Answer: $\frac{11}{21}$
Solution: We either have exactly one or three female members. Therefore, the required probability is
$$
\frac{\binom{4}{1}\binom{5}{2}+\binom{4}{3}}{\binom{9}{3}}=\frac{44}{84}=\frac{11}{21}
$$
2. A committee of three is to be selected from a pool of candidates consisting of five men and four women. If all the candidates are equally likely to be chosen, what is the probability that the committee will have an odd number of female members?