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)}
⇒n(S)=12
(i) Let A be the event in which the sum of numbers that turn up is 3
⇒A={(1,2)}
∴P(A)=NumberofoutcomesfavourabletoATotalnumberofpossibleoutcomes=n(A)n(S)=112
(ii) Let B be the event in which the sum of numbers that turn up is 12
⇒B={(6,6)}
∴P(B)=NumberofoutcomesfavourabletoBTotalnumberofpossibleoutcomes=n(B)n(S)=112