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

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

There are 2 numbers divisible by 3 (3 and 6) and 4 numbers not divisible by 3 (1, 2, 4, 5) on a die. The probability that one die shows a number divisible by 3 and the other does not is:

\(P(A) = \frac{1}{3} \times \frac{2}{3} + \frac{2}{3} \times \frac{1}{3} = \frac{4}{9}\)

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

The product is even if at least one number is even. There are 3 even numbers (2, 4, 6) on a die. The probability that the product is even is:

\(P(B) = 1 - \left(\frac{1}{2} \times \frac{1}{2}\right) = \frac{3}{4}\)

Now, calculate \(P(A \cap B)\):

\(P(A \cap B) = \frac{12}{36} = \frac{1}{3}\)

Check independence:

\(P(A) \times P(B) = \frac{4}{9} \times \frac{3}{4} = \frac{1}{3}\)

Since \(P(A \cap B) = P(A) \times P(B)\), events A and B are independent.

For mutual exclusivity, events are mutually exclusive if \(P(A \cap B) = 0\). Since \(P(A \cap B) = \frac{1}{3} \neq 0\), events A and B are not mutually exclusive.

The probability of event A, where the white marble is in either box L or M, is given by:

\(P(A) = \frac{1}{2}\)

The probability of event B, where the red marble is to the left of both the green marble and the yellow marble, is given by:

\(P(B) = \frac{8}{24} = \frac{1}{3}\)

The probability of both events A and B occurring together is:

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

To check for independence, we verify if:

\(P(A) \times P(B) = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6}\)

Since \(P(A \cap B) = P(A) \times P(B)\), events A and B are independent.

(iii) To determine if events \(R\) and \(S\) are independent, we check if \(P(R \cap S) = P(R) \times P(S)\).

Given: \(P(R) = 0.5\), \(P(S) = 0.4\), and \(P(R \cap S) = \frac{120}{720} = \frac{1}{6}\).

Calculate \(P(R) \times P(S) = 0.5 \times 0.4 = 0.2\).

Since \(P(R \cap S) = \frac{1}{6} \neq 0.2\), events \(R\) and \(S\) are not independent.

(iv) Events \(R\) and \(S\) are exclusive if \(P(R \cap S) = 0\). However, \(P(R \cap S) = \frac{1}{6} \neq 0\), indicating there is an overlap between \(R\) and \(S\) (e.g., the combination 3, 4, 4).

Therefore, events \(R\) and \(S\) are not exclusive.

The probability of event Q, getting a total of 5, can occur with the following pairs: (1,4), (2,3), (3,2), (4,1). Thus,

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

The probability of event S, getting one low score and one high score, can occur with the following pairs: (1,4), (1,5), (1,6), (2,4), (2,5), (2,6), (3,4), (3,5), (3,6), (4,1), (5,1), (6,1), (4,2), (5,2), (6,2), (4,3), (5,3), (6,3). Thus,

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

The probability of both events Q and S occurring, i.e., the intersection \(Q \cap S\), is given by the pairs: (1,4), (2,3), (3,2), (4,1). Thus,

\(P(Q \cap S) = \frac{4}{36} = \frac{1}{9}\).

To check for independence, we verify if \(P(Q \cap S) = P(Q) \times P(S)\).

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

Since \(P(Q \cap S) = \frac{1}{9}\) and \(P(Q) \times P(S) = \frac{1}{18}\), the events are independent.

(i) The total number of countries is 170. The number of countries with a medium GDP is \(20 + 42 + 12 = 74\). Therefore, the probability is \(\frac{74}{170} = \frac{37}{85} \approx 0.435\).

(ii) The number of countries that do not have a medium GDP is \(170 - 74 = 96\). The number of countries with a low birth rate and not a medium GDP is \(3 + 35 = 38\). Therefore, the probability is \(\frac{38}{96} = \frac{19}{49} \approx 0.396\).

(iii) The probability of a country having both a high GDP and a high birth rate is 0, as there are no such countries. Therefore, the events are exclusive.

(iv) The probability of choosing a country with a medium GDP is \(\frac{74}{170} = \frac{37}{85}\). The probability of choosing a country with a medium birth rate is \(\frac{42}{96} = \frac{7}{16}\). The combined probability is \(\frac{37}{85} \times \frac{7}{16} = \frac{287}{1360} \approx 0.211\).