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.=
