PROBLEMA DE BERNOULLI (4ta EVIDENCIA)

** OBTENGA LA RESOLUCION DE BERNOULLI SI SE TIRAN A LO MAS 8 LANZAMIENTOS DE UNA MONEDA **

DATOS:
n=8
p=1/2=50%=0.5
q=0.5
x=0,1,2,3,4,5,6,7,8

PROCEDIMIENTO:

(8/0)8C0= 8!/(8-0)!(0!)=8!/(8!)(0!)=40320/(40320)(1)=40320/40320=1
f(0)= (8C0)(0.5^0)(0.5^8-0)
f(0)= (1)(1)(0.00390625)
f(0)= 0.00390625

(8/1)8C1= 8!/(8-1)!(1!)=8!/(7!)(1!)=40320/(5040)(1)=40320/5040=8
f(1)= (8C1)(0.5^1)(0.5^8-1)
f(1)= (8)(0.5)(0.0078125)
f(1)= 0.03125

(8/2)8C2= 8!/(8-2)!(2!)=8!/(6!)(2!)=40320/(720)(2)=40320/1440=28
f(2)= (8C2)(0.5^2)(0.5^8-2)
f(2)= (8)(0.25)(0.015625)
f(2)= 0.109375

(8/3)8C3= 8!/(8-3)!(3!)=8!/(5!)(3!)=40320/(120)(6)=40320/720=56
f(3)= (8C3)(0.5^3)(0.5^8-3)
f(3)= (56)(0.125)(0.03125)
f(3)= 0.21875

(8/4)8C4= 8!/(8-4)!(4!)=8!/(4!)(4!)=40320/(24)(24)=40320/576=70
f(4)= (8C4)(0.5^4)(0.5^8-4)
f(4)= (70)(0.0625)(0.0625)
f(4)= 0.2734375

(8/5)8C5= 8!/(8-5)!(5!)=8!/(3!)(5!)=40320/(6)(120)=40320/720=56
f(5)= (8C5)(0.5^5)(0.5^8-5)
f(5)= (56)(0.03125)(0.125)
f(5)= 0.21875

(8/6)8C6= 8!/(8-6)!(6!)=8!/(2!)(6!)=40320/(2)(720)=40320/1440=28
f(6)= (8C6)(0.5^6)(0.5^8-6)
f(6)= (28)(0.015625)(0.25)
f(6)= 0.109375

(8/7)8C7= 8!/(8-7)!(7!)=8!/(1!)(7!)=40320/(1)(5040)=40320/5040=8
f(7)= (8C7)(0.5^7)(0.5^8-7)
f(7)= (8)(0.0078125)(0.5)
f(7)= 0.03125

(8/8)8C8= 8!/(8-8)!(8!)=8!/(0!)(8!)=40320/(1)(40320)=40320/40320=1
f(8)= (8C8)(0.5^8)(0.5^8-8)
f(8)= (1)(0.00390625)(1)
f(8)= 0.00390625

suma de todos los resultados:
0.00390625+0.03125+0.1093756+0.21875+0.2734375+0.21875+0.109375+0.03125+0.00390625=
1 (resultado total)