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:

Let us denote the 11 consecutive natural numbers by $$1,2,3,\dots ,11$$. Because only the relative positions matter, any other block of 11 consecutive numbers would give exactly the same count, so this labelling is allowed.

The total number of unordered selections of three numbers from these 11 is given by the combination formula

$$^nC_r=\frac{n!}{r!\,(n-r)!},$$

where $$n=11$$ and $$r=3$$. Substituting, we obtain

$$^{11}C_3=\frac{11!}{3!\,8!}=\frac{11\times10\times9}{3\times2\times1}=165.$$

Now we must count how many of those 165 triplets form an increasing arithmetic progression (A.P.) with positive common difference. Suppose the first term of such a progression is $$x$$ and the common difference is $$d$$, where $$d\ge 1$$. Then the three terms are

$$x,\;x+d,\;x+2d.$$

All three terms must lie in the set $$\{1,2,\dots ,11\}$$, so the largest term must satisfy

$$x+2d\le 11.$$

For a given positive integer $$d$$, the smallest permissible value of $$x$$ is $$1$$, and the largest is $$11-2d$$. Hence, for that particular $$d$$, the number of admissible values of $$x$$ equals

$$11-2d.$$

This quantity is positive only as long as $$11-2d\ge 1\; \Longrightarrow\; 2d\le 10\; \Longrightarrow\; d\le 5.$$ Therefore $$d$$ can take the integer values $$1,2,3,4,5$$. We now list the counts for each $$d$$ and add them.

For $$d=1$$: admissible $$x=1\text{ to }9$$, giving $$11-2(1)=9$$ progressions.
For $$d=2$$: admissible $$x=1\text{ to }7$$, giving $$11-2(2)=7$$ progressions.
For $$d=3$$: admissible $$x=1\text{ to }5$$, giving $$11-2(3)=5$$ progressions.
For $$d=4$$: admissible $$x=1\text{ to }3$$, giving $$11-2(4)=3$$ progressions.
For $$d=5$$: admissible $$x=1\text{ to }1$$, giving $$11-2(5)=1$$ progression.

Adding these gives the total number of favourable triplets:

$$9+7+5+3+1 = 25.$$

The required probability is therefore

$$\text{Probability}=\frac{\text{favourable cases}}{\text{total cases}}=\frac{25}{165}.$$

Simplifying the fraction, we divide numerator and denominator by their highest common factor, which is $$5$$:

$$\frac{25}{165}=\frac{25\div5}{165\div5}=\frac{5}{33}.$$

Hence, the correct answer is Option C.