# Ejercicios Resueltos

Páginas: 5 (1228 palabras) Publicado: 10 de febrero de 2013
200212422_T1_P1

TAREA 1

PROBLEMA 1
> m1<-array(0,dim=c(2,2))
> m1
[,1] [,2]
[1,] 0 0
[2,] 0 0
> m1[1,1]<-0.8
> m1[1,2]<-0.2
> m1[2,1]<-0.7
> m1[2,2]<-0.3
> A <- rbind(c(1,1), c(m1[1,1]-1,m1[2,1]))
> b <- c(1,0)
> m1E <- solve(A,b)
> m1E
[1] 0.7777778 0.2222222
>PROBLEMA 2
> m1<-array(0,dim=c(2,2))
> m1
[,1] [,2]
[1,] 0 0
[2,] 0 0
> m1[1,1]<-0.9
> m1[1,2]<-0.1
> m1[2,1]<-0.2
> m1[2,2]<-0.8
> A <- rbind(c(1,1), c(m1[1,1]-1,m1[2,1]))
> b <- c(1,0)
> m1E <- solve(A,b)
> m1E
[1] 0.6666667 0.3333333
>
La probabilidad q compren la cola 1 a lo largo del tiempo es 0.666PROBLEMA 4
> m1<-array(0,dim=c(2,2))
> m1
[,1] [,2]
[1,] 0 0
[2,] 0 0
> m1[1,1]<-0.9
> m1[1,2]<-0.1
> m1[2,1]<-0.2
> m1[2,2]<-0.8
> A <- rbind(c(1,1), c(m1[1,1]-1,m1[2,1]))
> b <- c(1,0)
> m1E <- solve(A,b)
> m1E
[1] 0.6666667 0.3333333
>
b)Si compro Prensa Libre la probabilidad que compre prensa libre pasadas 3 compras es 0.781
c)Des pues de un largo tiempo los periodicos estan distribuidos así:
P S
P 0.6666 0.3333
S 0.6666 0.3333
d) No deben contratar a la empresa de publicidad, por que la inversión no equivale a las ganancias

PROBLEMA 5
>m1<-array(0,dim=c(2,2))
> m1
[,1] [,2]
[1,] 0 0
[2,] 0 0
> m1[1,1]<-0.9
> m1[1,2]<-0.1
> m1[2,1]<-0.65
> m1[2,2]<-0.35
> A <- rbind(c(1,1), c(m1[1,1]-1,m1[2,1]))
> b <- c(1,0)
> m1E <- solve(A,b)
> m1E
[1] 0.8666667 0.1333333
>
TAREA 1.1
Problema 1

>m1<-array(0,dim=c(6,6))
> m1
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 0 0 0 0 0 0
[2,] 0 0 0 0 0 0
[3,] 0 0 0 0 0 0
[4,] 0 0 0 0 0 0
[5,] 0 0 0 0 0 0
[6,] 0 0 0 0 0 0> colnames(m1)=c("E1","E2","E3","E4","E5","E6")
> rownames(m1)=c("E1","E2","E3","E4","E5","E6")
> m1[1,]=c(0,0.5,0,0.5,0,0)
> m1[2,]=c(0.5,0,0.5,0,0,0)
> m1[3,]=c(0,0,1,0,0,0)
> m1[4,]=c(0.5,0,0,0,0.5,0)
> m1[5,]=c(0,0,0,0.5,0,0.5)
>m1[6,]=c(0,0,0,0,0,1)
> m1
E1 E2 E3 E4 E5 E6
E1 0.0 0.5 0.0 0.5 0.0 0.0
E2 0.5 0.0 0.5 0.0 0.0 0.0
E3 0.0 0.0 1.0 0.0 0.0 0.0
E4 0.5 0.0 0.0 0.0 0.5 0.0
E5 0.0 0.0 0.0 0.5 0.0 0.5
E6 0.0 0.0 0.0 0.0 0.0 1.0
> Q<-m1[-3,-3]> Q<-Q[-5,-5]
> m1
E1 E2 E3 E4 E5 E6
E1 0.0 0.5 0.0 0.5 0.0 0.0
E2 0.5 0.0 0.5 0.0 0.0 0.0
E3 0.0 0.0 1.0 0.0 0.0 0.0
E4 0.5 0.0 0.0 0.0 0.5 0.0
E5 0.0 0.0 0.0 0.5 0.0 0.5
E6 0.0 0.0 0.0 0.0 0.0 1.0
> QE1 E2 E4 E5
E1 0.0 0.5 0.5 0.0
E2 0.5 0.0 0.0 0.0
E4 0.5 0.0 0.0 0.5
E5 0.0 0.0 0.5 0.0
> R<-m1[-3,]
> R
E1 E2 E3 E4 E5 E6
E1 0.0 0.5 0.0 0.5 0.0 0.0
E2 0.5 0.0 0.5 0.0 0.0 0.0
E4 0.5 0.0 0.0 0.0 0.5 0.0...

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