Let $$S = \{1, 2, 3, \ldots, 2022\}$$. Then the probability, that a randomly chosen number n from the set S such that $$HCF(n, 2022) = 1$$, is

We need to find the probability that a randomly chosen number $$n$$ from $$S = \{1, 2, 3, \ldots, 2022\}$$ satisfies $$\gcd(n, 2022) = 1$$.

We first factorise 2022: $$2022 = 2 \times 1011 = 2 \times 3 \times 337$$. We verify that 337 is prime: it is not divisible by 2, 3, 5, 7, 11, 13, 17, or 19 (and $$19^2 = 361 > 337$$), so 337 is indeed prime.

The count of numbers from 1 to $$N$$ that are coprime to $$N$$ is given by Euler's totient function $$\phi(N)$$. Since $$2022 = 2 \times 3 \times 337$$:

$$\phi(2022) = 2022\left(1 - \frac{1}{2}\right)\left(1 - \frac{1}{3}\right)\left(1 - \frac{1}{337}\right) = 2022 \times \frac{1}{2} \times \frac{2}{3} \times \frac{336}{337}$$

We compute step by step: $$2022 \times \frac{1}{2} = 1011$$, then $$1011 \times \frac{2}{3} = 674$$, then $$674 \times \frac{336}{337} = \frac{674 \times 336}{337} = \frac{2 \times 337 \times 336}{337} = 2 \times 336 = 672$$.

So $$\phi(2022) = 672$$, and the required probability is $$\frac{672}{2022}$$. We simplify: $$\gcd(672, 2022) = 6$$ (since $$672 = 6 \times 112$$ and $$2022 = 6 \times 337$$), giving $$\frac{672}{2022} = \frac{112}{337}$$.

Hence, the correct answer is Option D.