Is there some catch, otherwise this is a simple question? The probability will be
1-Pr(H1,H2,H3) and under unbiasedness and, I presume, independence we have that Pr(H1,H2,H3)=Pr(H1)*Pr(H2)*Pr(H3)=1/8. Hence, the probability you ask for is 7/8.
of course, both prevoius answers are OK. In general, very often it is much easier and quicker to count the probability of the opposite (or complement) event.
If A is any event, and A/ denotes the opposite one, then P(A) + P(A/) = 1