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. 

Dear Parag,

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

Greetings, Tadeusz

It is 7/8

you can use binomial distribution also in calculating this probability but it is preferable to use if number of trials is more than 6. Paul's answer is the simplest one but only useful till 4 trials or four coins.