See explanation.

If we mark “heads” with ##H## and “tails” with ##T## we can write that:

##Omega={HHH,HHT,HTH,HT T,THH,THT,T TH,T T T}##

So there are ##8## events in the sample space. If we assume that the coin is fair then each event is equally possible, so all events mentioned in the sample space have probability of

##p=1/(bar(bar(Omega)))=1/8##