Q.Two dice are rolled. Find the probability that the sum of the numbers on the two dice is

Each die is independent, so there are $6\times 6=36$ equally likely ordered pairs; count the pairs giving each sum.

(i) $\dfrac{6}{36}=\dfrac{1}{6}$; (ii) $3+2+1=6$ pairs, $\dfrac{6}{36}=\dfrac{1}{6}$.

When two dice are rolled, an outcome is an ordered pair $(a,b)$ with $a,b\in\{1,\ldots,6\}$, so $n(S)=6\times 6=36$, all equally likely.

(i) Sum $=7$. Favourable pairs: $(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$ — that is $6$ pairs.

(ii) Sum at least $10$ means sum $=10, 11,$ or $12$:

• sum $10$: $(4,6),(5,5),(6,4)$ — $3$ pairs;
• sum $11$: $(5,6),(6,5)$ — $2$ pairs;
• sum $12$: $(6,6)$ — $1$ pair.

Total favourable $=3+2+1=6$, so

Independent check: the sums $10,11,12$ are mutually exclusive, so adding their separate probabilities $\dfrac{3}{36}+\dfrac{2}{36}+\dfrac{1}{36}=\dfrac{6}{36}=\dfrac{1}{6}$ (the addition theorem for mutually exclusive events) gives the same answer.

List the favourable pairs systematically by fixing the first die and finding the required second die; this guarantees none are missed or double-counted.

Treating $(3,4)$ and $(4,3)$ as one outcome — with two distinguishable dice they are different ordered pairs, so both count. Using $n(S)=21$ (unordered pairs) instead of $36$ breaks the equally-likely assumption and gives wrong probabilities.