The sample space of the given experiment is given by, S = {1, 2, 3, 4, 5, 6}
(i) Let A be the event of the occurrence of a prime number ⇒ A = {2, 3, 5}
$\therefore \mathrm{P}\left(\mathrm{A}\right)=\frac{\text{ Number of outcomes favourable to }\mathrm{A}}{\text{ Total number of possible outcomes }}=\frac{\mathrm{n}\left(\mathrm{A}\right)}{\mathrm{n}\left(\mathrm{S}\right)}=\frac{3}{6}=\frac{1}{2}$
(ii) Let B be the event of the occurrence of a number greater than or equal to 3 ⇒ B = {3, 4, 5, 6}
$\therefore \mathrm{P}\left(\mathrm{B}\right)=\frac{\text{ Number of outcomes favourableto }\mathrm{B}}{\text{ Total number of possible outcomes }}=\frac{\mathrm{n}\left(\mathrm{B}\right)}{\mathrm{n}\left(\mathrm{S}\right)}=\frac{4}{6}=\frac{2}{3}$
(iii) Let C be the event of the occurrence of a number less than or equal to one ⇒ C = {1}
$\therefore \mathrm{P}\left(\mathrm{C}\right)=\frac{\text{ Number of outcomes favourable to }\mathrm{C}}{\text{ Total number of possible outcomes }}=\frac{\mathrm{n}\left(\mathrm{C}\right)}{\mathrm{n}\left(\mathrm{S}\right)}=\frac{1}{6}$
(iv) Let D be the event of the occurrence of a number greater than 6 ⇒ D = ϕ
$\therefore \mathrm{P}\left(\mathrm{D}\right)=\frac{\text{ Number of outcomes favourableto }\mathrm{D}}{\text{ Total number of possible outcomes }}=\frac{\mathrm{n}\left(\mathrm{D}\right)}{\mathrm{n}\left(\mathrm{S}\right)}=\frac{0}{6}=0$
 (v) Let E be the event of the occurrence of a number less than 6 ⇒ E = {1, 2, 3, 4, 5}
$\therefore \mathrm{P}\left(\mathrm{E}\right)=\frac{\text{ Number of outcomes favourable to }\mathrm{E}}{\text{ Total number of possible outcomes }}=\frac{\mathrm{n}\left(\mathrm{E}\right)}{\mathrm{n}\left(\mathrm{S}\right)}=\frac{5}{6}$