The probability that A hits a target is $\frac14$ and that of B is $\frac13$. If they fire at once and one hits the target, find $P(\text{A hits})$

$$ \begin{align} P(\mbox{target is hit once}) &= P(\mbox{A hitting}) \cdot P(\mbox{B not hitting}) + P(\mbox{A not hitting}) \cdot P(\mbox{B hitting}) \\ &= \frac{1}{4}\cdot\frac{2}{3} + \frac{3}{4}\cdot\frac{1}{3} \\ &= \frac{5}{12} \end{align} $$

So, $$P(\mbox{A hitting | target is hit once}) = \frac{P(\mbox{A hitting}) \cdot P(\mbox{B not hitting})}{P(\mbox{target is hit once})} = \dfrac{\frac{1}{6}}{\frac{5}{12}} = \frac{2}{5}.$$

Your answer is not correct because you did not account for the case where only B hits, which has probability $\frac13×\frac34=\frac14$. Then the required probability is $$\frac{\frac16}{\frac14+\frac16}=\frac25$$ as the book gives.

The answer is indeed 2/5 I believe.

\begin{align} \mathbb{P}[\text{A hit | only one hit}] &= \frac{\mathbb{P}[\text{A hit} \,\cap\, \text{only one hit}]}{\mathbb{P}[\text{only one hit}]} \\ &= \frac{\mathbb{P}[\text{A hit}\,\cap\,\text{B didn't hit}]}{\mathbb{P}[\text{A hit}\,\cap\, \text{B didn't hit}] + \mathbb{P}[\text{A didn't hit}\,\cap\, \text{B hit}]} \\ &= \frac{1/4 \cdot 2/3}{1/4 \cdot 2/3 + 3/4 \cdot 1/3} \\ &=\frac{2}{5} \end{align}