<PRE><B>Question 102610</B>
If three coins are tossed what is the probability of tossing two heads and one tail?
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There are three ways to place the tail so those three are mutually 
exclusive events.  The probability of each event is (1/2)^2*(1/2)
because each coin is independent = 1/8
So the answer is 1/8 + 1/8 + 1/8 = 3/8
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Using the binomial formula you would get 3C2(1/2)^2(1/2) = 3(1/8) = 3/8

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Cheers,
Stan H.</PRE>