Probability that after 10,000 steps (+-1) you'll end up at the origin. How to use Central Limit Theorem? Starting at the origin and taking one step left or right with equal probability, what is the probability that you'll end up at 0 after 10,000 steps?
I figured it'd be $\frac1{2^{10000}}\binom{10000}{5000}$ since you will be taking half of the steps in one direction and half in the other in no particular order and then divide the number of all possible paths that land you at 0 by the total possible number of paths.
I got probability of about 0.008.

But how do I get this result using central limit theorem?


As noted above, we can also use the De Moivre-Laplace theorem to approximate the central term of the binomial.  This is a special case of the lattice form of the central limit theorem (Papoulis 8-5) which is obtained simply by substitution of the pmf for the binomial distribution into the normal pdf sampled at the points on which the probabilities must fall.  We shall consider Bernoulli random variables $X_i$ that take on the value $1$ with probability $1/2$ corresponding to a step to the right at step i, and the value $0$ with probability $1/2$ corresponding to a step to the left at step i.  Define
$$Y = \sum_{i=1}^{1000}X_i.$$
We wish to approximate 
$$P(Y = 5000) = {10000 \choose 5000}\frac{1}{2^{10000}}$$
the probability that we end up at $0$ after $10000$ steps.  Demoivre-Laplace tells us we can approximate this as
$$P(Y=5000) \approx \frac{1}{\sqrt{2\pi(2500)}} \approx 0.00797885$$
where the 2500 came from the variance of the binomial npq with n=10000, p=q=1/2.  Compare to the exact answer which to 8 decimal places is  0.00797864 and to my previous approximation of 0.00797871 using the cumulative normal.  We may verify that for even n from $2$ to $1$ million, the cumulative normal is always more accurate, with largest differences occurring for a small n.
Note that my earlier answer which subtracted 2 values of the cumulative normal has also been referred to as the De Moivre-Laplace limit theorem (Feller) to distinguish it from the version in this answer.
