1. Total sample space

When two dice are thrown independently, the outcomes are ordered pairs: $S=\{(i,j): i,j\in\{1,2,3,4,5,6\}\}$ So, $|S|=6\times 6=36.$

2. Define the events

• $A$: first die $<$ second die
• $B$: first die even, second die odd
• $C$: first die odd, second die even

3. Count favourable cases for $A$

We need ordered pairs $(i,j)$ such that $i<j$.

Count systematically:

• If first die is $1$: second die can be $2,3,4,5,6$ $\Rightarrow 5$
• If first die is $2$: second die can be $3,4,5,6$ $\Rightarrow 4$
• If first die is $3$: second die can be $4,5,6$ $\Rightarrow 3$
• If first die is $4$: second die can be $5,6$ $\Rightarrow 2$
• If first die is $5$: second die can be $6$ $\Rightarrow 1$
• If first die is $6$: none $\Rightarrow 0$

Hence, $|A|=5+4+3+2+1=15.$

4. Count favourable cases for $B$

First die even: $\{2,4,6\}$ (3 choices)

Second die odd: $\{1,3,5\}$ (3 choices)

Thus, $|B|=3\times 3=9.$

5. Count favourable cases for $C$

First die odd: $\{1,3,5\}$ (3 choices)

Second die even: $\{2,4,6\}$ (3 choices)

Thus, $|C|=3\times 3=9.$

6. Check option A: "A and B are mutually exclusive"

We test whether $A\cap B=\varnothing$.

Event $B$ contains outcomes with first even and second odd: $B=\{(2,1),(2,3),(2,5),(4,1),(4,3),(4,5),(6,1),(6,3),(6,5)\}.$

Now among these, those satisfying first $<$ second are: $(2,3),(2,5),(4,5).$ So, $A\cap B\neq \varnothing.$ Hence, A is false.

7. Check option B: "the number of favourable cases of A, B and C are 15, 6 and 6 respectively"

We found: $|A|=15,\quad |B|=9,\quad |C|=9.$ So this statement is false.

8. Check option C: "B and C are independent"

Notice:

• In $B$, first die is even and second die is odd.
• In $C$, first die is odd and second die is even.

These cannot happen together. Therefore, $B\cap C=\varnothing \implies P(B\cap C)=0.$

Now, $P(B)=\frac{9}{36}=\frac14,\qquad P(C)=\frac{9}{36}=\frac14.$ Thus, $P(B)P(C)=\frac14\cdot \frac14=\frac{1}{16}\neq 0.$

Since $P(B\cap C)\neq P(B)P(C),$ $B$ and $C$ are not independent.

So C is false.

9. Check option D: number of favourable cases of $(A\cup B)\cap C$

Since $B$ and $C$ are disjoint, $B\cap C=\varnothing.$ Therefore, $(A\cup B)\cap C=(A\cap C)\cup(B\cap C)=A\cap C.$

So we only count outcomes in $C$ that also satisfy $A$.

Event $C$ means first odd, second even: $C=\{(1,2),(1,4),(1,6),(3,2),(3,4),(3,6),(5,2),(5,4),(5,6)\}.$

Now impose $i<j$:

• $(1,2),(1,4),(1,6)$
• $(3,4),(3,6)$
• $(5,6)$

Hence, $|(A\cup B)\cap C|=|A\cap C|=6.$

So D is true.

10. Final conclusion

Among the given options, only D is correct.