Define events:
\begin{align} A &= \text{At least one female in capsule } A \\\ B &= \text{A female in capsule } B. \end{align}
The probability required is:
\begin{align} P(B\mid A) &= \dfrac{P(A\mid B)P(B)}{P(A)} \qquad\qquad\qquad\qquad\qquad\text{by Bayes Theorem} \\\ &= \dfrac{(1-P(A^c\mid B))P(B)}{1-P(A^c)} \qquad\qquad\qquad\qquad (1)\\\ \end{align}
Each of these required probabilities is easily calculated:
\begin{align} P(A^c\mid B) &= \binom{3}{2} \bigg/ \binom{4}{2} = \dfrac{1}{2} \\\ P(B) &= \dfrac{2}{5} \\\ P(A^c) &= \binom{3}{2} \bigg/ \binom{5}{2} = \dfrac{3}{10}. \end{align}
Substituting into $(1)$ gives:
$$P(B\mid A) = \dfrac{2}{7}.$$