«Banaketa binomial»: berrikuspenen arteko aldeak

t
 
::<math>\begin{align}
E[X] & = \sum_{ix=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\\
& = \frac{n!}{1!(n-1)!}pq^{n-1}+\frac{2 \cdot n!}{2!(n-1)!}p^2q^{n-1}+\frac{3 \cdot n!}{3!(n-3)!}p^3q^{n-3}+\ldots+\frac{(n-1) \cdot n!}{(n-1)!1!}p^{n-1}q^1+np^n\\
48.195

edits