Out of 11 consecutive natural numbers if three numbers are selected at random (without repetition), then the probability that they are in A.P. with positive common difference, is:

Consider 11 consecutive natural numbers 1, 2, 3, ………,113 numbers are chosen (without repetition)⇒nS=C3   11E be the event that the selected numbers are in A.P.  Number of A.P's with common difference 1=9 Number of A.P's with common difference 2=7 Number of A.P's with common difference 3=5 Number of A.P's with common difference 4=3 Number of A.P's with common difference 5=1n(E)=9+7+5+3+1=25P(E)=25C3   11=25×611×10×9=533