Probability Part 1

Out of 11 consecutive natural numbers either 6 even and 5 odd numbers or 5 even and 6 odd numbers
when 3 numbers are selected at random then total cases =‌11C3
since these 3 numbers are in A.P. Let no's are a,b,c
2b ⇒ even number
a+c⇒(

even+even

odd+odd

)
so favourable cases =‌6C2+‌5C2 =15+10=25
P(3‌numbers‌are‌in‌A.P.=

25

11C3

=

25

165

=

5

33

)