Let the probability of choosing a rope of length 3 m be \(p = \frac{4}{5}\) and the probability of choosing a rope of length 5 m be \(1 - p = \frac{1}{5}\).

(ii) The probability that two ropes have different lengths is given by:

\(P(3, 5) + P(5, 3) = p \times (1-p) + (1-p) \times p\)

\(= \frac{4}{5} \times \frac{1}{5} + \frac{1}{5} \times \frac{4}{5}\)

\(= \frac{8}{25}\) (0.32)

(iii) The probability that three ropes have a total length of 11 m can occur in the following ways: (3, 3, 5), (3, 5, 3), (5, 3, 3). The probability for each arrangement is:

\(P(3, 3, 5) = p \times p \times (1-p)\)

\(= \frac{4}{5} \times \frac{4}{5} \times \frac{1}{5}\)

\(= \frac{16}{125}\)

Since there are 3 such arrangements, the total probability is:

\(3 \times \frac{16}{125} = \frac{48}{125}\) (0.384)