(a) To find the probability that all 3 eggs chosen contain the same colour sweet, consider the combinations:

- All red: \(\frac{3}{12} \times \frac{2}{11} \times \frac{1}{10} = \frac{6}{1320}\)

- All orange: \(\frac{4}{12} \times \frac{3}{11} \times \frac{2}{10} = \frac{24}{1320}\)

- All yellow: \(\frac{5}{12} \times \frac{4}{11} \times \frac{3}{10} = \frac{60}{1320}\)

Total probability: \(\frac{6 + 24 + 60}{1320} = \frac{90}{1320} = 0.0682\)

(b) Given that all three children have the same colour sweet, the probability that all 3 eggs contain a yellow sweet is:

\(\frac{\frac{60}{1320}}{\frac{90}{1320}} = \frac{60}{90} = \frac{2}{3}\) or 0.667

(c) To find the probability that at least one child chooses an egg with an orange sweet, use the complement rule:

Probability of no orange sweets: \(\frac{8}{12} \times \frac{7}{11} \times \frac{6}{10} = \frac{336}{1320} = \frac{14}{55}\)

Probability of at least one orange sweet: \(1 - \frac{14}{55} = \frac{41}{55}\) or 0.745