(i) The tree diagram is completed as follows:

• First ball: Probability of red (R) = \(\frac{3}{8}\), Probability of blue (B) = \(\frac{5}{8}\).
• Second ball after red: Probability of red (R) = \(\frac{2}{8}\), Probability of blue (B) = \(\frac{5}{8}\), Probability of yellow (Y) = \(\frac{1}{8}\).
• Second ball after blue: Probability of red (R) = \(\frac{3}{8}\), Probability of blue (B) = \(\frac{4}{8}\), Probability of yellow (Y) = \(\frac{1}{8}\).

(ii) Probability that the two balls taken are the same colour:

\(P(RR) = \frac{3}{8} \times \frac{2}{8} = \frac{3}{32}\)

\(P(BB) = \frac{5}{8} \times \frac{4}{8} = \frac{5}{16}\)

Total probability = \(\frac{3}{32} + \frac{5}{16} = \frac{13}{32}\) or 0.406

(iii) Probability that the first ball taken is red, given that the second ball taken is blue:

\(P(RB) = \frac{3}{8} \times \frac{5}{8} = \frac{15}{64}\)

\(P(B) = \frac{3}{8} \times \frac{5}{8} + \frac{5}{8} \times \frac{4}{8} = \frac{35}{64}\)

\(P(R|B) = \frac{P(RB)}{P(B)} = \frac{15/64}{35/64} = \frac{3}{7}\) or 0.429