Harold tosses a coin four times. The probability that he gets at least as many heads as tails is
$\text{(A)}\ \frac{5}{16}\qquad\text{(B)}\ \frac{3}{8}\qquad\text{(C)}\ \frac{1}{2}\qquad\text{(D)}\ \frac{5}{8}\qquad\text{(E)}\ \frac{11}{16}$
Case 1: There are two heads and two tails. There are $\binom{4}{2} = 6$ ways to choose which two tosses are heads, and the other two must be tails.
Case 2: There are three heads, one tail. There are $\binom{4}{1} = 4$ ways to choose which of the four tosses is a tail.
Case 3: There are four heads and no tails. This can only happen $1$ way.
There are a total of $2^4=16$ possible configurations, giving a probability of $\frac{6+4+1}{16} = \boxed{\text{(E)}\ \frac{11}{16}}$.