Three unbiased coins are tossed. What is the probability of getting at most two heads?

With three coins there are 8 equally likely outcomes, and only one of them (all three heads) has more than two heads. So $P(\text{at most two heads}) = 1 - \frac{1}{8} = \frac{7}{8}$; $\frac{3}{8}$ is the probability of exactly two heads, not at most two.