The possible values of the random variable X are 0, 1, and 2.

To find the probability distribution, consider the following cases:

1. P(X = 0): No chocolates are taken. This occurs if a toffee or boiled sweet is taken from Susan's bag and then a toffee or boiled sweet is taken from Ahmad's bag.

\(P(T, B) + P(T, T) = \frac{5}{12} \times \frac{2}{10} + \frac{5}{12} \times \frac{5}{10} = \frac{7}{24} \approx 0.292\)

\(2. P(X = 2): Two chocolates are taken. This occurs if a chocolate is taken from Susan's bag and then a chocolate is taken from Ahmad's bag.\)

\(P(C, C) = \frac{7}{12} \times \frac{4}{10} = \frac{28}{120} = \frac{7}{30} \approx 0.233\)

\(3. P(X = 1): One chocolate is taken. This is the complement of the other two probabilities.\)

\(P(X = 1) = 1 - \frac{7}{24} - \frac{7}{30} = \frac{19}{40} \approx 0.475\)

Thus, the probability distribution is:

(i) To find \(P(X = 9)\), we consider the combinations of scores that sum to 9: (1, 4, 4), (2, 3, 4), and (3, 3, 3). The probabilities are calculated as follows:

\(P(1, 4, 4) = \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} \times 3 = \frac{3}{64}\)

\(P(2, 3, 4) = \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} \times 6 = \frac{6}{64}\)

\(P(3, 3, 3) = \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} = \frac{1}{64}\)

Adding these probabilities gives \(P(X = 9) = \frac{3}{64} + \frac{6}{64} + \frac{1}{64} = \frac{10}{64}\).

(ii) The completed probability distribution table is:

(iii) To determine if \(R\) and \(S\) are independent, we calculate:

\(P(S) = \frac{6}{64}\)

\(P(R \cap S) = \frac{3}{64}\)

Since \(P(R \cap S) \neq P(R) \times P(S)\), events \(R\) and \(S\) are not independent.

To find the probability distribution for the number of green pens Rod takes, we calculate the probabilities for 0, 1, 2, and 3 green pens.

Probability of 0 green pens:

\(P(0) = \frac{7}{10} \times \frac{6}{9} \times \frac{5}{8} = \frac{210}{720}\)

Probability of 1 green pen:

\(P(1) = \frac{3}{10} \times \frac{7}{9} \times \frac{6}{8} \times \binom{3}{1} = \frac{378}{720}\)

Probability of 2 green pens:

\(P(2) = \frac{3}{10} \times \frac{2}{9} \times \frac{7}{8} \times \binom{3}{2} = \frac{126}{720}\)

Probability of 3 green pens:

\(P(3) = \frac{3}{10} \times \frac{2}{9} \times \frac{1}{8} = \frac{6}{720}\)

These probabilities sum to 1, confirming the distribution is correct.

(i) To find the mean of W, calculate the expected value:

Mean = \(\frac{1+1+1+2+3+3}{6} = \frac{11}{6} \approx 1.83\)

For the standard deviation, use:

SD = \(\sqrt{\frac{(1-1.83)^2 + (1-1.83)^2 + (1-1.83)^2 + (2-1.83)^2 + (3-1.83)^2 + (3-1.83)^2}{6}}\)

SD = \(\sqrt{\frac{29}{6}} \approx 0.898\)

(ii) The probability distribution table for X is calculated by considering all possible sums of two throws:

(iii) Given \(E(Y) = 8\), and \(Y\) follows a binomial distribution with \(p = \frac{1}{3}\), we have:

\(np = 8 \Rightarrow n = 24\)

Variance is given by \(Var(Y) = np(1-p) = 24 \times \frac{1}{3} \times \frac{2}{3} = \frac{16}{3} \approx 5.33\)

(i) To find the probability distribution of Y, consider the possible outcomes when two values of X are chosen:

• Y = 0: This occurs when both values are the same. The probabilities are:
  • P(2, 2) = 0.5 × 0.5 = 0.25
  • P(4, 4) = 0.4 × 0.4 = 0.16
  • P(6, 6) = 0.1 × 0.1 = 0.01
  Total probability for Y = 0 is 0.25 + 0.16 + 0.01 = 0.42.
• Y = 2: This occurs when the values are (2, 4), (4, 2), (4, 6), or (6, 4). The probabilities are:
  • P(2, 4) = 0.5 × 0.4 = 0.2
  • P(4, 2) = 0.4 × 0.5 = 0.2
  • P(4, 6) = 0.4 × 0.1 = 0.04
  • P(6, 4) = 0.1 × 0.4 = 0.04
  Total probability for Y = 2 is 0.2 + 0.2 + 0.04 + 0.04 = 0.48.
• Y = 4: This occurs when the values are (2, 6) or (6, 2). The probabilities are:
  • P(2, 6) = 0.5 × 0.1 = 0.05
  • P(6, 2) = 0.1 × 0.5 = 0.05
  Total probability for Y = 4 is 0.05 + 0.05 = 0.1.

The probability distribution table for Y is:

(ii) The expected value of Y is calculated as:

\(E(Y) = (0 imes 0.42) + (2 imes 0.48) + (4 imes 0.1) = 0 + 0.96 + 0.4 = 1.36\)