A fair coin with 1 marked on one face and 6 on the other and a fair die are both tossed.

Since the fair coin has 1 marked on one face and 6 on the other, and the die has six faces that are numbered 1, 2, 3, 4, 5, and 6, the sample space is given by

S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

Accordingly, $n(S)=12$

(i) Let A be the event in which the sum of numbers that turn up is 3.

Accordingly, $A=\{(1,2)\}$

$\therefore \mathrm{P}(\mathrm{A})=\frac{\text { Number of outcomes favourable to } \mathrm{A}}{\text { Total number of possible outcomes }}=\frac{n(\mathrm{~A})}{n(\mathrm{~S})}=\frac{1}{12}$

(ii) Let B be the event in which the sum of numbers that turn up is 12.

Accordingly, B $=\{(6,6)\}$

$\therefore \mathrm{P}(\mathrm{B})=\frac{\text { Number of outcomes favourable to } \mathrm{B}}{\text { Total number of possible outcomes }}=\frac{n(\mathrm{~B})}{n(\mathrm{~S})}=\frac{1}{12}$