Out of 11 consecutive natural number if three numbers are selected...

$2b = a + c ⇒$sum of $a\text{\text{ and }}c$ is even and $b$ depends on $a\text{\text{ and }}c$

Thus, we can not choose all three independently, but we can choose $a\text{\text{ and }}c$

Moreover, $a\text{and}c$ should be both either even or odd as their sum is even.

Out of $11$ consecutive natural numbers, either five are even and six are odd or six are even and five are odd.

Hence, $P = \dfrac{1}{2}\dfrac{C2 5 + C2 6}{C3 11} + \dfrac{1}{2}\dfrac{C2 6 + C2 5}{C3 11} = \dfrac{5}{33}$