jmc

Case 1: There are two heads and two tails. There are $\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6$ 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 $\left(\genfrac{}{}{0ex}{}{4}{1}\right)=4$ 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 $\frac{6+4+1}{16}=\boxed{\frac{11}{16}}$.