Let $p$ and $q$ represents the probability of getting head and tail respectively on tossing a coin.
So, $p=q=\frac{1}{2}$
So, required probability $=\displaystyle\sum_{r=0}^{3}\left[{ }^{3} C_{,}\left(\frac{1}{2}\right)^{3}\right]^{2}$
$=\left(\frac{1}{2}\right)^{6}\left[1^{2}+3^{2}+3^{2}+1^{2}\right]$
$=\frac{20}{64}=\frac{5}{16}$