jmc

We compute the probability that Mahmoud flips three tails and subtract from 1. The probability that Mahmoud flips one tail with one coin is $\frac{1}{2}$, so the probability that he flips three tails is ${\left(\frac{1}{2}\right)}^3=\frac{1}{8}$. The probability that he gets at least one head is $1-\frac{1}{8}=\boxed{\frac{7}{8}}$.