First, calculate the probability of event R (product is 12). The possible pairs are (3, 4), (4, 3), (2, 6), and (6, 2). Thus,

\(P(R) = \frac{4}{36} = \frac{1}{9}\).

Next, calculate the probability of event T (one number is odd, one is even). There are 18 such pairs out of 36 total possibilities, so

\(P(T) = \frac{18}{36} = \frac{1}{2}\).

Now, calculate the probability of both events occurring together, \(P(R \cap T)\). The pairs (3, 4) and (4, 3) satisfy both conditions, so

\(P(R \cap T) = \frac{2}{36} = \frac{1}{18}\).

Check for independence by verifying if \(P(R) \times P(T) = P(R \cap T)\):

\(\frac{1}{9} \times \frac{1}{2} = \frac{1}{18}\).

Since \(P(R) \times P(T) = P(R \cap T)\), the events are independent.

(i) The number of red or silver cars manufactured in Korea is \(34 + 30 = 64\). The probability is \(\frac{64}{250} = 0.256\).

(ii) The number of cars not manufactured in Japan is \(250 - (26 + 22 + 12) = 190\). The probability is \(\frac{190}{250} = 0.76\).

(iii) Calculate \(P(X)\), \(P(Y)\), and \(P(X \cap Y)\):

\(P(X) = \frac{80}{250} = \frac{8}{25}\)

\(P(Y) = \frac{100}{250} = \frac{2}{5}\)

\(P(X \cap Y) = \frac{32}{250} = \frac{16}{125}\)

Check independence: \(P(X) \times P(Y) = \frac{8}{25} \times \frac{2}{5} = \frac{16}{125}\)

Since \(P(X) \times P(Y) = P(X \cap Y)\), events X and Y are independent.

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

• Total number of adults = 30
• Right-handed adults = 25
• Adults who wear spectacles = 8
• Right-handed and do not wear spectacles = 19

From this, we can fill in the table:

• Right-handed and wears spectacles = 25 - 19 = 6
• Not right-handed = 30 - 25 = 5
• Not right-handed and wears spectacles = 8 - 6 = 2
• Not right-handed and does not wear spectacles = 5 - 2 = 3

The completed table is:

(ii) To determine if events X and Y are independent, we check if:

\(P(X \cap Y) = P(X) \times P(Y)\)

Calculate the probabilities:

• \(P(X) = \frac{25}{30}\)
• \(P(Y) = \frac{8}{30}\)
• \(P(X \cap Y) = \frac{6}{30}\)

Check independence:

\(P(X) \times P(Y) = \frac{25}{30} \times \frac{8}{30} = \frac{200}{900} = \frac{2}{9}\)

\(P(X \cap Y) = \frac{6}{30} = \frac{1}{5}\)

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

(i) To find the probability of event R, we list the pairs where one score is exactly 3 greater than the other: \((1, 4), (2, 5), (3, 6), (4, 7), (5, 8)\). Each pair can occur in two ways (e.g., \((1, 4)\) and \((4, 1)\)). Thus, there are \(5 \times 2 = 10\) favorable outcomes. The total number of outcomes when the die is thrown twice is \(8 \times 8 = 64\). Therefore, \(P(R) = \frac{10}{64}\).

(ii) For event S, we need the product of the scores to be more than 19. The pairs satisfying this condition are \((3, 8), (3, 7), (4, 8), (4, 7), (4, 6), (4, 5), (5, 8), (5, 7), (5, 6), (6, 8), (6, 7), (7, 8)\). Additionally, \((5, 5), (6, 6), (7, 7), (8, 8)\) also satisfy the condition. The total number of favorable outcomes is \(24 + 4 = 28\). Thus, \(P(S) = \frac{28}{64}\).

(iii) To determine if R and S are independent, we check if \(P(R \cap S) = P(R) \times P(S)\). From the mark scheme, \(P(R \cap S) = \frac{4}{64}\). Calculating \(P(R) \times P(S) = \frac{10}{64} \times \frac{28}{64} = \frac{280}{4096}\), which simplifies to \(\frac{35}{512}\), not equal to \(\frac{4}{64}\). Therefore, events R and S are not independent.

To determine if events S and T are independent, we need to check if \(P(S \cap T) = P(S) \times P(T)\).

First, calculate \(P(S)\):

The possible outcomes for the sum to be 4 are (1,3), (2,2), and (3,1). Thus, \(P(S) = \frac{3}{16}\).

Next, calculate \(P(T)\):

The product is odd if both numbers are odd. The possible outcomes are (1,1), (1,3), (3,1), and (3,3). Thus, \(P(T) = \frac{4}{16}\).

Now, calculate \(P(S \cap T)\):

The outcomes that satisfy both events are (1,3) and (3,1). Thus, \(P(S \cap T) = \frac{2}{16}\).

Check independence:

\(P(S) \times P(T) = \frac{3}{16} \times \frac{4}{16} = \frac{12}{256} = \frac{3}{64}\).

Since \(\frac{3}{64} \neq \frac{2}{16}\), events S and T are not independent.

For exclusivity, events are exclusive if \(P(S \cap T) = 0\).

Since \(P(S \cap T) = \frac{2}{16} \neq 0\), events S and T are not exclusive.