First, calculate the probability of event S (sum is even). The possible sums when two dice are thrown range from 2 to 12. The even sums are 2, 4, 6, 8, 10, and 12. The number of outcomes for each sum is:

• 2: (1,1)
• 4: (1,3), (2,2), (3,1)
• 6: (1,5), (2,4), (3,3), (4,2), (5,1)
• 8: (2,6), (3,5), (4,4), (5,3), (6,2)
• 10: (4,6), (5,5), (6,4)
• 12: (6,6)

Total even outcomes = 18. Therefore, \(P(S) = \frac{18}{36} = \frac{1}{2}\).

Next, calculate the probability of event T (sum is less than 6 or a multiple of 4). The sums less than 6 are 2, 3, 4, 5, and the multiples of 4 are 4, 8, 12. The number of outcomes for each sum is:

• 2: (1,1)
• 3: (1,2), (2,1)
• 4: (1,3), (2,2), (3,1)
• 5: (1,4), (2,3), (3,2), (4,1)
• 8: (2,6), (3,5), (4,4), (5,3), (6,2)
• 12: (6,6)

Total outcomes for T = 16. Therefore, \(P(T) = \frac{16}{36} = \frac{4}{9}\).

Now, calculate \(P(S \cap T)\) (sum is even and either less than 6 or a multiple of 4). The even sums that satisfy T are 2, 4, 8, 12. The number of outcomes for each sum is:

• 2: (1,1)
• 4: (1,3), (2,2), (3,1)
• 8: (2,6), (3,5), (4,4), (5,3), (6,2)
• 12: (6,6)

Total outcomes for \(S \cap T\) = 10. Therefore, \(P(S \cap T) = \frac{10}{36} = \frac{5}{18}\).

For events S and T to be independent, \(P(S) \times P(T) = P(S \cap T)\). Calculate \(P(S) \times P(T) = \frac{1}{2} \times \frac{4}{9} = \frac{4}{18} = \frac{2}{9}\).

Since \(\frac{2}{9} \neq \frac{5}{18}\), the events S and T are not independent.