Solution
when the 3 cion are tossed the total number of cases are = $$2^3$$ = 8
and at least two heads are there in the following = HHT, HTH, THH, HHH = 4
then the probility of getting at least two heads = number of favourable cases \ total number of cases
then the probility of getting at least two heads = 4\8 = 1\2 Answer
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