Q.Two dice are thrown together, and the sum of the numbers appearing is observed. Let $A =$ 'the sum is a multiple of $3$'. Find $P(A)$ and $P(A')$.

[!TLDR] Count every two-dice outcome whose sum is $3$, $6$, $9$ or $12$, then apply the complement rule. [! …

The possible sums that are multiples of $3$ (between $2$ and $12$) are $3, 6, 9, 12$. Counting outcomes for each: sum $=3$: $(1,2),(2,1)$ — $2$ outcomes. Sum $=6$: $(1,5),(2,4),(3,3),(4,2),(5,1)$ — $5$ outcomes. Sum $=9$: $(3,6),(4,5),(5,4),(6,3)$ — $4$ outcomes. Sum $=12$: $(6,6)$ — $1$ outcome. Total favourable outcomes: $2+5+4+1=12$. So $P(A) = \df …

List the outcomes for each qualifying sum ($3,6,9,12$) systematically, total them over $36$, simplify, …

A common mistake is forgetting the sum $=12$ outcome $(6,6)$, or double-counting sum $=6$'s $(3,3)$ — recount each s …