«Banaketa binomial»: berrikuspenen arteko aldeak

 
 
::<math>\begin{alignarray}{rcl}
E[X] & = \sum_{x=0}^nxp(x)\\
& =0{n \choose 0}q^n+1{n \choose 1}pq^{n-1}+2{n \choose 2}p^2q^{n-1}+3{n \choose 3}p^3q^{n-2}+\ldots+(n-1){n \choose n-1}p^{n-1}q^1+n{n \choose n}p^n\\
& = np(p+q)^{n-1}\ \ \ ; \ \ \textstyle{Newtonen\ binomioaz}\\
& = np\ \ \ ; \ \ \textstyle{p+q=1\ betetzen\ baita.}
\end{alignarray}
</math>
}}{{kaxa zabalkorra
 
 
::<math>\begin{alignarray}{rcl}
E[Y] & = E[X_1]+E[X_2]+\ldots+E[X_n]\\
& =p+p+\ldots+p\\
& = np\\
\end{alignarray}
</math>
}}
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