Suppose a girl throws a die. If she gets 1 or 2, she tosses a coin three times and notes the number of tails. If she gets 3, 4, 5 or 6, she tosses a coin once and notes whether a ‘head’ or ‘tail’ is obtained. If she obtained exactly one ‘tail’, what is the probability that she threw 3, 4, 5 or 6 with the die?

From the given question,
Let,
E1= the event that girl gets 1 or 2 on the roll
E2= the event that girl gets 3, 4, 5, or 6 on the roll of a die.
A= the event that she gets exactly one tail
$\therefore \begin{array}{l}\mathrm{P}\left({\mathrm{E}}_1\right)=\frac{2}{6}=\frac{1}{3}\\ \mathrm{P}\left({\mathrm{E}}_2\right)=\frac{4}{6}=\frac{2}{3}\end{array}$
Now, If she tossed coin 3 times and gets exactly one tail then possible outcomes are HTH, HHT, THH.
$\therefore \mathrm{P}\left(\mathrm{A}/{\mathrm{E}}_1\right)=\frac{3}{8}$
If she tossed coin only once and exactly one tail shows
Then, $\mathrm{P}\left(\mathrm{A}/{\mathrm{E}}_2\right)=\frac{1}{2}$
$\begin{array}{l}\therefore \mathrm{P}\left({\mathrm{E}}_2/\mathrm{A}\right)=\frac{\mathrm{P}\left({\mathrm{E}}_2\right)\cdot \mathrm{P}\left(\mathrm{A}/{\mathrm{E}}_2\right)}{\mathrm{P}\left({\mathrm{E}}_1\right)\cdot \mathrm{P}\left(\mathrm{A}/{\mathrm{E}}_1\right)+\mathrm{P}\left({\mathrm{E}}_2\right)\cdot \mathrm{P}\left(\mathrm{A}/{\mathrm{E}}_2\right)}\\ =\frac{\frac{2}{3}\times \frac{1}{2}}{\frac{1}{3}\times \frac{3}{8}+\frac{2}{3}\times \frac{1}{2}}=\frac{\frac{1}{3}}{\frac{1}{8}+\frac{1}{3}}=\frac{\frac{1}{3}}{\frac{11}{24}}\\ =\frac{8}{11}\end{array}$
Therefore, the required probability is $\frac{8}{11}$.