AP EAPCET 23-Aug-2021 Shift 1 Paper

Two integers are drawn from {5,6,...,35}
Given, integer are 31
Total ways of choosing 2 integer from 31 integers =‌31C2
Two integer difference will be odd only when one is even and other is odd.
Let A be the event of selecting two integers whose difference is odd.
Now, P(A)+P(A)=1
P(A)= Probability of selecting two integers whose difference is not odd.
Case I When two integers are even.
Total ways of choosing 2 integer (even) from 15 even integers =‌15C2
Case II when two integers are odd
Total ways =‌16C2
P(A)=

‌15C2+‌16C2

‌31C2

=

15×14 2 + 16×15 2

31×302

P(A)=

(15×14)+(16×15)

31×30

=

14+16

31×2

=

30

2×31

=

15

31

∴P(A)=1−P(A) =1−

15

31

=

31−15

31

=

16

31