Probability of at least one king in a 13-card hand? What is the probability of drawing 13 cards and having at least one king?

Here is what I came up with:

$$1 - \frac{\begin{pmatrix}
48\\
13
\end{pmatrix}}{\begin{pmatrix}
52\\
13
\end{pmatrix}}$$

Is this correct?
Thanks.
Your answer is indeed correct.

There are $\binom{52}{13}$ hands with $13$ cards that can be selected from a standard deck.

The number with no kings is $\binom{48}{13}$ since we must select $13$ of the $52 - 4 = 48$ cards that are not kings.  

Thus, the probability that no kings are selected is 
$$\Pr(\text{no kings}) = \frac{\dbinom{48}{13}}{\dbinom{52}{13}}$$
Since the probability that at least one king is selected is found by subtracting the probability of no kings from $1$, we obtain your answer
$$\Pr(\text{at least one king}) = 1 - \frac{\dbinom{48}{13}}{\dbinom{52}{13}}$$
$\begin{pmatrix}52 \\ 13\end{pmatrix}$ is the number of 13 card hands that can be dealt from a 52 card deck.  $\begin{pmatrix}52 \\ 48\end{pmatrix}$ is the number of 13 card hands that can be dealt from a 48 card deck- with the kings missing.  So your $\frac{\begin{pmatrix}52 \\ 13\end{pmatrix}}{\begin{pmatrix}52 \\ 48\end{pmatrix}}$ is the probability that a 13 card hand randomly dealt from a 52 card deck has no kings and, yes, $1- \frac{\begin{pmatrix}52 \\ 13\end{pmatrix}}{\begin{pmatrix}52 \\ 48\end{pmatrix}}$ is the probability that such a hand does contain at least one king.