Chapter 7 - Section 7.1 - An Introduction to Discrete Probability - Exercises - Page 451: 12

$\frac{3243}{10829}$

There are $\binom{52}{5} = \frac{52!}{47!5!}$ equally likely poker hands. For the number of hands that contain exactly one ace, there are $4\binom{48}{4} = 4\frac{48!}{44!4!}$ choices. So the probability is $$4\frac{48!}{44!4!} \frac{47!5!}{52!} = 4\frac{47(46)(45)(5)}{52(51)(50)(49)} = \frac{3243}{10829}$$