Number of possible microscopic states and their probabilities
Some notes I had easily at hand may help. You have already calculated the probability of obtaining the chance with a number $k$ of type of ball (or molecule) out of a total on $n$ is
$$\displaystyle p=\frac{n!}{k!(n-k)!}\frac{1}{2^n} \tag{25c}$$
This distribution is a maximum when $k=n/2$. This can be seen with a straightforward argument. The factorial terms are symmetric, $k!(n-k)!=(n-k)!k!$ and always positive.  When $k=0$ or $k=n$ the probability is very small tending to zero; i.e $1/2^n$ is small so there must be a maximum somewhere in the range $0\to n$. The symmetric nature ensures that this will be at $k = n/2$. This can also be determined by differentiation. It is not possible to differentiate a factorial, as $n$ is discrete, but replacing $n!$ with the Sterling approximation $(n!=n^ne^n)$ and setting the derivative in $n$ to zero allows the maximum to be found.
The second feature of this probability distribution is that it becomes extremely narrow as $n$ increases. This is shown in the figure which shows the normalised probability $p/p_{max}$ vs $k/n$ which is a fraction between $0$ to $1$. At large $n$ the distribution becomes so narrow, or peaked at $n/2$, that almost all the probability is described by that at $n/2$, i.e. this value is so great compared to others that $p_{max}$ is in effect greater than all others combined.

The normalised probability function  $\displaystyle p=\frac{n!}{k!(n-k)!}\frac{1}{2^n}$ vs $k/n$ for different $n$. The the data only exists for integer values as shown as circles for $n=20$, at larger $n$ a line is drawn as the points become too congested. The curve for $n=10000$ is in red. At large $n$ the distribution becomes very narrow and the maximum value at $n/2$ is greater than all others combined and the distribution can be approximated by its maximum value.
You next need to choose $k/n$ corresponding to 2% either side of $k/n=0.5$ and calculate $p/p_{max}$.