using central limit theorem I recently got a tute question which I don't know how to proceed with and I believe that the tutor won't provide solution... The question is 

Pick a real number randomly (according to the uniform measure) in the interval $[0, 2]$. Do this one million times and let $S$ be the sum of all the numbers. What, approximately, is the probability that
  a) $S\ge1,$
  b) $S\ge0.001,$
  c) $S\ge0$?
  Express as a definite integral of the function $e^\frac{-x^2}{2}$.

Can anyone show me how to do it? It is in fact from a Fourier analysis course but I guess I need some basic result from statistcs which I am not familiar with at all..
 A: You have $1\, 000\, 000$ independent random variables $X_j$ with uniform distribution on the interval $[0,2]$.  Such a random variable has mean $\mu = 1$ and variance $\sigma^2 = 1/3$.  The central limit theorem says that if $S_N$ is the sum of $N$ independent random variables
with the same distribution, having mean $\mu$ and variance $\sigma^2$, then in the limit as $N \to \infty$, the distribution of $(S_N - N \mu)/(\sqrt{N} \sigma)$ approaches the standard normal distribution.  Thus, for any $z$,
$$\text{Prob} \left( \frac{S_N - N\mu}{\sqrt{N} \sigma} \ge z \right) \to \int_z^\infty \frac{e^{-t^2/2}}{\sqrt{2\pi}} dt$$
A: Let's call $S_n$ the sum of the first $n$ terms.  Then for $0 \le x \le 1$ it can be shown by induction that  $\Pr(S_n \le x) = \dfrac{x^n}{2^n \; n!}$ 
So the exact answers are 
a) $1 - \dfrac{1}{2^{1000000}  \times  1000000!}$
b) $1 - \dfrac{1}{2000^{1000000}  \times  1000000!}$
c) $1$
The first two are extremely close to 1; the third is 1.  The central limit theorem will not produce helpful approximations here, so you may have misquoted the question. 
