(i) Complete the table:

(ii) The probability of choosing the pair of low-heeled shoes with designer labels is \(\frac{1}{20}\).

(iii) The probability of choosing a pair of sports shoes is \(\frac{10}{20} = \frac{1}{2}\).

(iv) The probability of choosing a pair of high-heeled shoes given that it has a designer label is \(\frac{2}{8} = \frac{1}{4}\).

(v) To check independence, calculate:

\(P(D) = \frac{8}{20} = 0.4\)

\(P(S) = \frac{10}{20} = 0.5\)

\(P(D \cap S) = \frac{5}{20} = 0.25\)

Since \(P(D) \times P(S) = 0.4 \times 0.5 = 0.2 \neq 0.25 = P(D \cap S)\), the events are not independent.

(i) Let \(F\) be the event that a person is female, and \(W\) be the event that a person watches 'Kops are Kids'.

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

\(P(W) = \frac{16}{30} = 0.533\)

\(P(F \cap W') = \frac{5}{30} = 0.167\)

Using the formula for the union of two events:

\(P(F \cup W) = P(F) + P(W) - P(F \cap W')\)

\(P(F \cup W) = \frac{12}{30} + \frac{16}{30} - \frac{5}{30} = \frac{5}{6} = 0.833\)

(ii) Let \(M\) be the event that a person is male.

\(P(M) = \frac{18}{30} = 0.6\)

\(P(W) = \frac{16}{30} = 0.533\)

\(P(M \cap W) = \frac{13}{30} = 0.433\)

Check independence by comparing \(P(M \cap W)\) and \(P(M) \times P(W)\):

\(P(M) \times P(W) = \frac{18}{30} \times \frac{16}{30} = \frac{8}{25} = 0.32\)

Since \(P(M \cap W) \neq P(M) \times P(W)\), the events are not independent.

(i) Calculate the probability for each scenario where exactly two balls have the same number:

- Probability of selecting 2 from A, 2 from B, and any from C: \(\frac{1}{4} \times \frac{1}{5} \times \frac{1}{7} = \frac{1}{140}\)

- Probability of selecting 8 from A, 8 from B, and any from C: \(\frac{1}{4} \times \frac{2}{5} \times \frac{3}{7} = \frac{3}{70}\)

- Probability of selecting 8 from A, any from B, and 8 from C: \(\frac{1}{4} \times \frac{3}{5} \times \frac{4}{7} = \frac{3}{35}\)

- Probability of selecting any from A, 8 from B, and 8 from C: \(\frac{3}{4} \times \frac{2}{5} \times \frac{4}{7} = \frac{6}{35}\)

Summing these probabilities gives \(\frac{47}{140}\).

(ii) Given that exactly two of the selected balls have the same number, the probability that they are both numbered 2 is:

\(\frac{\frac{1}{28}}{\frac{47}{140}} = \frac{5}{47}\)

(iii) Calculate \(P(X) = \frac{47}{140}\) and \(P(Y) = \frac{1}{4}\). Check independence:

\(P(X \cap Y) = \frac{1}{28}\)

Since \(P(X \cap Y) \neq P(X) \times P(Y)\), events X and Y are not independent.

(i) (a) To get a final score of 12, the second throw must be 6 and the first throw must also be 6, since \(6 + 6 = 12\). The probability of this is \(\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}\).

(b) To get a final score of 5, the possible outcomes are (1,4), (2,3), (3,2), (4,1) where the second throw is not 5. The probability is \(\frac{4}{36} = \frac{1}{9}\).

(ii) Calculate probabilities:

\(P(A) = \frac{1}{6}\)

\(P(B) = \frac{5}{36}\) for pairs (1,5), (2,4), (3,3), (4,2), (5,1)

\(P(C) = 1 - P(O, O) = \frac{3}{4}\)

Check independence:

\(P(A \cap B) = \frac{1}{36}\) which is not equal to \(P(A) \times P(B)\)

\(P(A \cap C) = \frac{3}{36}\) which is not equal to \(P(A) \times P(C)\)

\(P(B \cap C) = \frac{2}{36}\) which is not equal to \(P(B) \times P(C)\)

None of the events are independent.

(i) To find \(P(Q)\), we need the pairs \((a, b)\) such that \(a \times b = 24\). The possible pairs are \((2, 12), (3, 8), (4, 6), (6, 4), (8, 3), (12, 2)\). There are 6 favorable outcomes. The total number of outcomes when two dice are thrown is \(12 \times 12 = 144\). Thus, \(P(Q) = \frac{6}{144} = \frac{1}{24}\).

(ii) For \(P(R)\), both numbers must be greater than 8. The numbers greater than 8 are 9, 10, 11, and 12. Thus, there are 4 choices for each die, giving \(4 \times 4 = 16\) favorable outcomes. Therefore, \(P(R) = \frac{16}{144} = \frac{1}{9}\).

(iii) Events \(Q\) and \(R\) are exclusive if they cannot occur simultaneously. Since \(R\) requires both numbers to be greater than 8, and none of the pairs for \(Q\) satisfy this, \(P(R \cap Q) = 0\). Thus, \(R\) and \(Q\) are exclusive.

(iv) Events \(Q\) and \(R\) are independent if \(P(R \cap Q) = P(R) \times P(Q)\). Since \(P(R \cap Q) = 0\) and \(P(R) \times P(Q) \neq 0\), they are not independent.