Let ${\mathit{E}}_1$ be the event that the outcome on the die is 1 or 2
P $\left({\mathit{E}}_1\right)$ = $\frac{2}{6}$ = $\frac{1}{3}$
let ${\mathit{E}}_2$ be the event that the outcome on the die is 3,4,5 or 6
P $\left({\mathit{E}}_2\right)$ = $\frac{4}{6}$ = $\frac{2}{3}$
let A be the event that exactly one tail comes
So
P $\left(\frac{\mathit{A}}{{\mathit{E}}_1}\right)$ = $\frac{2}{8}$ (A occurs given that ${\mathit{E}}_1$ had already occured, coin tossed thrice)
P $\left(\frac{\mathit{P}}{{\mathit{E}}_2}\right)$ = $\frac{1}{2}$ (A occures given that ${\mathit{E}}_2$ had already occured, coin tossed once
By Bayes's theorem, (A had already occured, probability that ${\mathit{E}}_2$ had occured)
P $\left(\frac{{\mathit{E}}_2}{\mathit{A}}\right)$ =
= $\frac{\frac{2}{3}\times \frac{1}{2}}{\frac{2}{3}\times \frac{1}{2}+\frac{1}{3}\times \frac{3}{8}}$
= $\frac{8}{11}$