Expectation with respect to the binomial distribution

Statement

Let $n \ge 1$ be an integer, let $p \in [0, 1]$, and let $X$ have the binomial distribution $\operatorname{Bin}(n, p)$. Then $$E[X] = \sum_{k=0}^{n} k \binom{n}{k} p^k (1 - p)^{n - k} = np.$$

Remarks

A short proof: $X$ has the same law as $Y_1 + \dots + Y_n$ with independent Bernoulli $Y_i$ with $E[Y_i] = p$. Then linearity of expectation gives $np$. A direct proof uses $k \binom{n}{k} = n \binom{n-1}{k-1}$ and the binomial theorem. The variance of $X$ is $np(1 - p)$.

Explanations

L1 human

In $n$ trials with success probability $p$, you expect $np$ successes. For $100$ tosses of a fair coin, the expected number of heads is $50$.