I'll assume that the intended question is this: Given the length of a sequence and the counts of all four nucleotides in this sequence (as opposed to their frequency in sequences in general), what is the probability that a sequence drawn randomly uniformly from all sequences fulfilling that description would contain exactly a certain number $k$ of CG pairs?
Denote the counts of the nucleotides by $\def\n#1{n_{\text #1}}\n A$, $\n C$, $\n G$ and $\n T$ and their sum, the length of the sequence, by $n$. Then we can form $k$ CG pairs and distribute these $k$ pairs and the remaining $n-2k$ individual nucleotides in
$ \binom{n-k}{\n A,\n T,\n C-k,\n G-k,k} $
different ways (see multinomial coefficients). But this overcounts, since we're allowing the remaining C and G nucleotides to form pairs. Every combination with $m$ pairs is being counted $\binom mk$ times, where it shouldn't be counted at all. Making use of
$ \sum_{j=k}^m\binom mj\binom jk(-1)^{j-k}=\delta_{km}\;, $
we can correct for the overcounting and calculate the desired count of sequences fulfilling the description as
$ \begin{align} &\sum_{j=k}^\infty\binom{n-j}{\n A,\n T,\n C-j,\n G-j,j}\binom jk(-1)^{j-k}\\=&\sum_{j=k}^\infty\binom{n-j}{\n A,\n T,\n C-j,\n G-j,j-k,k}(-1)^{j-k}\;, \end{align} $
where the sum actually only runs up to $\min(\n C,\n G)$ and the remaining terms are zero. This count has to be divided by the total number of sequences fulfilling the description, which is
$ \binom n{\n A,\n T,\n C,\n G}\;. $
In your example, with $\n A=3$, $\n C=\n G=\n T=4$, $n=15$ and $k=3$, the result would be
$ \binom{15}{3,4,4,4}^{-1}\left(\binom{12}{3,4,1,1,3}-\binom{11}{3,4,1,3}\right)=\frac{44}{1365}\approx3\%\;. $
[Edit in response to comment:]
If you want to count the sequences with at least $k$ pairs, we still need to correct for overcounting, since each of the sequences with more than $k$ pairs is counted more than once, but the correction is slightly different. The required binomial coefficient identity is
$ \sum_{j=k}^m\binom mj\binom{j-1}{k-1}(-1)^{j-k}=1\;, $
and the resulting sum is
$ \sum_{j=k}^\infty\binom{n-j}{\n A,\n T,\n C-j,\n G-j,j}\binom{j-1}{k-1}(-1)^{j-k}\;. $
In your example, with $\n A=3$, $\n C=\n G=\n T=4$, $n=15$ and $k=3$, the result would be
$ \binom{15}{3,4,4,4}^{-1}\left(\binom{12}{3,4,1,1,3}\binom22-\binom{11}{3,4,4}\binom32\right)=\frac{3}{91}\approx3\%\;. $
The change relative to the result for exactly $3$ pairs is less than one tenth of a percent. The difference in the counts,
$ \left(\binom{12}{3,4,1,1,3}\binom22-\binom{11}{3,4,4}\binom32\right) - \left(\binom{12}{3,4,1,1,3}-\binom{11}{3,4,1,3}\right) = \binom{11}{3,4,4} \;, $
is precisely the number of sequences with $4$ CG pairs.