Quick ways for approximating $\sum_{k=a_1}^{k=a_2}C_{100}^k(\frac{1}{2})^k(\frac{1}{2})^{100-k}$? Consider the following problem:

A fair coin is to be tossed 100 times, with each toss resulting in a head or a tail. Let
  $$H:=\textrm{the total number of heads}$$
  and 
  $$T:=\textrm{the total number of tails},$$
  which of the following events has the greatest probability?
A. $H=50$
B. $T\geq 60$
C. $51\leq H\leq 55$
D. $H\geq 48$ and $T\geq 48$
E. $H\leq 5$ or $H\geq 95$

What I can think is the direct calculation:
$$P(a_1\leq H\leq a_2)=\sum_{k=a_1}^{k=a_2}C_{100}^k(\frac{1}{2})^k(\frac{1}{2})^{100-k}$$
Here is my question:
Is there any quick way to solve this problem except the direct calculation?

In short: no. But you can cut a few things out immediately. For example, A is nonsense. A is eaten by C. And C is eaten by D.
So you need only to check B, D, and E. Of course, depending on your intuition, you might have a 'feel' for how unlikely E is as well. But that isn't as certain.
