A fair coin is tossed 3 times. What is the probability of at least two consecutive heads?
$\textbf{(A)}\frac{1}{8} \qquad \textbf{(B)}\frac{1}{4} \qquad \textbf{(C)}\frac{3}{8} \qquad \textbf{(D)}\frac{1}{2} \qquad \textbf{(E)}\frac{3}{4}$
There are $2^3 = 8$ ways to flip the coins, in order.
There are two ways to get exactly two consecutive heads: HHT and THH.
There is only one way to get three consecutive heads: HHH.
Therefore, the probability of flipping at least two consecutive heads is $\boxed{\textbf{(C)}\frac{3}{8}}$.