(i) To find \(P(X = 3)\), we consider the sequences where the process stops after 3 apples. The sequences are GRR and RGR. The probability for GRR is \(\frac{2}{4} \times \frac{2}{3} \times \frac{1}{2} = \frac{1}{6}\). The probability for RGR is \(\frac{2}{4} \times \frac{2}{3} \times \frac{1}{2} = \frac{1}{6}\). Therefore, \(P(X = 3) = \frac{1}{6} + \frac{1}{6} = \frac{1}{3}\).

(ii) The probability distribution table for \(X\) is:

(iii) To find the probability that all 3 peppers are orange given that at least 2 are orange, we use conditional probability:

\(P(3 \text{ orange } | \text{ at least 2 orange }) = \frac{P(3 \text{ orange })}{P(\text{at least 2 orange})}\).

\(P(3 \text{ orange }) = \frac{5}{7} \times \frac{4}{6} \times \frac{3}{5} = \frac{2}{7}\).

\(P(\text{at least 2 orange}) = P(2 \text{ orange, 1 yellow}) + P(3 \text{ orange})\).

\(P(2 \text{ orange, 1 yellow}) = \frac{5}{7} \times \frac{4}{6} \times \frac{2}{5} + \frac{5}{7} \times \frac{2}{6} \times \frac{4}{5} + \frac{2}{7} \times \frac{5}{6} \times \frac{4}{5} = \frac{6}{7}\).

Thus, \(P(3 \text{ orange } | \text{ at least 2 orange }) = \frac{\frac{2}{7}}{\frac{6}{7}} = \frac{1}{3}\).