> [!NOTE] Theorem > Let $n\in \mathbb{N}$ and $p\in[0,1].$ Let $X$ be a [[Discrete random variables|discrete real-valued random]] that has a [[Binomial Distribution|binomial distribution]] with parameters $n$ and $p.$ Then the [[Expectation of Discrete Real-Valued Random Variable|expectation]] of $X$ is $np.$ **Proof**: We have $\begin{align} \mathbb{E}[X] &= \sum_{k=0}^{\infty} p_{X}(k)\cdot k = \sum_{k=0}^{\infty} k {n \choose k} p^{k} (1-p)^{n-k} \\ &= \sum_{k=1}^{\infty} k {n \choose k} p^{k} (1-p)^{n-k} \\ \end{align}$since for $k=0,$ $k {n \choose k} p^{k} (1-p)^{n-k}=0.$ By [[Factors of Binomial Coefficient]] $k {n \choose k}=n{n-1 \choose k-1}$, thus $\begin{align} \mathbb{E}[X]&= \sum_{k=1}^{\infty}n {n-1 \choose k-1} p^{k}(1-p)^{n-k} \\ &= np \sum_{k=1}^{\infty}{n-1 \choose k-1} p^{k-1}(1-p)^{(n-1)-(k-1)} \\ &= np \sum_{j=0}^{\infty} {m \choose j} p^{j}(1-p)^{m-j} \\ &= np \end{align}$using [[Binomial Theorem]] and $p+1-p=1.$ # Applications By [[Variance of Binomial Distribution]], $\text{Var}(X)=np(p-1).$