We are given a set S = {0, 1, 2, ..., 10}. We need to choose two different numbers from this set. The total number of ways to choose 2 different numbers from 11 elements is given by the combination formula: $C_{11}^2=\frac{11\times 10}{2}=55$. So, total number of elementary events = 55.

We want the probability that for the chosen two numbers (say a and b, with a > b without loss of generality), both (a + b) and (a - b) are multiples of 4.

Note: Since a and b are integers, (a + b) and (a - b) are either both even or both odd. But for both to be multiples of 4, they must be even. Actually, we require:

$a+b\equiv 0(mod4)$ and $a-b\equiv 0(mod4)$.

Adding these two congruences: (a+b) + (a-b) = 2a ≡ 0 (mod 4) ⇒ a ≡ 0 (mod 2). So a is even.
Subtracting: (a+b) - (a-b) = 2b ≡ 0 (mod 4) ⇒ b ≡ 0 (mod 2). So b is even.

Thus, both a and b must be even. So we only consider even numbers from the set: {0, 2, 4, 6, 8, 10}. There are 6 even numbers.

Now, let a and b be even. Write a = 2a₁, b = 2b₁, where a₁ and b₁ are integers. Then:

$a+b=2(a_1+b_1)$ and $a-b=2(a_1-b_1)$.

For these to be divisible by 4, we need (a₁ + b₁) ≡ 0 (mod 2) and (a₁ - b₁) ≡ 0 (mod 2). But note: (a₁ + b₁) and (a₁ - b₁) have the same parity. So both must be even. This implies that a₁ and b₁ are both even or both odd.

So, we need to choose two even numbers from {0,2,4,6,8,10} such that when divided by 2, the resulting integers are either both even or both odd.

List the even numbers with their half-values:

• 0 → 0 (even)
• 2 → 1 (odd)
• 4 → 2 (even)
• 6 → 3 (odd)
• 8 → 4 (even)
• 10 → 5 (odd)

So, the even numbers that yield even half: {0,4,8} (3 numbers)
The even numbers that yield odd half: {2,6,10} (3 numbers)

Now, we choose two numbers from the even set. For the condition to hold:

Case 1: Both numbers from the first group (both halves even). Number of ways = $C_3^2=3$.

Case 2: Both numbers from the second group (both halves odd). Number of ways = $C_3^2=3$.

So, total favorable outcomes = 3 + 3 = 6.

Therefore, the required probability = favorable outcomes / total outcomes = $\frac{6}{55}$.

Related Topics

This problem involves concepts from probability and number theory (modular arithmetic). Specifically, it requires understanding of:

• Combinations: Counting ways to choose elements from a set.
• Modular arithmetic: Checking divisibility conditions.
• Parity: Properties of even and odd numbers.

Formulae

Key formulae used:

• Number of ways to choose r items from n: $C_n^r=\frac{n!}{r!(n-r)!}$
• Congruence: a ≡ b (mod m) means m divides (a-b).
• Probability = (Number of favorable outcomes) / (Total number of outcomes)