Suppose a girl throws a die. If she gets a 5 or 6,

Let E1 be the event that the outcome on the die is 5 or 6 and E2 be the event that the outcome on the die is 1, 2, 3, or 4.

$\therefore \mathrm{P}\left(\mathrm{E}_{1}\right)=\frac{2}{6}=\frac{1}{3}$ and $\mathrm{P}\left(\mathrm{E}_{2}\right)=\frac{4}{6}=\frac{2}{3}$

Let A be the event of getting exactly one head.

$P\left(A \mid E_{1}\right)=$ Probability of getting exactly one head by tossing the coin three times if she gets 5 or $6=\frac{3}{8}$

$P\left(A \mid E_{2}\right)=$ Probability of getting exactly one head in a single throw of coin if she gets $1,2,3$, or $4=\frac{1}{2}$

The probability that the girl threw $1,2,3$, or 4 with the die, if she obtained exactly one head, is given by $P\left(E_{2} \mid A\right)$.

By using Bayes’ theorem, we obtain

$P\left(E_{2} \mid A\right)=\frac{P\left(E_{2}\right) \cdot P\left(A \mid E_{2}\right)}{P\left(E_{1}\right) \cdot P\left(A \mid E_{1}\right)+P\left(E_{2}\right) \cdot P\left(A \mid E_{2}\right)}$

$=\frac{\frac{2}{3} \cdot \frac{1}{2}}{\frac{1}{3} \cdot \frac{3}{8}+\frac{2}{3} \cdot \frac{1}{2}}$

$=\frac{\frac{1}{3}}{\frac{1}{3}\left(\frac{3}{8}+1\right)}$

$=\frac{1}{\frac{11}{8}}$

$=\frac{8}{11}$