The probability of heads is 1/2, so the probability of not-heads is 1/2.
Then the probability of 2 heads and a tails is (1/2)²·(1/2) = 1/8.
There are 3 choose 2 = (3·2)/(2·1) = 3 ways that the pair of heads may appear among the three tosses. Thus the probability of 2 heads in 3 tosses is ...
... 3·(1/8) = 3/8
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Or you can simply count the favorable outcomes among the possible outcomes:
TTT TTH THT THH HTT HTHHHT HHH . . . . 3 of 8 are favorable.
There are 8 possibilities: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. Of these 4 have at least two heads. Assuming a fair coin, the probability of tossing at least two heads is 4/8 or 1/2. The Answer would probably be 3/8 though.
