Let $E_1$ be the event that the outcome on the die is 1 or 2 P $(E_1)$ = $\frac{2}{6}$ = $\frac{1}{3}$ let $E_2$ be the event that the outcome on the die is 3,4,5 or 6 P $(E_2)$ = $\frac{4}{6}$ = $\frac{2}{3}$ let A be the event that exactly one tail comes So P $\left(\frac{A}{E_1}\right)$ = $\frac{2}{8}$ (A occurs given that $E_1$ had already occured, coin tossed thrice) P $\left(\frac{P}{E_2}\right)$ = $\frac{1}{2}$ (A occures given that $E_2$ had already occured, coin tossed once By Bayes's theorem, (A had already occured, probability that $E_2$ had occured) P $\left(\frac{E_2}{A}\right)$ = = $\frac{\frac{2}{3} \times \frac{1}{2}}{\frac{2}{3} \times \frac{1}{2} + \frac{1}{3} \times \frac{3}{8}}$ = $\frac{8}{11}$