A fair coin has etqual probabilitiers for heads and tails. Therefore the probability can simply be found by counting.
There are eight different possibilities for outcomes of three tosses:
- Head, Head, Head
- Head, Head, Tail
- Head, Tail, Head
- Head, Tail, Tail
- Tail, Head, Head
- Tail, Head, Tail
- Tail, Tail, Head
- Tail, Tail, Tail
Possibilities 1 to 7 have at least 1 head, while possibility 8 doesn't. Therefore remove that last one.
From the 7 remaining possibilities, 1 has all three heads, so the probability is one out of seven, or $\frac17$.
OK, but what to do if there are too many cases to explicitly list them? Well, it is easy to see that you cannot have three heads without having at least one head. So your probability is $p_{\text{3 heads if at least 1 head}} = \frac{\text{Number of cases with 3 heads}}{\text{Number of cases with at least 1 head}}$ However you know that $p_{\text{3 heads}} = \frac{\text{Number of cases with 3 heads}}{\text{Number of all cases}}$ and $p_{\text{at least 1 head}} = \frac{\text{Number of cases with at least 1 head}}{\text{Number of all cases}}$ From this, it is not hard to see that $p_{\text{3 heads if at least 1 head}} =\frac{p_{\text{3 heads}}}{p_{\text{at least 1 head}}}$ Now you surely know that $p_{\text{3 heads}}=\left(\frac12\right)^3=\frac18$ and $p_{\text{at least 1 head}} = 1-p_{\text{3 tails}}=\frac78$. Inserting then again gives $p_{\text{3 heads if at least 1 head}}=\frac18/\frac78=\frac17$.
Indeed, that latter formula works even for arbitrary probabilities (i.e. not only fair coins). Note however that it still depends on the fact that having 3 heads implies having at least 1 head.