Answer
$P(E)=\frac{1}{3}$
Work Step by Step
All the possible outcomes (sample space):
$S=[(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)]$
We have that the total number of possible outcomes in the sample space is:
$N(S)=36$
We want the sum to be odd and no more than 7 (the event):
$E=[(1,2),(1,4), (1,6),(2,1),(2,3),(2,5),(3,2)(3,4),(4,1),(4,3),(5,2),(6,1)]$
$N(E)=12$
$P(E)=\frac{N(E)}{N(S)}=\frac{12}{36}=\frac{1}{3}$
