(i) Calculate the probability for each scenario where exactly two balls have the same number:

- Probability of selecting 2 from A, 2 from B, and any from C: \(\frac{1}{4} \times \frac{1}{5} \times \frac{1}{7} = \frac{1}{140}\)

- Probability of selecting 8 from A, 8 from B, and any from C: \(\frac{1}{4} \times \frac{2}{5} \times \frac{3}{7} = \frac{3}{70}\)

- Probability of selecting 8 from A, any from B, and 8 from C: \(\frac{1}{4} \times \frac{3}{5} \times \frac{4}{7} = \frac{3}{35}\)

- Probability of selecting any from A, 8 from B, and 8 from C: \(\frac{3}{4} \times \frac{2}{5} \times \frac{4}{7} = \frac{6}{35}\)

Summing these probabilities gives \(\frac{47}{140}\).

(ii) Given that exactly two of the selected balls have the same number, the probability that they are both numbered 2 is:

\(\frac{\frac{1}{28}}{\frac{47}{140}} = \frac{5}{47}\)

(iii) Calculate \(P(X) = \frac{47}{140}\) and \(P(Y) = \frac{1}{4}\). Check independence:

\(P(X \cap Y) = \frac{1}{28}\)

Since \(P(X \cap Y) \neq P(X) \times P(Y)\), events X and Y are not independent.