Two dice are thrown independently. Let $$A$$ be the event that the number appeared on the 1st die is less than the number appeared on the 2nd die, $$B$$ be the event that the number appeared on the 1st die is even and that on the second die is odd, and $$C$$ be the event that the number appeared on the 1st die is odd and that on the 2nd is even. Then

Two dice are thrown. Let us enumerate the events.

Event A (1st die < 2nd die): 15 outcomes

$$\{(1,2),(1,3),(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6),(4,5),(4,6),(5,6)\}$$

Event B (1st even, 2nd odd): 9 outcomes

$$\{(2,1),(2,3),(2,5),(4,1),(4,3),(4,5),(6,1),(6,3),(6,5)\}$$

Event C (1st odd, 2nd even): 9 outcomes

$$\{(1,2),(1,4),(1,6),(3,2),(3,4),(3,6),(5,2),(5,4),(5,6)\}$$

Now check each option:

Option A: Favourable cases of $$(A \cup B) \cap C$$.

$$A \cap C = \{(1,2),(1,4),(1,6),(3,4),(3,6),(5,6)\}$$ — 6 cases (where 1st < 2nd AND 1st odd, 2nd even)

$$B \cap C = \emptyset$$ (1st die cannot be both even and odd)

$$(A \cup B) \cap C = (A \cap C) \cup (B \cap C) = 6 + 0 = 6$$ ✓

Option B: A and B mutually exclusive?

$$A \cap B$$ includes $$(2,3), (2,5), (4,5)$$ — not empty ✗

Option C: Favourable cases of A, B, C are 15, 6, 6?

A has 15 ✓, but B has 9 ✗ and C has 9 ✗

Option D: B and C independent?

$$P(B) = 9/36 = 1/4$$, $$P(C) = 9/36 = 1/4$$

$$P(B \cap C) = 0 \neq P(B) \cdot P(C) = 1/16$$ ✗

The answer is Option A: The number of favourable cases of $$(A \cup B) \cap C$$ is 6.