Q.A die is rolled once. Find the probability of getting

With a fair die all six faces are equally likely, so use $P=\dfrac{n(A)}{n(S)}$ with $n(S)=6$.

(i) $\dfrac{2}{6}=\dfrac{1}{3}$; (ii) multiples of $3$ are $\{3,6\}$, so $P=1-\dfrac{2}{6}=\dfrac{2}{3}$.

Here $S=\{1,2,3,4,5,6\}$, so $n(S)=6$ and all outcomes are equally likely.

(i) Number greater than $4$. Favourable outcomes: $\{5,6\}$, so $n(A)=2$.

(ii) Number that is not a multiple of $3$. The multiples of $3$ on a die are $\{3,6\}$, so $P(\text{multiple of }3)=\dfrac{2}{6}=\dfrac{1}{3}$. By the complement rule,

Independent check for (ii): the non-multiples of $3$ are $\{1,2,4,5\}$, which is $4$ outcomes, giving $\dfrac{4}{6}=\dfrac{2}{3}$ directly — matches.

Part (ii) can be done by direct count of $\{1,2,4,5\}$ ($=4/6$) instead of the complement; both give $\tfrac{2}{3}$.

Counting "greater than $4$" as $\{4,5,6\}$ — but $4$ is not greater than $4$, so it is only $\{5,6\}$. In (ii), forgetting that $6$ is also a multiple of $3$ and counting only $\{3\}$.