Mostrando entradas con la etiqueta matemàtiques-probabilitats. Mostrar todas las entradas
Mostrando entradas con la etiqueta matemàtiques-probabilitats. Mostrar todas las entradas

viernes, 13 de septiembre de 2019

distribucions de probabilitat en series

f(k) = (x^{k}/k!)·e^{-x}
f(k) = k·(x^{k}/k!)·(1/x)·e^{-x}
f(k) = k(k+(-1))·(x^{k}/k!)·(1/x^{2})·e^{-x}
f(k) = k(k+(-1))·...(m)...·(k+(-m))·(x^{k}/k!)·(1/x^{m+1})·e^{-x}


f(k) = (x^{2k}/(2k)!)·(1/cosh(x))
f(k) = (x^{2k+1}/(2k+1)!)·(1/sinh(x))


f(k) = (2k)(x^{2k}/(2k)!)·(1/x)·(1/sinh(x))
f(k) = (2k+1)(x^{2k+1}/(2k+1)!)·(1/x)·(1/cosh(x))

distribucions de probabilitat

f(k) = [ n // k ]·2^{(-n)}


f(0)+...+f(n) = ( [ n // 0 ]+...+[ n // n ] )·2^{(-n)}=2^{n}·2^{(-n)}=1


f(k) = [ n // k ]·p^{(n+(-k))}·(1+(-p))^{k}


f(0)+...+f(n) = ( [ n // 0 ]p^{n}+...+[ n // n ](1+(-p))^{n} )=( p+(1+(-p)) )^{n}=1


f(k) = (1/p^{n})·( p^{k}+(-1)·p^{(k+(-1))} )


f(0)+...+f(n) = ( (1/p^{n})·( ( (p^{n+1}+(-1))/(p+(-1)) )+(-1)·( (p^{n}+(-1))/(p+(-1)) ) )=...
...(1/p^{n})·p^{n}·( (p+(-1))/(p+(-1)) )=1