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, ………,11

3 numbers are chosen (without repetition)

$\Rightarrow n\left(S\right)={}^{11}C_3$

E be the event that the selected numbers are in A.P. 

$\text{ Number of A.P's with common difference }1=9$

$\text{ Number of A.P's with common difference }2=7$

$\text{ Number of A.P's with common difference }3=5$

$\text{ Number of A.P's with common difference }4=3$

$\begin{array}{l}\text{ Number of A.P's with common difference }5=1\\ \mathrm{n}\left(\mathrm{E}\right)=9+7+5+3+1=25\\ \mathrm{P}\left(\mathrm{E}\right)=\frac{25}{{}^{11}C_3}=\left(\frac{25\times 6}{11\times 10\times 9}\right)=\frac{5}{33}\end{array}$