# What is the probability that exactly $i$ heads will result from $n$ coin tosses?

It seems like an extremely easy question, but I can't solve it. All I know is that it is a $\frac{1}{2^n}$ chance that all tosses will result in heads and $\sum_{k=0}^{n} f(k, n)$ is equal to 1 given the function.

Even though school has just started for me, this is not a homework question.

-

This probability is equal to $f(i, n) = \binom{n}{i}2^{-n}$, where $\binom{n}{i} = \frac{n!}{i! (n-i)!}$ is binomial coefficient. This is true because there are $\binom{n}{i}$ ways to choose which coins will be heads up and each such combinaton of $i$ heads up coins has the same probability $2^{-n}$.
Flipping a coin is an example of a Bernoulli random variable—a variable which only accepts one of two possibilities such as YES/NO, TRUE/FALSE, HEADS/TAILS, etc. The distribution that represents the outcome of a group of individual Bernoulli random variables is, as mentioned above, the binomial distribution. If each Bernoulli has a probability $p$ of success and $q = 1 - p$, the binomial probability mass function for having exactly $i$ successes in $n$ trials is: $${n\choose i}p^iq^{n-i}$$
In your case, $p=q=\frac{1}{2}$ so the answer would be: $${n \choose i}\left(\frac{1}{2}\right)^i\left(\frac{1}{2}\right)^{n-i} = {n \choose i}2^{-n}$$ as said above.