This is my attempt at starting, am I going in the right direction? -
We know at least $k$ are H, so the probability of $k$ being H is
$\begin{pmatrix} 30\\k \end{pmatrix}\frac{1}{2^k} \frac{1}{2^{30-k}}= \begin{pmatrix} 30\\k \end{pmatrix}\frac{1}{2^{30}}$
Now $k$ of the 30 flips have been taken up, so out of the remaining $30-k$ flips, we need $k$ to be T, and the probability of this is:
$\begin{pmatrix} 30-k\\k \end{pmatrix}\frac{1}{2^{30}}$
So multiplying should give the result?
Have I thought about this correctly?
The probability of exactly $j$ heads in $30$ flips is $$ \binom{30}{j}\Bigl(\frac{1}{2}\Bigr)^{30} $$ hence for any fixed $k\in\{0,...,15\}$, the probability that in $30$ flips you get at least $k$ heads and at least $k$ tails is $$ \sum_{j=k}^{30-k} \binom{30}{j}\Bigl(\frac{1}{2}\Bigr)^{30} = \Bigl(\frac{1}{2}\Bigr)^{30}\left(\sum_{j=k}^{30-k}\binom{30}{j}\right) $$ where the lower bound $j=k\;$is to guarantee at least $k$ heads, and the upper bound $j=30-k\;$is to guarantee at least $k$ tails.
By symmetry, the result can be simplified to $$ \Bigl(\frac{1}{2}\Bigr)^{30} \left(\left(2\sum_{j=k}^{14} \binom{30}{j}\right) + \binom{30}{15}\right) $$ Equivalently, by complementary counting, the result can be rewritten as $$ 1-\Bigl(\frac{1}{2}\Bigr)^{29}\left(\sum_{j=0}^{k-1} \binom{30}{j}\right) $$ which requires less work if $k$ is small.