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.