Two integers $$x$$ and $$y$$ are chosen with replacement from the set $$\{0, 1, 2, 3, \ldots, 10\}$$. Then the probability that $$|x - y| > 5$$ is :

Two integers $$x, y$$ are chosen from $$\{0, 1, 2, ..., 10\}$$. Total outcomes = $$11 \times 11 = 121$$.

We need $$P(|x - y| > 5)$$, i.e., $$|x - y| \geq 6$$.

Count pairs where $$x - y \geq 6$$: For each difference $$d = 6, 7, 8, 9, 10$$:

$$d = 6$$: $$(6,0),(7,1),(8,2),(9,3),(10,4)$$ → 5 pairs

$$d = 7$$: 4 pairs

$$d = 8$$: 3 pairs

$$d = 9$$: 2 pairs

$$d = 10$$: 1 pair

Total with $$x - y \geq 6$$: $$5 + 4 + 3 + 2 + 1 = 15$$.

By symmetry, pairs with $$y - x \geq 6$$ = 15.

Total favorable = $$15 + 15 = 30$$.

$$P = \frac{30}{121}$$.

The answer is Option (1): $$\boxed{\frac{30}{121}}$$.