(i) To find the probability that Noor chooses exactly one T-shirt, we use combinations. The total number of ways to choose 3 items from 12 is given by \(\binom{12}{3}\).

The number of ways to choose 1 T-shirt from 3 is \(\binom{3}{1}\), and the number of ways to choose 2 items from the remaining 9 (blouses and jumpers) is \(\binom{9}{2}\).

Thus, the probability is:

\(\frac{\binom{3}{1} \times \binom{9}{2}}{\binom{12}{3}} = \frac{3 \times 36}{220} = \frac{108}{220} = \frac{27}{55}\)

(ii) The probability distribution table for X is constructed by considering the possible values of X (0, 1, 2, 3) and their respective probabilities:

To find the expectation of the difference between the two scores, we first determine the probability distribution of the differences.

The expectation is calculated as:

\(E(X) = \sum (\text{difference} \times \text{probability})\)

\(E(X) = (0 \times \frac{6}{36}) + (1 \times \frac{10}{36}) + (2 \times \frac{8}{36}) + (3 \times \frac{6}{36}) + (4 \times \frac{4}{36}) + (5 \times \frac{2}{36})\)

\(E(X) = \frac{0 + 10 + 16 + 18 + 16 + 10}{36}\)

\(E(X) = \frac{70}{36}\)

\(E(X) \approx 1.94\)

(i) To find the probability distribution, consider all possible outcomes when two dice are thrown. There are 36 possible outcomes (6 sides on the first die and 6 sides on the second die).

- The score is 0 when both dice show the same number. There are 6 such outcomes: (1,1), (2,2), (3,3), (4,4), (5,5), (6,6). So, P(0) = \(\frac{6}{36}\).

- For scores 1 to 5, count the outcomes where the smaller number is 1, 2, 3, 4, or 5, respectively:

• Score 1: (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (3,1), (4,1), (5,1), (6,1) - 10 outcomes, so P(1) = \(\frac{10}{36}\).
• Score 2: (2,3), (2,4), (2,5), (2,6), (3,2), (4,2), (5,2), (6,2) - 8 outcomes, so P(2) = \(\frac{8}{36}\).
• Score 3: (3,4), (3,5), (3,6), (4,3), (5,3), (6,3) - 6 outcomes, so P(3) = \(\frac{6}{36}\).
• Score 4: (4,5), (4,6), (5,4), (6,4) - 4 outcomes, so P(4) = \(\frac{4}{36}\).
• Score 5: (5,6), (6,5) - 2 outcomes, so P(5) = \(\frac{2}{36}\).

(ii) The expected score is calculated using the formula for expected value: \(E(X) = \sum x_i p_i\).

\(E(X) = 0 \times \frac{6}{36} + 1 \times \frac{10}{36} + 2 \times \frac{8}{36} + 3 \times \frac{6}{36} + 4 \times \frac{4}{36} + 5 \times \frac{2}{36}\)

\(E(X) = \frac{0 + 10 + 16 + 18 + 16 + 10}{36} = \frac{70}{36} \approx 1.94\)

(a) To find \(P(X = 2)\), we consider the scenarios where the highest number is 2. The possible outcomes are (2,2,2), (2,2,1), (2,1,2), (1,2,2), (2,1,1), (1,2,1), and (1,1,2). There are 7 such outcomes.

The total number of outcomes when spinning three 4-sided spinners is \(4 \times 4 \times 4 = 64\).

Thus, \(P(X = 2) = \frac{7}{64}\).

(b) To complete the probability distribution table, we need to find \(P(X = 1)\) and \(P(X = 4)\).

\(P(X = 1)\) occurs only when all spinners show 1, which is (1,1,1). Thus, \(P(X = 1) = \frac{1}{64}\).

For \(P(X = 4)\), we use the fact that the probabilities must sum to 1:

\(P(X = 4) = 1 - \left( \frac{1}{64} + \frac{7}{64} + \frac{19}{64} \right) = \frac{37}{64}\).

(i) The probability distribution table is:

Since the total probability must equal 1, we have:

\(k + 2k + 3k + 4k = 1\)

\(10k = 1\)

Thus, \(k = \frac{1}{10}\).

(ii) The mean \(E(X)\) is calculated as:

\(E(X) = 1 \times \frac{1}{10} + 2 \times \frac{2}{10} + 3 \times \frac{3}{10} + 4 \times \frac{4}{10}\)

\(E(X) = \frac{1}{10} + \frac{4}{10} + \frac{9}{10} + \frac{16}{10} = 3\)

The variance \(\text{Var}(X)\) is calculated as:

\(\text{Var}(X) = \left(1^2 \times \frac{1}{10} + 2^2 \times \frac{2}{10} + 3^2 \times \frac{3}{10} + 4^2 \times \frac{4}{10}\right) - E(X)^2\)

\(\text{Var}(X) = \left(\frac{1}{10} + \frac{8}{10} + \frac{27}{10} + \frac{64}{10}\right) - 3^2\)

\(\text{Var}(X) = \frac{100}{10} - 9 = 1\)