What is the probability that a five card poker hand contains at most 1 ace?
I know that if it is at least $1$:
$$\frac{C(4,1) \cdot C(51,4)}{C(52,5)}$$
I also know that if it is exactly $1$, then it is:
$$\frac{C(4,1) \cdot C(48,4)}{C(52,5)}$$
What would at most be:
$$1 - \frac{C(4,1) \cdot C(51,4)}{C(52,5)}$$
and if so what is non complement representation?
If a hand has at most one ace, it has either no aces or one ace.
The number of five-card hands with no aces is $$\binom{4}{0}\binom{48}{5} = \binom{48}{5}$$ since all five cards must be chosen from the $52 - 4 = 48$ cards that are not aces.
The number of five-card hands with exactly one ace is $$\binom{4}{1}\binom{48}{4}$$ since one of the four aces and four of the other $48$ cards in the deck must be selected.
Since there are $$\binom{52}{5}$$ possible five-card hands, the probability of selecting at most one ace is $$\Pr(\text{at most one ace}) = \frac{\dbinom{48}{5} + \dbinom{4}{1}\binom{48}{4}}{\dbinom{52}{5}}$$
What errors did you make?
The number of hands with at least one ace is found by subtracting the number of hands with no aces from the total number of hands, which is $$\binom{52}{5} - \binom{48}{5}$$ Alternatively, you could add the number of hands with exactly $k$ aces, where $1 \leq k \leq 5$. $$\sum_{k = 1}^{4} \binom{4}{k}\binom{48}{5 - k} = \binom{4}{1}\binom{48}{4} + \binom{4}{2}\binom{48}{3} + \binom{4}{3}\binom{48}{2} + \binom{4}{4}\binom{48}{1}$$
Also, at most $1$ is the complement of at least $2$.