Let $E_1$ be the event of showing 1, 2, 3 or 4
$E_2$ be the event showing 5 or 6
$P(E_1) = \frac{4}{6} = \frac{2}{3}; P(E_2) = \frac{2}{6} =\frac{1}{3}$
Let A be the event of showing exactly one head.
$\therefore P(A/E_1) = \frac{1}{2} $ and $P(A/E_2) = \frac{3}{8}$
$\begin{align*}Required\; probability & = \frac{ P( E_1) . P(A/E_1)}{P(E_1). P(A/E_1) + P(E_2). P(A/E_2)} \\ & = \frac{\frac{2}{3} . \frac{1}{2}}{\frac{2}{3} . \frac{1}{2} + \frac{1}{3}. \frac{3}{8}} \\ & = \frac{8}{11}\end{align*}$