(i) To find the probability that the socks taken are of different colours, we can use the formula:

\(P(\text{different colours}) = P(RB) + P(BR)\)

Where:

• \(P(RB) = \frac{4}{12} \times \frac{8}{11}\)
• \(P(BR) = \frac{8}{12} \times \frac{4}{11}\)

Thus,

\(P(\text{different colours}) = \frac{4}{12} \times \frac{8}{11} + \frac{8}{12} \times \frac{4}{11} = \frac{16}{33}\)

(ii) The probability distribution table for \(X\), the number of red socks taken, is:

(iii) To find \(E(X)\), the expected value of \(X\), we use:

\(E(X) = 0 \times \frac{14}{33} + 1 \times \frac{16}{33} + 2 \times \frac{3}{33}\)

\(E(X) = \frac{16}{33} + \frac{6}{33} = \frac{22}{33} = \frac{2}{3}\)

First, calculate the possible values of X by squaring each face value and subtracting 4:

• For face 1: \(1^2 - 4 = -3\)
• For face 2: \(2^2 - 4 = 0\)
• For face 3: \(3^2 - 4 = 5\)
• For face 6: \(6^2 - 4 = 32\)

The probability distribution table is:

To find \(E(X)\):

\(E(X) = (-3) \cdot \frac{1}{6} + 0 \cdot \frac{1}{2} + 5 \cdot \frac{1}{6} + 32 \cdot \frac{1}{6} = \frac{-3}{6} + \frac{5}{6} + \frac{32}{6} = \frac{34}{6} = \frac{17}{3} \approx 5.67\)

To find \(\text{Var}(X)\):

\(\text{Var}(X) = \frac{(-3)^2}{6} + \frac{0^2}{2} + \frac{5^2}{6} + \frac{32^2}{6} - \left(\frac{34}{6}\right)^2\)

\(= \frac{9}{6} + \frac{25}{6} + \frac{1024}{6} - \left(\frac{34}{6}\right)^2\)

\(= 144 \left( \frac{1298}{9} \right)\)

(i) To find \(P(X = 3)\), we need the probability of drawing two red discs followed by a blue disc, i.e., \(P(RRB)\).

The probability of drawing the first red disc is \(\frac{2}{6}\), the second red disc is \(\frac{1}{5}\), and the first blue disc is \(\frac{4}{4}\).

Thus, \(P(X = 3) = \frac{2}{6} \times \frac{1}{5} \times \frac{4}{4} = \frac{1}{15}\).

(ii) The probability distribution table for \(X\) is as follows:

\(P(X = 1)\) is the probability of drawing a blue disc first, which is \(\frac{4}{6} = \frac{2}{3}\).

\(P(X = 2)\) is the probability of drawing one red disc followed by a blue disc, i.e., \(P(RB) = \frac{2}{6} \times \frac{4}{5} = \frac{4}{15}\).

\(P(X = 3)\) is already calculated as \(\frac{1}{15}\).

(i) The probability \(P(X = -2) = k(-2)^2 = 4k\) and \(P(X = 2) = k(2)^2 = 4k\). Thus, \(P(X = -2) = P(X = 2)\).

(ii) The probability distribution table is:

Sum of probabilities: \(4k + k + 4k + 16k = 1\)

\(25k = 1\)

\(k = \frac{1}{25}\)

(iii) \(E(X) = (-2)(4k) + (-1)(k) + (2)(4k) + (4)(16k)\)

\(E(X) = -8k - k + 8k + 64k = 63k\)

Substitute \(k = \frac{1}{25}\):

\(E(X) = 63 \times \frac{1}{25} = \frac{63}{25}\)

(i) To find \(P(X = 2)\), we need the combinations where the sum of the numbers from pack A and pack B equals 2. The possible combinations are:

• Card from A is 0 and card from B is 2: Probability = \(\frac{2}{10} \times \frac{4}{6} = \frac{8}{60} = \frac{2}{15}\)

Thus, \(P(X = 2) = \frac{2}{15}\).

(ii) The probability distribution table for \(X\) is constructed by considering all possible sums:

(iii) Given \(X = 3\), we find the probability that the card from pack A is 1. The combinations for \(X = 3\) are:

• Card from A is 1 and card from B is 2: Probability = \(\frac{5}{10} \times \frac{4}{6} = \frac{20}{60} = \frac{1}{3}\)
• Card from A is 3 and card from B is 0: Probability = \(\frac{4}{10} \times \frac{2}{6} = \frac{8}{60} = \frac{2}{15}\)

Total probability for \(X = 3\) is \(\frac{13}{30}\). The probability that the card from A is 1 given \(X = 3\) is:

\(P(A1 | \text{Sum 3}) = \frac{\frac{5}{10} \times \frac{4}{6}}{\frac{13}{30}} = \frac{10}{13}\)