When two dice are rolled:

Possible outcomes when two dice are rolled :

$\mathrm{S}=\{(1,1),(1,2),(1,3),(1,4), \cdots,(6,5),(6,6)\}$

Therefore, the number of possible outcomes in the sample space is 36 .

(i) The outcomes for the event that the total is odd:

E $=\{(1,2),(1,4),(1,6),(2,1),(2,3),(2,5),(3,2),(3,4),(3,6),(4,1),(4,3),(4,5),(5,2),(5,4),(5,6),(6,1),(6,3),(6,5)\}$

(ii) The number of favourable outcomes is 18 .

$\therefore \mathrm{P}(\mathrm{E})=\frac{18}{36}=\frac{1}{2}$

(iii) The outcomes for the event that total is less than 5 :

$\mathrm{B}=\{(1,1),(1,2),(1,3),(2,1),(2,2),(3,1)\}$

(iv) The number of favourable outcomes is 6 .

$\therefore P(B)=\frac{6}{36}=\frac{1}{6}$