(i) The probability of getting a green robot from one packet is \(\frac{1}{4}\) since there are four equally likely colours.

(ii) The probability that Nick gets his first green robot on the fifth packet means he does not get a green robot in the first four packets and then gets a green robot on the fifth. The probability of not getting a green robot in one packet is \(\frac{3}{4}\). Therefore, the probability is \(\left( \frac{3}{4} \right)^4 \times \frac{1}{4} = \frac{81}{1024} = 0.0791\).

(iii) The probability that the first four packets Amos opens all contain different coloured robots is calculated by considering the permutations of the four colours. The probability is \(\frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} \times 4! = \frac{3}{32}\).

There are 3 even numbers (2, 4, 6) and 2 odd numbers (1, 7) in the box.

The total number of ways to choose 3 discs from 5 is given by \(\binom{5}{3} = 10\).

To find the number of favorable outcomes (2 even and 1 odd), we can choose 2 even numbers from 3 and 1 odd number from 2:

\(\binom{3}{2} \times \binom{2}{1} = 3 \times 2 = 6\).

Thus, the probability is:

\(\frac{6}{10} = 0.6\).

The tree diagram is constructed as follows:

1. First choice: Sharik has a probability of \(\frac{1}{3}\) of choosing the correct answer (C) and \(\frac{2}{3}\) of choosing a wrong answer (W).
2. Second choice (if first was wrong): From the remaining two answers, the probability of choosing the correct answer is \(\frac{1}{2}\) and the wrong answer is \(\frac{1}{2}\).
3. Third choice (if second was also wrong): The remaining answer must be correct, so the probability is 1.

The tree diagram shows these probabilities and the paths leading to the correct answer.

(i) To find the probability that there is a winner after exactly two sets, we consider two scenarios: Roger wins both sets (RR) or Andy wins both sets (AA).

The probability that Roger wins both sets (RR) is:

\(P(RR) = 0.6 \times 0.7 = 0.42\)

The probability that Andy wins both sets (AA) is:

\(P(AA) = 0.4 \times 0.75 = 0.3\)

Thus, the probability that there is a winner after exactly two sets is:

\(P(2 \text{ sets in match}) = P(RR) + P(AA) = 0.42 + 0.3 = 0.72\)

(ii) To find the probability that Andy wins the match given that there is a winner after exactly two sets, we use conditional probability:

\(P(A \text{ wins and 2 sets}) = P(AA) = 0.3\)

\(P(\text{2 sets}) = 0.72\)

Thus, the probability is:

\(\frac{P(AA)}{P(\text{2 sets})} = \frac{0.3}{0.72} = \frac{5}{12}\)

To find the probability that Nur chooses a play-house, we calculate the probability for each playground and sum them up.

For Playground X, the probability of choosing a play-house is:

\(P(X \text{ and } P) = \frac{1}{4} \times \frac{4}{9} = \frac{1}{9}\)

For Playground Y, the probability of choosing a play-house is:

\(P(Y \text{ and } P) = \frac{1}{4} \times \frac{2}{12} = \frac{1}{24}\)

For Playground Z, the probability of choosing a play-house is:

\(P(Z \text{ and } P) = \frac{1}{2} \times \frac{1}{16} = \frac{1}{32}\)

Summing these probabilities gives:

\(P(P) = \frac{1}{9} + \frac{1}{24} + \frac{1}{32} = \frac{53}{288} \approx 0.184\)