(a) The tree diagram shows the following probabilities:

• Choosing bag X or Y: each with probability \(\frac{1}{2}\).
• For bag X: Removing a red marble \(\frac{7}{10}\), blue marble \(\frac{3}{10}\).
• For bag Y: Removing a red marble \(\frac{4}{5}\), blue marble \(\frac{1}{5}\).
• After adding a red marble, for bag X: Removing a red marble \(\frac{8}{10}\), blue marble \(\frac{2}{10}\).
• After adding a red marble, for bag Y: Removing a red marble \(\frac{5}{6}\), blue marble \(\frac{1}{6}\).

(b) Probability that both marbles are the same colour:

\(P(\text{both same colour}) = P(BB) + P(RR) = P(XBB) + P(XRR) + P(YRR)\)

\(= \frac{1}{2} \times \frac{3}{10} \times \frac{2}{10} + \frac{1}{2} \times \frac{7}{10} \times \frac{7}{10} + \frac{1}{2} \times \frac{4}{5} \times \frac{4}{5}\)

\(= \frac{6}{200} + \frac{49}{200} + \frac{16}{50}\)

\(= 0.03 + 0.245 + 0.32 = 0.595\)

(c) Probability that bag Y is chosen given different colours:

\(P(\text{bag } Y | \text{different colours}) = \frac{P(\text{bag } Y \cap \text{different colours})}{P(\text{different colours})}\)

\(= \frac{\frac{1}{2} \times \frac{4}{5} \times \frac{1}{6} + \frac{1}{2} \times \frac{1}{5} \times \frac{5}{6}}{1 - 0.595}\)

\(= \frac{\frac{4}{50} + \frac{1}{12}}{0.405}\)

\(= \frac{9}{81} = 0.444\)