(i) To complete the table, we use the given information:

• There are 12 biscuits wrapped in gold foil, 7 of which are chocolate-covered. Therefore, 5 are not chocolate-covered.
• There are 17 chocolate-covered biscuits in total, so 10 must be unwrapped.
• There are 30 biscuits in total, so 13 are not chocolate-covered. Of these, 5 are wrapped, so 8 are unwrapped.

(ii) The probability that a biscuit is wrapped in gold foil is given by:

\(\frac{12}{30} = 0.4\)

(iii) The probability that an unwrapped biscuit is chocolate-covered is:

\(\frac{10}{18} = \frac{5}{9} \approx 0.556\)

(iv) The probability that a biscuit is unwrapped, given that it is chocolate-covered, is:

\(\frac{10}{17} \approx 0.588\)

(v) The probability that Nasir takes exactly 2 wrapped biscuits is calculated using combinations:

\(\frac{\binom{12}{2} \times \binom{18}{2}}{\binom{30}{4}}\)

Calculating each part:

• \(\binom{12}{2} = 66\)
• \(\binom{18}{2} = 153\)
• \(\binom{30}{4} = 27,405\)

Thus, the probability is:

\(\frac{66 \times 153}{27,405} = 0.368\)

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)

Let the probability of landing on the blue side be \(p\).

Then the probability of landing on the red side is \(4p\), and the probability of landing on the green side is \(3p\).

Since the total probability must equal 1, we have:

\(4p + p + 3p = 1\)

\(8p = 1\)

\(p = \frac{1}{8}\)

Thus, the probability of landing on the blue side is \(\frac{1}{8}\).

For part (ii), the probabilities are:

\(P(R) = \frac{1}{2}\), \(P(B) = \frac{1}{8}\), \(P(G) = \frac{3}{8}\).

The probability that the spinner lands on a different coloured side each time is:

\(P(\text{all different}) = \frac{1}{2} \times \frac{1}{8} \times \frac{3}{8} \times 3!\)

\(= \frac{9}{64}\)

Therefore, the probability is \(\frac{9}{64}\) or approximately 0.141.

To find the probability that Ronnie wins, we consider the scenarios where Ronnie draws the first yellow ball.

1. Probability that Julie draws a green ball first and Ronnie draws a yellow ball second:

\(P(GY) = \frac{5}{8} \times \frac{3}{7} = \frac{15}{56}\)

2. Probability that both Julie and Ronnie draw green balls first, and Julie draws a green ball again, then Ronnie draws a yellow ball:

\(P(GGGY) = \frac{5}{8} \times \frac{4}{7} \times \frac{3}{6} \times \frac{3}{5} = \frac{3}{28}\)

3. Probability that the first five draws are green balls, and Ronnie draws a yellow ball on his third turn:

\(P(GGGGGY) = \frac{5}{8} \times \frac{4}{7} \times \frac{3}{6} \times \frac{2}{5} \times \frac{1}{4} \times \frac{3}{3} = \frac{1}{56}\)

Adding these probabilities gives the total probability that Ronnie wins:

\(P(R \text{ wins}) = \frac{15}{56} + \frac{3}{28} + \frac{1}{56} = \frac{11}{28}\)

To find the probability that Mrs Lin sits directly behind a student and Mrs Brown sits in the front row, we can use the following approach:

There are 14 seats, and 12 passengers are seated randomly. We need to calculate the probability of the specific seating arrangement.

First, calculate the total number of ways to arrange 12 passengers in 14 seats: \(^{14}P_{12}\).

Next, calculate the favorable outcomes:

1. Choose 3 seats for Mrs Brown in the front row: 3 ways.
2. Choose 10 seats for Mrs Lin to sit directly behind a student: 10 ways.
3. Choose 5 students to sit in front of Mrs Lin: 5 ways.

The number of favorable outcomes is: \(3 \times 10 \times 5 \times ^{11}P_9\).

The probability is then given by: \(\frac{3 \times 10 \times 5 \times ^{11}P_9}{^{14}P_{12}}\).

Calculating this gives: \(\frac{150}{2184} = 0.0687\).