EAMCET Engineering 2013 Solved Paper

Case I When smaller of the two numbers is 1.
Then, total number of cases
=1×‌7C1=7
Case II When smaller of two numbers is 2.
Then, total number of cases
=1×‌6C1=6
Case III When smaller of two numbers is 3.
Then, total number of cases
=1×‌5C1=5
Total favourable cases =7+6+5=18
Total case =‌8C2=28
∴ Required probability =

18

28

=

9

14