Asignacion 5

\[ \Omega = {4+1, 3+2, 2+3, 1+4}=4\] Formula mxn \[mn={(6)(6)=36}\] \[ P(A)= \frac{4}{36}=0.1\ \] \[mn={(0.1)(100)=11.11}\]

\[ \Omega = {(1,2),(1,3),(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6),(4,5),(4,6),(5,6)}=15\] \[ P(A)= \frac{15}{36}=(0.416)(100)=41.66\ \]

\[ P(A)= \frac{64}{4}=(0.0625)(100)=6.25\ \] * ¿Cual es la probabilidad de que el segundo numero sea mayor que el primero?

\[ \Omega =64\] \[ \Omega =28\]

\[ P(A)= \frac{28}{64}=(0.437)(100)=43.7\ \]