Probability of getting exactly $N/2$ heads when flipping a coin $N$ times

Question

I'm really stuck on this probability question. Any hints to help me get going would be great.

Let $N$ be $1346$. Suppose you flip a fair coin $N$ times and let $H$ be the number of heads you get.

$(a)$ What is the probability that $H$ is exactly $\frac{N}{2}$?

Answer 1

This is an example of a Binomial Probability Distribution. You can see more info on it here. In this case, you are looking for the probability of $\frac{1346}{2}=673$ successes out of 1346 trials with probability of success $p=0.5$, which gives us $\binom{1346}{673}(\frac{1}{2})^{1346}\approx0.02174.$

Answer 2

The total number of combinations is $\color\red{2^N}$

The number of combinations with $\frac12N$ heads is $\color\green{\dbinom{N}{\frac12N}=\dfrac{N!}{(\frac12N)!\cdot(\frac12N)!}}$

So the probability to obtain $\frac12N$ heads is $\dfrac{\color\green{\dfrac{N!}{(\frac12N)!\cdot(\frac12N)!}}}{\color\red{2^N}}=\dfrac{N!}{(\frac12N)!\cdot(\frac12N)!\cdot2^N}$