二項分布

${\displaystyle f\left(x\right)={}_{n}C{}_x{p}^x{q}^{n-x}}$    ${\displaystyle \left(p>0,q>0,p+q=1;x=0,1,2,\cdot \cdot \cdot ,n\right)}$

${\displaystyle E\left(X\right)={\displaystyle \sum _{x=0}^{n}xf\left(x\right)}}$　（平均を参照）

${\displaystyle ={\displaystyle \sum _{x=0}^{n}x{}_{n}C{}_x{p}^x{q}^{n-x}}}$

$x=0$ のとき，${\displaystyle xf\left(x\right)=0}$ となるので，$x=1$から始めても$\sum$の値はかわらない．よって

${\displaystyle ={\displaystyle \sum _{x=1}^{n}x{}_{n}C{}_x{p}^x{q}^{n-x}}}$

${\displaystyle ={\displaystyle \sum _{x=1}^n\frac{xn!}{x!\left(n-x\right)!}{p}^x{q}^{n-x}}}$　（組合わせ ${\displaystyle {}_{n}C_r}$を参照）

${\displaystyle ={\displaystyle \sum _{x=1}^n\frac{xn\left(n-1\right)!}{x!\left(n-x\right)!}p\cdot p^{x-1}{q}^{n-x}}}$

${\displaystyle =np{\displaystyle \sum _{x=1}^n\frac{\left(n-1\right)!}{\left(x-1\right)!\left(n-x\right)!}{p}^{x-1}{q}^{n-x}}}$

${\displaystyle x-1=y}$ とおいて式を書きかえる．

${\displaystyle =np{\displaystyle \sum _{y=0}^{n-1}\frac{\left(n-1\right)!}{y!\left\{n-\left(y+1\right)\right\}!}{p}^y{q}^{n-\left(y+1\right)}}}$

${\displaystyle n-1=m}$とおいて式を書きかえる．

${\displaystyle =np{\displaystyle \sum _{y=0}^m\frac{m!}{y!\left(m-y\right)!}{p}^y{q}^{m-y}}}$

${\displaystyle =np{\left(p+q\right)}^m}$　（二項定理を参照）

${\displaystyle p+q=1}$ より

${\displaystyle =np}$

${\displaystyle V\left(X\right)=E\left(X^2\right)-{\left\{E\left(X\right)\right\}}^2}$　（分散を参照）

${\displaystyle =E\left(X\left(X-1\right)+X\right)-{\left\{E\left(X\right)\right\}}^2}$

${\displaystyle =E\left(X\left(X-1\right)\right)+E\left(X\right)-{\left\{E\left(X\right)\right\}}^2}$

${\displaystyle ={\displaystyle \sum _{x=0}^{n}x\left(x-1\right)f\left(x\right)}+np-{\left(np\right)}^2}$

${\displaystyle ={\displaystyle \sum _{x=0}^{n}x\left(x-1\right){}_{n}C{}_x{p}^x{q}^{n-x}}}$${\displaystyle +np-{\left(np\right)}^2}$

$x=0$，$x=1$ のとき，$x\left(x-1\right)f\left(x\right)=0$ となるので，$x=2$から始めても$\sum$の値はかわらない．よって

${\displaystyle ={\displaystyle \sum _{x=2}^{n}x\left(x-1\right){}_{n}C{}_x{p}^x{q}^{n-x}}}$${\displaystyle +np-{\left(np\right)}^2}$

${\displaystyle ={\displaystyle \sum _{x=2}^n\frac{x\left(x-1\right)n!}{x!\left(n-x\right)!}{p}^x{q}^{n-x}}+np-{\left(np\right)}^2}$　（組合わせ ${\displaystyle {}_{n}C_r}$を参照）

${\displaystyle ={\displaystyle \sum _{x=2}^n\frac{n\left(n-1\right)\left(n-2\right)!}{\left(x-2\right)!\left(n-x\right)!}{p}^2{p}^{x-2}{q}^{n-x}}}$${\displaystyle +np-{\left(np\right)}^2}$

${\displaystyle =n\left(n-1\right)p^2{\displaystyle \sum _{x=2}^n\frac{\left(n-2\right)!}{\left(x-2\right)!\left(n-x\right)!}{p}^{x-2}{q}^{n-x}}}$${\displaystyle +np-{\left(np\right)}^2}$

${\displaystyle x-2=y}$ とおいて式を書きかえる．

${\displaystyle =n\left(n-1\right)p^2{\displaystyle \sum _{y=0}^{n-2}\frac{\left(n-2\right)!}{y!\left\{n-\left(y+2\right)\right\}!}{p}^y{q}^{n-\left(y+2\right)}}}$${\displaystyle +np-{\left(np\right)}^2}$

${\displaystyle n-2=l}$ とおいて式を書きかえる．

${\displaystyle =n\left(n-1\right)p^2{\displaystyle \sum _{y=0}^{n-2}\frac{l!}{y!\left(l-y\right)!}{p}^y{q}^{l-y}}}$${\displaystyle +np-{\left(np\right)}^2}$

${\displaystyle =n\left(n-1\right)p^2{\left(p+q\right)}^l+np-{\left(np\right)}^2}$　（二項定理を参照）

${\displaystyle p+q=1}$ より

${\displaystyle =n\left(n-1\right)p^2+np-{\left(np\right)}^2}$

${\displaystyle ={\left(np\right)}^2-n{p}^2+np-{\left(np\right)}^2}$

${\displaystyle =np\left(1-p\right)}$

${\displaystyle =npq}$