When three coins are tossed at once, the sample space is {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}.
I.e., n(S) = 8
Let E be the event of getting at least two tails.
So, the favourable outcomes are {TTH, THT, HTT, TTT}
Thus, n(E) = 4
We know that the probability of an event = Number of favourable outcomes/Total number of outcomes.
Hence, the probability of getting at least two tails = 4/8 = ½.