A die is thrown. Let $$A$$ be the event that the number obtained is greater than 3. Let $$B$$ be the event that the number obtained is less than 5. Then $$P(A \cup B)$$ is

The sample space for a single, fair die throw is $$S=\{1,2,3,4,5,6\}$$, so $$n(S)=6$$.

Event $$A$$: “number obtained is greater than 3”
  ⇒ $$A=\{4,5,6\}$$, so $$n(A)=3$$.

Event $$B$$: “number obtained is less than 5”
  ⇒ $$B=\{1,2,3,4\}$$, so $$n(B)=4$$.

The union $$A\cup B$$ consists of all outcomes that are in $$A$$ or in $$B$$ (or in both).
Combine the two sets: $$A\cup B=\{1,2,3,4,5,6\}=S$$, so $$n(A\cup B)=6$$.

Using the classical definition of probability,
$$P(A\cup B)=\frac{n(A\cup B)}{n(S)}=\frac{6}{6}=1.$$

Therefore, $$P(A \cup B)=1.$$

Option C which is: 1