$55$ of registered voters favor incumbent mayor. Find probability that the race ends in a tie. Fifty-five percent of the registered voters in Sheridanville favor their incumbent mayor in her bid for re-election. If four hundred voters go to the polls, approximate the probability that:
(a) the race ends in a tie.
Attempt: For part (a) given $n = 400$ and $p = 0.55$ and $(1-p) = 0.45$ we recognized this is a binomial representation.
Thus if the race ends in a tie, then $X = k = 200$.
Thus $P(X = k = 200) = \binom{400}{200}(0.55)^{200}(0.45)^{400-200}$. 
However, I don't know how to simplify, when I plug in the value for a calculator I get error since it is to big. The answer is $0.0053$
(b) the challenger scores an upset victory
Attempt: For part b) we  have $P(X < 200) = {\Sigma_{k = 0}^{199} }{\binom{400}{k}(0.55^k)(0.45^{400 - k}}$.
The answer is $0.0197$ at the back of the book.
can someone please help me simplify or if someone knows a better way to approach it? Any feedback/help would be appreciated.
Thanks in advance.
 A: Although I'm sure you're not looking for the answer anymore, here's the solution, for future searchers. The secret for both of these questions is the continuity correction. 
a) We are looking for P(X = 200), as you noted, but since it's basically impossible to calculate this using a calculator, we will instead use a normal distribution using the binomial and continuity correction. So, we will look for P(200 <= X <= 200). This makes sense, since we will apply the correction, and come out to:
F_z [(200.5-(0.55*400))/√(0.55*400*(1-0.55))] - F_z [(199.5-(0.55*400))/√(0.55*400*(1-0.55))]
Calculating the inside of those functions gives:
F_z [-1.960]-F_z [-2.060]
After looking these up in the normal distribution table in the appendix of your textbook (or online) you'll notice the solution is 0.0250 -0.0197 = 0.0053, same as the back of the book.
b) This question is easy once you have a). For the competitor to win, you're just looking for P(X < 200). This can easily be attained from values gotten from a) : Simply do the following:
P(X < 200) = P(X <= 200) - P(X = 200)
We know both of these values from a, so P(X <= 200) = 0.0250 and P(X = 200) = 0.0053 so the answer to b) is 0.0197, the answer in the back of the book. Alternatively, you know that P(X <= 199) also gives the solution, so taking our lower bound from earlier and making it our upper bound, we can get
F_z [(199.5-(0.55*400))/√(0.55*400*(1-0.55))], which we already found to be 0.0197.
Sorry for not formatting, didn't have much time. 
A: In part a, you should probably compute the logarithm of the probability, since the logarithm of the product of terms is the sum of logarithms, and the logarithm of a term with an exponent is just that exponent times the log of a term. When you do this, you should use the Stirling approximation to estimate the different factorial terms appearing in the binomial coefficient
$\ln(N!) \approx N\ln(N) - N \, \mbox{ for $N$ large}$
For part b, there are several ways to proceed. One way is to note that the binomial distribution for large $n$ will look like a Gaussian distribution, according to the de Moivre-Laplace theorem (which also uses the Stirling approximation...). Alternatively, you could consider the amount of votes the challenger gets $Y = 400 - X$, so that you are interested in $P[Y > 200]$, which is an unlikely or 'extreme' event. You can bound the probability of such extreme events using the Chebyshev inequality, which stated simply says that for positive random variables ($Y$ is positive...)
$P[Y > a] \leq \frac{E[Y]}{a}$
