jmc

There are $2^6=64$ possible outcomes, since each of the 6 coins has 2 possibilities. If we do not get at least 2 heads, then we get either no heads or one heads. There is only 1 way to get 0 heads, and $\left(\genfrac{}{}{0ex}{}{6}{1}\right)=6$ ways to get 1 head, so the probability of getting at most one head is $\frac{7}{64}$. Therefore, the probability of getting at least 2 heads is $1-\frac{7}{64}=\boxed{\frac{57}{64}}$.