1. Total number of ways to select 3 numbers

From 11 consecutive natural numbers, the number of ways to choose any 3 distinct numbers is

$\binom{11}{3} = \frac{11\cdot 10\cdot 9}{3\cdot 2\cdot 1} = 165.$

2. Condition for three selected numbers to be in A.P.

Let the three selected numbers be $a<b<c$. For them to be in arithmetic progression with positive common difference, we must have

$b-a = c-b,$

so equivalently,

$2b=a+c.$

Thus, the middle term must be exactly the average of the other two.

3. Count all 3-term A.P.s from 11 consecutive numbers

Take the numbers as

$1,2,3,\dots,11.$

A 3-term A.P. is determined by its middle term and common difference $d>0$:

$m-d,\ m,\ m+d.$

We count possible choices of $m$ and $d$ such that all terms lie between 1 and 11.

• For $m=2$, possible $d$: $1$ → 1 A.P.
• For $m=3$, possible $d$: $1,2$ → 2 A.P.s
• For $m=4$, possible $d$: $1,2,3$ → 3 A.P.s
• For $m=5$, possible $d$: $1,2,3,4$ → 4 A.P.s
• For $m=6$, possible $d$: $1,2,3,4,5$ → 5 A.P.s
• For $m=7$, possible $d$: $1,2,3,4$ → 4 A.P.s
• For $m=8$, possible $d$: $1,2,3$ → 3 A.P.s
• For $m=9$, possible $d$: $1,2$ → 2 A.P.s
• For $m=10$, possible $d$: $1$ → 1 A.P.

Total favorable selections:

$1+2+3+4+5+4+3+2+1 = 25.$

4. Required probability

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

5. Compare with options

$\frac{5}{33}$ corresponds to Option B.

6. Comparison with stored correct answer

Stored correct answer is B, which matches our result.