(i) To complete the table, we use the given probabilities and the fact that the total probability is 1.

The probability that the meal is served on time is \(\frac{1}{5}\), so the probability that it is not served on time is \(1 - \frac{1}{5} = \frac{4}{5}\).

The probability that the kitchen is left in a mess is \(\frac{3}{5}\), so the probability that it is not left in a mess is \(1 - \frac{3}{5} = \frac{2}{5}\).

We know that the probability that the meal is not served on time and the kitchen is not left in a mess is \(\frac{3}{10}\).

Using these, we can fill in the table:

(ii) To find the probability that the meal is not served on time given that the kitchen is left in a mess, we use conditional probability:

\(P(\text{Not on time} | \text{Kitchen mess}) = \frac{P(\text{Not on time and Kitchen mess})}{P(\text{Kitchen mess})}\)

From the table, \(P(\text{Not on time and Kitchen mess}) = \frac{1}{2}\) and \(P(\text{Kitchen mess}) = \frac{3}{5}\).

Thus, \(P(\text{Not on time} | \text{Kitchen mess}) = \frac{\frac{1}{2}}{\frac{3}{5}} = \frac{1}{2} \times \frac{5}{3} = \frac{5}{6}\)